{"id":7042,"date":"2023-12-28T09:35:36","date_gmt":"2023-12-28T04:05:36","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=7042"},"modified":"2023-12-28T17:21:49","modified_gmt":"2023-12-28T11:51:49","slug":"samacheer-kalvi-12th-physics-solutions-chapter-6","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-12th-physics-solutions-chapter-6\/","title":{"rendered":"Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics"},"content":{"rendered":"

Students who wish to prepare the Samacheer Kalvi Class 12th Physics Solutions Subject Chapter 6 Optics can rely on the Tamilnadu State Board Solutions for Class 12th Physics Solutions Chapter 6 Optics Questions and Answers prevailing. Become perfect with the concepts of Samacheer Kalvi Class 12th Physics Solutions<\/a> Chapter 6 Optics Questions and Answers and score better grades in your exams. Detailed Solutions are provided to the concepts by experts keeping in mind the latest edition textbooks and syllabus.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics<\/h2>\n

Ace up your preparation by referring to the Samacheer Kalvi Class 12th Physics Solutions Chapter 6 Optics and learn all the topics within. Click on the topic you want to prepare from the Class 12th Chapter 6 Optics Questions and Answers and prepare it easily. You can understand your strengths and weaknesses by practicing the Questions in Samacheer Kalvi Class 12th Physics Solutions PDF.<\/p>\n

Samacheer Kalvi 12th Physics Optics Textual Evaluation Solved<\/h3>\n

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions\u00a0<\/span><\/strong><\/p>\n

Question 1.
\nThe speed of light in an isotropic medium depends on,
\n(a) its intensity
\n(b) its wavelength
\n(c) the nature of propagation
\n(d) the motion of the source w.r.to medium
\nAnswer:
\n(b) its wavelength<\/p>\n

Question 2.
\nA rod of length 10 cm lies along the principal axis of a concav e mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, (AIPMT Main 2012)
\n(a) 2.5 cm
\n(b) 5cm
\n(c) 10 cm
\n(d) 15cm
\nAnswer:
\n(b) 5cm
\nHint:
\n<\/span>By mirror formula, image distance of A
\n\\(\\frac { 1 }{ v }\\) + \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ f }\\) ; \\(\\frac { { 1 } }{ { v }_{ A } } \\) + \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ f }\\)
\n\\(\\frac { { 1 } }{ { v }_{ A } } \\) + \\(\\frac { 1 }{ u }\\) + \\(\\frac { 1 }{ (-30) }\\) = \\(\\frac { 1 }{ (-10) }\\)
\n\"Samacheer
\n\u2234 vA<\/sub> = – 15 cm
\nImage distance of C, vc<\/sub> = – 20 cm
\nThe length of image = |v<\/span>A<\/sub> – v<\/span>c<\/sub>|
\n<\/span>= |-15 + 20| = 5 cm<\/p>\n

Question 3.
\nAn object is placed in front of a convex mirror of focal length of\/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (IEE Main 2009)]
\n(a) 2\u0192 and c
\n(b) c and \u221e
\n(c) \u0192 and O
\n(d) None of these
\nAnswer:
\n(d) None of these
\nHint:
\nThere is no maximum-minimum object distance for the convex mirror to form a real and inverted image.<\/p>\n

Question 4.
\nFor light incident from the air onto a slab of refractive index 2. the maximum possible angle of refraction is,
\na) 30\u00b0
\nb) 45
\nc) 60\u00b0
\nd) 90\u00b0
\nAnswer:
\na) 30\u00b0
\nSolution:
\n\u00b5 = \\(\\frac{\\sin i}{\\sin r}\\)
\nsin r = \\(\\frac{\\sin \\mathrm{i}}{\\mu}\\)
\nsin r = 1\/2
\nr = \\(\\sin ^{-1}\\left(\\frac{1}{2}\\right)\\)
\nr = 30\u00b0<\/p>\n

Question 5.
\nIf the velocity and wavelength of light in air is Va<\/sub> and \u03bba<\/sub> and that in water is Va<\/sub> and \u03bbw<\/sub>, then the refractive index of water is,
\n(a) \\(\\frac { { V }_{ w } }{ { V }_{ a } } \\)
\n(b) \\(\\frac { { V }_{ a } }{ { V }_{ w } } \\)
\n(c) \\(\\frac { { \u03bb }_{ w } }{ { \u03bb }_{ a } } \\)
\n(d) \\(\\frac { { V }_{ a }{ \\lambda } }{ { V }_{ w }{ \\lambda }_{ w } } \\)
\nAnswer:
\n(b) \\(\\frac { { V }_{ a } }{ { V }_{ w } } \\)<\/p>\n

Question 6.
\nStars twinkle due to
\n(a) reflection
\n(b) total internal reflection
\n(c) refraction
\n(d) polarisation
\nAnswer:
\n(c) refraction<\/p>\n

Question 7.
\nWhen a biconvex lens of glass having a refractive index of 1.47 is dipped in a liquid, it acts as a plain sheet of glass. This implies that the liquid must have a refractive index,
\n(a) less than one
\n(b) less than that of glass
\n(c) greater than that of glass
\n(d) equal to that of glass
\nAnswer:
\n(d) equal to that of glass
\nHint:
\nAccording to len’s maker formula,
\n\\(\\frac { 1 }{ \u0192 } \\) = \\(\\left(\\frac{n_{2}}{n_{1}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\)
\nBiconvex lens dipped in a liquid, acts as a plane sheet of glass, \u0192 = \u221e; \\(\\frac { 1 }{ \u0192 } \\) = 0
\n\\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\) -1 = 0;\u00a0 n2<\/sub> = n1<\/sub><\/p>\n

Question 8.
\nThe radius of curvature of the curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be,
\n(a) 5 cm
\n(b) 10 cm
\n(c) 15 cm
\n(d) 20 cm
\nAnswer:
\n(b) 10 cm
\nHint:
\nAccording to len\u2019s maker formula,
\n\\(\\frac { 1 }{ \u0192 }\\) = (n – 1) \\(\\left( \\frac { 1 }{ { R }_{ 1 } } -\\frac { 1 }{ { R }_{ 2 } } \\right) \\)
\nR1<\/sub> = \u221e; R2<\/sub> = -6
\n\u2234 \u0192 = \\(\\frac { R }{ \\left( n – 1 \\right) } \\) (R= 10 cm; n= 1.5
\n\u0192 = \\(\\frac { 10}{ \\left( 1.5 – 1 \\right) } \\) =20 cm<\/p>\n

Question 9.
\nAn air bubble in a glass slab of refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
\na) 8 cm
\nb) 10 cm
\nc) 12 cm
\nd) 16 cm
\nAnswer:
\nc) 12 cm
\nSolution:
\nTotal apparent depth = 5 + 3 = 8 cm
\nReal depth = thickness of slab
\n\u00b5 = \\(\\frac{\\text { Real depth }}{\\text { Apparent depth }}\\)
\n1.5 = \\(\\frac{\\text { thickness }}{8}\\)
\nthickness = 1.5 \u00d7 8 = 12 cm<\/p>\n

Question 10.
\nA ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45\u00b0. The ray can undergo total internal reflection for the following n,
\n(a) n= 1.25
\n(b) n = 1.33
\n(c) n= 1.4
\n(d) n= 1.5
\nAnswer:
\n(d) n= 1.5
\nHint:
\nFor total internal reflection
\nsin i > sin c\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 where, i – angle of incidence
\nBut, sin c = 1\/n\u00a0 \u00a0 \u00a0 c – critical angle
\nsin i > 1\/n
\nn > \\(\\frac { 1 }{ sin i }\\)
\nn > \\(\\frac { 1 }{ sin 45 }\\) ; n > \u221a2 ; n > 1.414<\/p>\n

Question 11.
\nA plane glass is placed over various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
\n(a) red
\n(b) yellow
\n(c) green
\n(d) violet
\nAnswer:
\n(d) violet
\nHint:
\nLetters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.<\/p>\n

Question 12.
\nTwo-point white dots are 1 mm apart on black paper. They are viewed by the eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, \u03bb = 500 nm]
\n(a) 1 m
\n(b) 5 m
\n(c) 3 m
\n(d) 6m
\nAnswer:
\n(b) 5 m
\nHint:
\nResolution limit sin \u03b8 = \\(\\frac { Y }{ d }\\) = \\(\\frac { 1.22c }{ d }\\)
\n\"Samacheer<\/p>\n

Question 13.
\nIn Young\u2019s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to
\na) 2D
\nb) \\(\\frac{D}{2}\\)
\nc) \u221a2D
\nd) \\(\\frac{D}{\\sqrt{2}}\\)
\nAnswer:
\na) 2D
\nSolution:
\n\u03b2 = \\(\\frac{\\mathrm{D} \\lambda}{\\mathrm{d}}\\)
\n\u03b2’ = \\(\\frac{D^{\\prime} \\lambda}{(2 d)}\\)
\nfor \u03b2 = \u03b2’
\nD’ = 2D<\/p>\n

Question 14.
\nTwo coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
\n(a) 5I and I
\n(b) 5I and 3I
\n(c) 9I and I
\n(d) 9I and 3I
\nAnswer:
\n(c) 9I and I
\nHint:
\n\"Samacheer<\/p>\n

Question 15.
\nWhen light is incident on a soap film of thickness 5 x 10-5<\/sup>
\ncm, the wavelength of light reflected maximum in the visible region is 5320 A. Refractive index of the film will be,
\n(a) 1.22
\n(b) 1.33
\n(c) 1.51
\n(cl) 1.83
\nAnswer:
\n(b) 1.33
\nHint.
\nThe condition for constructive interference, (for reflection)
\n2\u00b5 tcos r = (2n +1) \\(\\frac { \u03bb }{ 2 }\\) [\u2234 cos r = 1]
\n\u00b5 = \\(\\frac { \\left( 2n+1 \\right) \\lambda }{ 4t } \\)
\nFor visible region, n = 2
\n\"Samacheer<\/p>\n

Question 16.
\nFirst diffraction minimum due to a single slit of width 1.0 x 10-5<\/sup> cm is at 30\u00b0. Then wavelength of light used is,
\n(a) 400 \u00c5
\n(b) 500 \u00c5
\n(c) 600 \u00c5
\nAnswer:
\n(b) 500 \u00c5
\nHint.
\nFor diffraction minima, d sin \u03b8 = n\u03bb
\n\u03bb = \\(\\frac { dsin\\theta }{ n } =\\frac { 1\\times { 10 }^{ -5 }\\times { 10 }^{ -2 }\\times sin30\u00b0 }{ 1 } \\) = 0.5 x 10-7<\/sup>
\n\u03bb = 500 \u00c5<\/p>\n

Question 17.
\nA ray of light strikes a glass plate at an angle 60\u00b0. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
\na) \u221a3
\nb) \\(\\frac{3}{2}\\)
\nc) \\(\\sqrt{\\frac{3}{2}}\\)
\nd) 2
\nAnswer:
\na) \u221a3
\nSolution:
\n\u00b5 = tan IP\u00a0<\/sub>= tan 60\u00b0
\n\u00b5 = \u221a3<\/p>\n

Question 18.
\nOne of Young\u2019s double slits is covered with a glass plate as shown in the figure. The position of central maximum will,
\n\"Samacheer
\n(a) get shifted downwards
\n(b) get shifted upwards
\n(c) will remain the same
\n(d) data insufficient to conclude
\nAnswer:
\n(b) get shifted upwards<\/p>\n

Question 19.
\nLight transmitted by Nicol prism is,
\n(a) partially polarised
\n(b) unpolarised
\n(c) plane polarised
\n(d) elliptically polarised
\nAnswer:
\n(c) plane polarised<\/p>\n

Question 20.
\nThe transverse nature of light is shown in,
\n(a) interference
\n(b) diffraction
\n(c) scattering
\n(d) polarisation
\nAnswer:
\n(d) polarisation<\/p>\n

Samacheer Kalvi 12th Physics Optics Short Answer Questions<\/span><\/strong><\/p>\n

Question 1.
\nState the laws of reflection.
\nAnswer:
\n(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar.
\n(b) Angle of incidence is equal to the angle of reflection.<\/p>\n

Question 2.
\nWhat is the angle of deviation due to reflection?
\nAnswer:
\nThe angle between the incident and deviated light ray is called the angle of deviation of the light ray. It is written as, d= 180 – (i + r).
\nAs i = r in reflection, we can write angle of deviation ‘ in reflection at the plane surface as, d = 180 – 2i<\/p>\n

Question 3.
\nGive the characteristics of the image formed by a plane mirror.
\nAnswer:
\ni) The image formed by a plane mirror by virtual, erect, and laterally inverted.
\nii) The size of the image is equal to the size of the object.
\niii) The image distance far behind the mirror is equal to the object distance in front of it.
\niv) If an object is placed between two plane mirrors inclined at an angle 0, then the number of images formed is
\nIf \\(\\left(\\frac{360}{\\theta}\\right)\\) is even, then n = \\(\\left(\\frac{360}{\\theta}\\right)\\) – 1
\nIf \\(\\left(\\frac{360}{\\theta}\\right)\\) is odd, then n = \\(\\left(\\frac{360}{\\theta}\\right)\\) – 1
\nFor objects placed symmetrically,
\nIf \\(\\left(\\frac{360}{\\theta}\\right)\\) is odd, then n = \\(\\left(\\frac{360}{\\theta}\\right)\\)
\nFor objects placed unsymmetrically.<\/p>\n

Question 4.
\nDerive the relation between \u0192 and R for a spherical mirror.
\nLet C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection. If MP is the perpendicular from M on the principal axis, then from the geometry, The angles \u2220MCP = i and \u2220MFP = 2i From right angle triangles \u2206MCP and \u2206MFP,
\n\"Samacheer
\n\\(\\frac { 1 }{ f }\\) = \\(\\left(\\frac{n_{2}}{n_{1}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\)
\ntan i = \\(\\frac { PM }{ PC }\\) and tan 2 i = \\(\\frac { PM }{ PF }\\)
\nAs the angles are small, tan i \u2248 i = \\(\\frac { PM }{ PC }\\) and tan 2 i = \\(\\frac { PM }{ PF }\\)
\nSimplifying further, 2 = \\(\\frac { PM }{ PC }\\) and tan = \\(\\frac { PM }{ PF }\\) ;2PF = PC
\nPF is focal length\/and PC is the radius of curvature R.
\n2 \u0192= R (or) \u0192 = \\(\\frac { R }{ 2 }\\)
\n\u0192 = \\(\\frac { R }{ 2 }\\) is the relation between \u0192 and R.<\/p>\n

Question 5.
\nWhat are the Cartesian sign conventions for a spherical mirror?
\nAnswer:<\/p>\n

    \n
  1. The Incident light is taken from left to right (i.e. object on the left of the mirror).<\/li>\n
  2. All the distances are measured from the pole of the mirror (pole is taken as origin).<\/li>\n
  3. The distances measured to the right of the pole along the principal axis are taken as positive.<\/li>\n
  4. The distances measured to the left of the pole along the principal axis are taken as negative.<\/li>\n
  5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.<\/li>\n
  6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.<\/li>\n<\/ol>\n

    Question 6.
    \nWhat is optical path? Obtain the equation for optical path of a medium of thickness d and refractive index n.
    \nAnswer:
    \nOptical path of a medium is defined as the distance d\u2019 light travels in vacuum in the same time it travels a distance d in the medium.
    \nLet us consider a medium of refractive index n and thickness d. Light travels with a speed v through the medium in a time t. Then we can write,
    \n\"Samacheer
    \nv = \\(\\frac { d }{ t }\\); rewritten as, t = \\(\\frac { d }{ v }\\)
    \nIn the same time, light can cover a greater distance d’
    \nin vacuum as it travels with greater speed c in vacuum.
    \nThen we have,
    \nc = v = \\(\\frac { d’ }{ t }\\); rewritten as, t = \\(\\frac { d’ }{ c }\\)
    \nAs the time taken in both the cases is the same, we can equate the time t as,
    \n\\(\\frac { d’ }{ c }\\) = \\(\\frac { d }{ v }\\)
    \nrewritten for the optical path d’ as d’ = \\(\\frac { c }{ v }\\)d
    \nAs, \\(\\frac { c }{ v }\\) = n ; The optical path d’ is, d’ = nd
    \nAs n is always greater than 1, the optical path d’ of the medium is always greater than d.<\/p>\n

    Question 7.
    \nState the laws of refraction.
    \nAnswer:
    \na) The incident ray, refracted ray and normal to the refracting surface are all coplanar (ie. lie in the same plane).
    \nb) The ratio of the angle of the incident (Sin i) in the first medium to the angle of reflection (Sin r) in the second medium is equal to the ratio of the refractive index of the second medium n2<\/sub> to that of the refractive index of the first medium n1<\/sub>
    \n\\(\\frac{\\sin i}{\\sin r}\\) = \\(\\frac{n_{2}}{n_{1}}\\)<\/p>\n

    Question 8.
    \nWhat is the angle of deviation due to refraction?
    \nAnswer:
    \nThe angle between the incident and deviated light is called the angle of deviation. When light travels from rarer to the denser medium it deviates towards normal. The angle of deviation, in this case, is, d = i – r<\/p>\n

    Question 9.
    \nWhat is the principle of reversibility?
    \nAnswer:
    \nLight will follow exactly the same path if its direction of travel is reversed.<\/p>\n

    Question 10.
    \nWhat is a relative refractive index?
    \nAnswer:
    \nSnell\u2019s law, the term \\(\\left( \\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\right) \\) is called relative refractive index of second medium with respect to the first medium which is denoted as (n21<\/sub>). n21<\/sub> = \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\)<\/p>\n

    Question 11.
    \nObtain the equation for apparent depth.
    \nAnswer:
    \nLight from the object O at the bottom of the tank passes from a denser medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is n1<\/sub> and rarer medium is n2<\/sub>. Here, n1<\/sub> > n2<\/sub>. The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines NN’ and OD are parallel. Thus angle \u2220DIB is also r. The angles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell\u2019s law in product form for this refraction is,
    \nn1 <\/sub>sin i = n2<\/sub> sin r
    \nAs the angles i and r are small, we can approximate, sin i \u2248 tan i;
    \nn1<\/sub> tan i = n2<\/sub> tan i
    \nIn triangles \u2206DOB and \u2206DIB,
    \ntan(i) = \\(\\frac { DB }{ DO }\\) and tan(r) = \\(\\frac { DB }{ DI }\\)
    \nn1<\/sub> = \\(\\frac { DB }{ DO }\\) n2<\/sub> = \\(\\frac { DB }{ DI }\\)
    \nDB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d’.
    \nRearranging the above equation for the apparent depth d\u2019,
    \nd’ = \\(\\left( \\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\right) \\)d
    \nAs the rarer medium is air and its refractive index n2<\/sub> can be taken as 1, (n2<\/sub> = 1). And the refractive index n1<\/sub> of denser medium could then be taken as n, (n1 <\/sub>= n).
    \nIn that case, the equation for apparent depth becomes,
    \nd = \\(\\frac { d }{ n }\\)<\/p>\n

    Question 12.
    \nWhy do stars twinkle?
    \nAnswer:
    \nThe stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.<\/p>\n

    Question 13.
    \nWhat are critical angle and total internal reflection?
    \nAnswer:
    \nThe angle of incidence in the denser medium for which the refracted ray grazes the boundary is called the critical angle.
    \nWhen light travels from denser medium to rarer medium and its angle of incidence becomes greater than its critical angle then the entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.<\/p>\n

    Question 14.
    \nObtain the equation for the critical angle.
    \nAnswer:
    \nSnell\u2019s law in the product form, equation for critical angle incidence becomes,
    \nn1<\/sub> sini ic<\/sub> = n2<\/sub> sin 90\u00b0
    \nn1<\/sub> sini ic<\/sub> = n2<\/sub> (\u2235 sin 90\u00b0 = 1)
    \nsini ic<\/sub> = \\(\\left( \\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\right) \\)
    \nHere, n1 <\/sub>> n2<\/sub>
    \nIf the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n2<\/sub> = 1) and (n1<\/sub>= n).
    \nsini ic<\/sub> = \\(\\frac { 1 }{ n }\\) (or) ic<\/sub> = sin-1<\/sup> \\(\\left( \\frac { 1 }{ n } \\right) \\)<\/p>\n

    Question 15.
    \nExplain the reason for the glittering of diamonds.
    \nAnswer:<\/p>\n

      \n
    1. The brilliancy of diamonds is due to total internal reflection.<\/li>\n
    2. It’s refractive index is very large and critical angle is very small about 24.4\u00b0.<\/li>\n
    3. By repeated total internal reflections, the diamond glitters, when seen in the direction of emergent light.<\/li>\n<\/ol>\n

      Question 16.
      \nWhat are mirage and looming?
      \nAnswer:
      \nMirage: Mirage takes place in hot regions. The light from distant objects appears to be reflected from the ground. For mirage to form refractive index goes on increasing as we go up. Looming: Looming takes place in cold regions. The light from distant objects appears to be flying. For looming to form refractive index goes on decreasing.<\/p>\n

      Question 17.
      \nWrite short notes on the prisms making use of total internal reflection.
      \nAnswer:
      \nPrisms can be designed to reflect light by 90\u00b0 or by 180\u00b0 by making use of total internal reflection. The critical angle ic for the material of the prism must be less than 45\u00b0. This is true for both crown glass and flint glass. Prisms are also used to invert images without changing their size.<\/p>\n

      Question 18.
      \nWhat is Snell\u2019s window?
      \nAnswer:
      \nWhen light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle. This restricted circular illuminated area is called Snell’s window.<\/p>\n

      Question 19.
      \nWrite a note on optical fibre.
      \nAnswer:
      \nTransmitting signals through optical fibres is possible due to the phenomenon of total internal reflection. Optical fibres consists of inner part called core and outer part called cladding (or) sleeving. The refractive index of the material of the core must be higher than that of the cladding for total internal reflection to happen. Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. Hence, it undergoes repeated total internal reflections along the length of the fibre without undergoing any refraction.<\/p>\n

      Question 20.
      \nExplain the working of an endoscope.
      \nAnswer:<\/p>\n

        \n
      1. It is an instrument used to see inside a patient’s body.<\/li>\n
      2. It consists of bundle of optical fibres which are inserted into the body through mouth, nose or a special hole made in body.<\/li>\n
      3. Even operations can be done by necessary instruments attached at their ends.<\/li>\n
      4. It works on the principle of total internal reflection.<\/li>\n<\/ol>\n

        Question 21.
        \nWhat are primary focus and secondary focus of convex lens?
        \nAnswer:
        \nPrimary focus: The primary focus F1<\/sub> is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
        \nSecondary focus: The secondary focus F2<\/sub> is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.<\/p>\n

        Question 22.
        \nWhat are the sign conventions followed for lenses?
        \nAnswer:
        \nThe sign conventions for thin lenses differ only in the signs followed for focal lengths.
        \n(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).
        \n(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.<\/p>\n

        Question 23.
        \nArrive at lens equation from lens maker\u2019s formula.
        \nAnswer:
        \nFrom refraction through a double convex lens, the relation between the object distance u, image distance v1<\/sub> and radius of curvature R1<\/sub> as
        \n\\(\\frac { { \\mu }_{ 2 } }{ { v }_{ 1 } } -\\frac { { \\mu }_{ 1 } }{ u } =\\frac { { \\mu }_{ 2 }-{ \\mu }_{ 1 } }{ { R }_{ 1 } } \\) …… (1)
        \nThe relation between the object distance image distance v1<\/sub> and radius of curvature R2<\/sub> can be
        \n\\(\\frac { { \\mu }_{ 1 } }{ { v } } -\\frac { { { \\mu }_{ 2 } } }{ v_{ 1 } } =\\frac { { \\mu }_{ 1 }-{ \\mu }_{ 2 } }{ { R }_{ 2 } } \\) …… (2)
        \nAdding equation (1) and (2)
        \n\"Samacheer
        \nIf the object is placed at infinity (\u00b5 = \u221e), the image will be formed at the focus. i.e.v = \u0192
        \n\"Samacheer
        \nThis is len\u2019s maker\u2019s formula. When the lens is placed in air \u00b51<\/sub> = 1 and \u00b52<\/sub> = \u00b5.
        \nEquation (4) becomes,
        \n\"Samacheer
        \nFrom equation (3) and (4), we have \\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\)
        \n=\\(\\frac { 1 }{ \u0192 }\\)
        \nThis is the len\u2019s equation.<\/p>\n

        Question 24.
        \nObtain the equation for lateral magnification for thin lens.
        \nAnswer:
        \nLateral magnification in terms of u and \u0192.
        \nThe thin lens formula is
        \n\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\)=\\(\\frac { 1 }{ \u0192 }\\)
        \nMultiplying on both sides by ‘u’
        \n\"Samacheer
        \nIn terms of v and \u0192 multiplying by v, we get
        \n\"Samacheer
        \nHence, lateral magnification for thin lens,
        \n\"Samacheer<\/p>\n

        Question 25.
        \nWhat is power of a lens?
        \nAnswer:
        \nThe power of a lens P is defined as the reciprocal of its focal length.
        \np = \\(\\frac { 1 }{ \u0192 }\\)
        \nThe unit of power is diopter D.<\/p>\n

        Question 26.
        \nDerive the equation for effective focal length for lenses in contact.
        \nAnswer:
        \nConsider a two thin lenses in contact. In the absence of second lens L2<\/sub>, the first lens L1<\/sub> will form a real image I’. Using thin lens formula.
        \n\\(\\frac { 1 }{ { f }_{ 2 } } \\)=\\(\\frac { 1 }{ v’ }\\)–\\(\\frac { 1 }{ u }\\) ….. (1)
        \nThe image I’ acts as a virtual object (u = v’) for the second lens L2<\/sub> which finally forms its real image I at distance v. Thus
        \n\\(\\frac { 1 }{ { f }_{ 2 } } \\)=\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ v’ }\\) ….. (2)
        \nAdding equation (1) and (2) we get,
        \n\\(\\frac { 1 }{ { f }_{ 1 } } \\)+\\(\\frac { 1 }{ { f }_{ 2 } } \\)=\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\) ….. (3)
        \nFor the combination of thin lenses in contact, if ‘f’ is the equivalent focal length, then
        \n\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\)=\\(\\frac { 1 }{ \u0192 }\\) ….. (4)
        \nFrom equations (3) and (4), the effective focal length for lenses in contact.
        \n\\(\\frac { 1 }{ \u0192 }\\)=\\(\\frac { 1 }{ { f }_{ 1 } } \\)+\\(\\frac { 1 }{ { f }_{ 2 } } \\)<\/p>\n

        Question 27.
        \nWhat is angle of minimum deviation?
        \nAnswer:
        \nThe minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation.<\/p>\n

        Question 28.
        \nWhat is dispersion?
        \nAnswer:
        \nThe phenomenon of spliting of white light into its component colours on passing through a refracting medium is called dispersion.<\/p>\n

        Question 29.
        \nHow are rainbows formed?
        \nAnswer:<\/p>\n

          \n
        1. Rainbows are formed by the dispersion of light by droplets of water.<\/li>\n
        2. When Sunlight falls on the water droplets suspended in air, it splits into its constituent seven colours.<\/li>\n
        3. The angle of view for violet to red in primary rainbow 40\u00b0 to 42\u00b0<\/li>\n
        4. The angle of view for red to violet in a secondary rainbow 52\u00b0 to 54\u00b0<\/li>\n<\/ol>\n

          Question 30.
          \nWhat is Rayleigh\u2019s scattering?
          \nAnswer:
          \nThe scattering of light by particles in a medium without a change in wavelength is called as Rayleigh\u2019s scattering.<\/p>\n

          Question 31.
          \nWhy does sky appear blue?
          \nAnswer:
          \nBlue colour of the sky is due to the scattering of sunlight by air molecules. According to Rayleigh\u2019s law, intensity of scattered light, I \u221d \\(\\frac { 1 }{ { \\lambda }^{ 4 } } \\) So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appears blue.<\/p>\n

          Question 32.
          \nWhat is the reason for reddish appearance of sky during sunset and sunrise?
          \nAnswer:<\/p>\n

            \n
          1. At the sunset or sunrise, the sun is near the horizon.<\/li>\n
          2. The rays travel more distance through the atmosphere than at noon.<\/li>\n
          3. Therefore, only red colour, which is least scattered, reaches our eyes.<\/li>\n
          4. For this reason, the sun looks reddish at the sunset and sunrise.<\/li>\n<\/ol>\n

            Question 33.
            \nWhy do clouds appear white?
            \nAnswer:
            \nClouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.<\/p>\n

            Question 34.
            \nWhat are the salient features of the corpuscular theory of light?
            \nAnswer:<\/p>\n

              \n
            1. Light is emitted as tiny, massless and perfectly elastic particles called corpuscles.<\/li>\n
            2. They are unaffected by gravity and then the path is straight line in a medium of uniform refractive index.<\/li>\n
            3. The energy of light is K.E of corpuscles.<\/li>\n
            4. Different size of corpuscles are reason for different colours.<\/li>\n
            5. When they approach a surface they are either attracted or repelled.<\/li>\n
            6. This theory could not explain the reason for lesser velocity of light in denser medium than rarer medium.<\/li>\n
            7. It does not explain the phenomenon of interference, diffraction and polarization.<\/li>\n<\/ol>\n

              Question 35.
              \nWhat is the wave theory of light?
              \nAnswer:
              \nLight is a disturbance from a source that travels as longitudinal mechanical waves through either medium that was presumed to pervade all space as mechanical wave requires medium for its propagation. The wave theory could successfully explain phenomena of reflection, refraction, interference and diffraction of light.<\/p>\n

              Question 36.
              \nWhat is the electromagnetic wave theory of light?
              \nAnswer:
              \nElectromagnetic wave theory:
              \nMaxwell (1864) proved that light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. He could also show that no medium is necessary for the propagation of electromagnetic waves. All the phenomena of light could be successfully explained by this theory.<\/p>\n

              Question 37.
              \nWrite a short note on the quantum theory of light.
              \nAnswer:
              \nAlbert Einstein (1905), endorsing the views of Max Plank (1900), was able to explain photoelectric effect in which light interacts with matter as photons to eject the electrons. A photon is a discrete packet of energy. Each photon has energy E of,
              \nE = hv
              \nWhere, h is Plank\u2019s constant (h = 6.625 x 10-34<\/sup>J s) and v is the frequency of the electromagnetic wave. As light has both wave as well as particle nature it is said to have dual nature. Thus, it is concluded that light propagates as a wave and interacts with matter as a particle.<\/p>\n

              Question 38.
              \nWhat is a wavefront?
              \nAnswer:
              \nA wavefront is defined as the continuous locus of such particles of the medium which are vibrating in the same phase.<\/p>\n

              Question 39.
              \nWhat is Huygens\u2019 principle?
              \nAnswer:
              \nEach point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called secondary wavelets.<\/p>\n

              Question 40.
              \nWhat is the interference of light?
              \nAnswer:
              \nThe phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.<\/p>\n

              Question 41.
              \nWhat is phase of a wave?
              \nAnswer:
              \nPhase is a particular point in time on the cycle of a waveform, measured as an angle in degrees.<\/p>\n

              Question 42.
              \nObtain the relation between phase difference and path difference.
              \nAnswer:
              \nPhase difference (\u03a6):
              \nIt is the difference expressed in degrees or radians between two waves having same frequency and referenced to same point in time.
              \nPath difference (\u03b4):
              \nIt is the difference between the lengths of two paths of the two different having same frequency and travelling at same velocity. \u03b4 =\\(\\frac { \\lambda }{ 2\\pi } \\) \u03a6<\/p>\n

              Question 43.
              \nWhat are coherent sources?
              \nAnswer:
              \nTwo sources are said to be coherent if they emit light waves of the same wavelength and start with same phase or have a constant phase difference.<\/p>\n

              Question 44.
              \nWhat is intensity division?
              \nAnswer:
              \nIntensity’ or amplitude division: If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams. They will be either in-phase or at constant phase difference.<\/p>\n

              Question 45.
              \nHow does wavefront division provide coherent sources?
              \nAnswer:
              \nWavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.<\/p>\n

              Question 46.
              \nHow do source and images behave as coherent sources?
              \nAnswer:
              \nSource and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel\u2019s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd\u2019s mirror uses a source and its virtual image as two coherent sources.<\/p>\n

              Question 47.
              \nWhat is bandwidth of interference pattern?
              \nAnswer:
              \nBandwidth is defined as the distance between any two consecutive bright or dark fringes.<\/p>\n

              Question 48.
              \nWhat is diffraction?
              \nAnswer:
              \nDiffraction is the bending of waves around sharp edges into the geometrically shadowed region.<\/p>\n

              Question 49.
              \nDifferentiate between Fresnel and Fraunhofer diffraction.
              \nAnswer:<\/p>\n\n\n\n\n\n\n\n
              S.No.<\/strong><\/td>\nFresnel diffraction<\/strong><\/td>\nFraunhofer diffraction<\/strong><\/td>\n<\/tr>\n
              1.<\/td>\nSpherical or cylindrical wavefront undergoes diffraction<\/td>\nPlane wavefront undergoes diffraction<\/td>\n<\/tr>\n
              2.<\/td>\nLight wave is from a source at finite distance<\/td>\nLight wave is from a source at infinity<\/td>\n<\/tr>\n
              3.<\/td>\nFor laboratory conditions, convex lenses need not be used<\/td>\nIn laboratory conditions, convex lenses are to be used<\/td>\n<\/tr>\n
              4.<\/td>\ndifficult to observe and analyse<\/td>\nEasy to observe and analyse<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Question\u00a050.
              \nDiscuss the special cases on first minimum in Fraunhofer diffraction.
              \nLet us consider the condition for first minimum with (n = 1). a sin \u03b8 = \u03bb
              \nThe first minimum has an angular spread of, sin \u03b8 = \\(\\frac { \\lambda }{ a } \\). Special cases to discuss on the condition.
              \n1. When a < \u03bb, the diffraction is not possible, because sin 0 can never be greater than 1.
              \n2. When a \u2265 \u03bb, the diffraction is possible.<\/p>\n

                \n
              • For a = \u03bb, sin \u03b8 = 1 i.e, \u03b8 = 90\u00b0. That means the first minimum is at 90\u00b0. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90\u00b0.<\/li>\n
              • For a >> \u03bb, sin \u03b8 << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.<\/li>\n<\/ul>\n

                3. When a > \u03bb and also comparable, say a = 2\u03bb, sin \u03b8 = \\(\\frac { \\lambda }{ a } \\) = \\(\\frac { \\lambda }{ { 2\\lambda } } \\) =\\(\\frac { 1 }{ 2 }\\); then \u03b8 = 30\u00b0. These are practical cases where diffraction could be observed effectively.<\/p>\n

                Question 51.
                \nWhat is Fresnel\u2019s distance? Obtain the equation for Fresnel\u2019s distance.
                \nAnswer:
                \nFresnel\u2019s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
                \nFresnel\u2019s distance z as, z = \\(\\frac { { a }^{ 2 } }{ 2\\lambda } \\).<\/p>\n

                Question 52.
                \nMention the differences between interference and diffraction.
                \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
                S.No.<\/strong><\/td>\nInterference<\/strong><\/td>\nDiffraction<\/strong><\/td>\n<\/tr>\n
                1.<\/td>\nSuperposition of two waves<\/td>\nBending of waves around edges<\/td>\n<\/tr>\n
                2.<\/td>\nSuperposition of waves from two coherent sources.<\/td>\nSuperposition wavefronts emitted from various points of the same wavefront.<\/td>\n<\/tr>\n
                3.<\/td>\nEqually spaced fringes.<\/td>\nUnequally spaced fringes<\/td>\n<\/tr>\n
                4.<\/td>\nIntensity of all the bright fringes is almost same<\/td>\nIntensity falls rapidly for higher orders<\/td>\n<\/tr>\n
                5.<\/td>\nLarge number of fringes are obtained<\/td>\nLess number of fringes are obtained<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Question 53.
                \nWhat is a diffraction grating?
                \nAnswer:<\/p>\n

                  \n
                1. Grating has multiple slits with equal widths of size comparable to the wavelength of diffracting light.<\/li>\n
                2. It is a plane sheet of transparent material on which opaque rulings are made with fine diamond pointer.<\/li>\n<\/ol>\n

                  Question 54.
                  \nWhat are resolution and resolving power?
                  \nAnswer:
                  \nOptical resolution describes the ability of an imaging system to resolve detail in the object that is being imaged. Resolving power is the ability of an optical instrument to resolve or separate the image of two nearby point objects so that they can be distinctly seen. It is equal to the reciprocal of the limit of resolution of the optical instrument.<\/p>\n

                  Question 55.
                  \nWhat is Rayleigh\u2019s criterion?
                  \nAnswer:<\/p>\n

                    \n
                  1. For two-point objects are just resolved, the minimum distance between their diffraction images must be in such a way that central maximum coincides with the first minimum of other and vice versa.<\/li>\n
                  2. Such an image is resolved image.<\/li>\n
                  3. It is the limit of resolution.<\/li>\n<\/ol>\n

                    Question 56.
                    \nWhat is polarisation?
                    \nAnswer:
                    \nThe phenomenon of restricting the vibrations of light (electric or magnetic field vector) to a particular direction perpendicular to the direction of wave propagation motion is called polarization of light.<\/p>\n

                    Question 57.
                    \nDifferentiate between polarised and unpolarised light.
                    \nAnswer:<\/p>\n\n\n\n\n\n\n
                    S.No.<\/strong><\/td>\nPolarised light<\/strong><\/td>\nUnpolarised light<\/strong><\/td>\n<\/tr>\n
                    1.<\/td>\nConsists of waves having their electric field vibrations in a single plane normal to the direction of ray.<\/td>\nConsists of waves having their electric field vibrations equally distributed in all directions normal to the direction of ray.<\/td>\n<\/tr>\n
                    2.<\/td>\nAsymmetrical about the ray direction<\/td>\nSymmetrical about the ray direction<\/td>\n<\/tr>\n
                    3.<\/td>\nIt is obtained from unpolarised light with the help of polarisers<\/td>\nProduced by conventional light sources.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                    Question 58.
                    \nDiscuss polarisation by selective absorption.
                    \nAnswer:
                    \nSelective absorption is the property of a material which transmits waves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism.<\/p>\n

                    Question 59.
                    \nWhat are polariser and analyser?
                    \nAnswer:<\/p>\n

                      \n
                    1. Tourmaline is a natural polarising material.<\/li>\n
                    2. The polaroid which plane polarises the unpolarised light passing through it is called polariser.<\/li>\n
                    3. The polaroid which is used to examine whether a beam of light is polarised or not is called an analyzer.<\/li>\n<\/ol>\n

                      Question 60.
                      \nWhat are plane polarised, unpolarised and partially polarised light?
                      \nAnswer:
                      \nPlane polarised:
                      \nIf the vibrations of a wave are present in only one direction in a plane perpendicular to the direction of propagation of the wave is said to be polarised or plane polarised light.<\/p>\n

                      Unpolarised:
                      \nA transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.<\/p>\n

                      Partially polarised light:
                      \nIf the intensity of light varies between maximum and minimum for every’ rotation of 90\u00b0 of the analyser, the light is said to be partially polarised light.<\/p>\n

                      Question 61.
                      \nState and obtain Malus\u2019 law.
                      \nAnswer:
                      \nWhen a beam of plane polarised light of intensity I0<\/sub> is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle 0 between the transmission axis of polariser and analyser. This is known as Malus\u2019 law. I = I0<\/sub> cos2<\/sup>\u03b8<\/p>\n

                      Question 62.
                      \nList the uses of polaroids.
                      \nAnswer:
                      \nUses of polaroids:<\/p>\n

                        \n
                      1. In goggles and cameras to avoid glares<\/li>\n
                      2. In 3 – D picture.<\/li>\n
                      3. To improve contrast in old oil paintings.<\/li>\n
                      4. In optical stress analysis.<\/li>\n
                      5. As window glass to adjust intensity.<\/li>\n
                      6. As needle to read\/write in CD (compact disc)<\/li>\n
                      7. In LCD (Liquid Crystal Display)<\/li>\n<\/ol>\n

                        Question 63.
                        \nState Brewster\u2019s law.
                        \nAnswer:
                        \nThe law states that the tangent of the polarising angle for a transparent medium is equal to its refractive index, tan i = n. This relation is known as Brewster\u2019s law.<\/p>\n

                        Question 64.
                        \nWhat is angle of polarisation and obtain the equation for angle of polarisation.
                        \nAnswer:
                        \nThe angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster\u2019s angle. It is denoted by ip<\/sub>
                        \nip<\/sub> = 90\u00b0 – Rp<\/sub><\/p>\n

                        Question 65.
                        \nDiscuss pile of plates.
                        \nAnswer:
                        \nThe phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of 33.7\u00b0 (90\u00b0 – 56.3\u00b0) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3\u00b0 which is the polarising angle for glass.<\/p>\n

                        The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser.<\/p>\n

                        Question 66.
                        \nWhat is double refraction?
                        \nAnswer:
                        \nWhen a ray of unpolarised light is incident on a calcite crystal, the two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.<\/p>\n

                        Question 67.
                        \nMention the types of optically active crystals for example.
                        \nAnswer:
                        \nTypes of optically active crystals:
                        \nUniaxial crystals:
                        \nCrystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.<\/p>\n

                        Biaxial crystals:
                        \nCrystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.<\/p>\n

                        Question 68.
                        \nDiscuss Nicol prism.
                        \nAnswer:
                        \nNicol prism is an optical device incorporated in optical instruments both for producing and analysing plane polarised light. The construction of a Nicol prism is based on the phenomenon of Double refraction. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.<\/p>\n

                        It is cut into two halves along the diagonal so that their face angles are 72\u00b0 and 108\u00b0. The two halves are joined i together by a layer of Canada balsam, a transparent cement.<\/p>\n

                        Question 69.
                        \nHow is polarisation of light obtained by scattering of light?
                        \nAnswer:<\/p>\n

                          \n
                        1. The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated.<\/li>\n
                        2. This is because of sunlight, which has changed its direction (having been scattered) on encountering the molecules of the earth’s atmosphere.<\/li>\n
                        3. The electric field of light interact with the electrons present in the air molecules.<\/li>\n
                        4. Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions.<\/li>\n
                        5. Charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component.<\/li>\n
                        6. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.<\/li>\n
                        7. This explains the reason for polarisation of sunlight by scattering.<\/li>\n<\/ol>\n

                          Question 70.
                          \nDiscuss simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
                          \nAnswer:
                          \nA simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.<\/p>\n

                            \n
                          1. Magnification in near point focusing m = 1 + \\(\\frac { D }{ f }\\)<\/li>\n
                          2. Magnification in normal focusing (angular magnification), m = \\(\\frac { D }{ f }\\)<\/li>\n<\/ol>\n

                            Question 71.
                            \nWhat are near point and normal focusing?
                            \nAnswer:<\/p>\n

                              \n
                            • Near point focusing:
                              \nThe image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.<\/li>\n
                            • Normal focusing:
                              \nThe image is formed at infinity. In this position the eye is most relaxed to view the image.<\/li>\n<\/ul>\n

                              Question 72.
                              \nWhy is oil-immersed objective preferred in a microscope?
                              \nAnswer:
                              \nIt is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.<\/p>\n

                              Question 73.
                              \nWhat are the advantages and disadvantages of using a reflecting telescope?
                              \nAnswer:
                              \nAdvantages:<\/p>\n

                                \n
                              1. Only one surface is too polished and maintain<\/li>\n
                              2. Mirrors weigh much less<\/li>\n
                              3. Support can be given by the entire back of the mirror.<\/li>\n<\/ol>\n

                                Disadvantages:<\/p>\n

                                  \n
                                1. The objective mirror would focus the light inside the telescope tube.<\/li>\n
                                2. One must have an eyepiece inside obstructing some light.<\/li>\n<\/ol>\n

                                  Question 74.
                                  \nWhat is the use of an erecting lens in a terrestrial telescope?
                                  \nAnswer:
                                  \nA terrestrial telescope has an additional erecting lens to make the final image erect.<\/p>\n

                                  Question 75.
                                  \nWhat is the use of a collimator?
                                  \nAnswer:
                                  \nIt is a device which filters a stream of rays so that only those travelling parallel to a specified direction are allowed through it.<\/p>\n

                                  Question 76.
                                  \nWhat are the uses of a spectrometer?
                                  \nAnswer:
                                  \nThe spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.<\/p>\n

                                  Question 77.
                                  \nWhat is myopia? What is its remedy?
                                  \nAnswer:
                                  \nMyopia (or) short-sightedness:
                                  \nIt is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.<\/p>\n

                                  Remedy (correction):
                                  \nA myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.<\/p>\n

                                  Question 78.
                                  \nWhat is hypermetropia? What is its remedy?
                                  \nAnswer:
                                  \nHypermetropia (or) Long-sightedness: It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly.
                                  \nRemedy (correction): A hypermetropic eye is corrected by using a convex lens of suitable focal length.<\/p>\n

                                  Question 79.
                                  \nWhat is presbyopia?
                                  \nAnswer:
                                  \nThe long-sightedness (farsightedness) defect of vision occurring with age is called presbyopia.<\/p>\n

                                  Question 80.
                                  \nWhat is astigmatism?
                                  \nAnswer:
                                  \nAstigmatism is the defect arising due to different curvatures along different planes in the eye lens. An astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.<\/p>\n

                                  Samacheer Kalvi 12th Physics Optics Long Answer Questions<\/span><\/strong><\/p>\n

                                  Question 1.
                                  \nDerive the mirror equation and the equation for lateral magnification.
                                  \nAnswer:
                                  \nThe mirror equation:
                                  \n1. The mirror equation establishes a relation among object distance u, image distance v and focal length\/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C’.
                                  \n2. Let us consider three paraxial rays from point B on the object.
                                  \n3. The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PBThe third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
                                  \n4. The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.
                                  \n\"Samacheer
                                  \nAs per law of reflection, the angle of incidence \u2220BPA is equal to the angle of reflection \u2220B’PA’. The triangles \u2206BPA and \u2206B’PA’ are similar. Thus, from the rule of similar triangles,
                                  \n\\(\\frac { A’B’ }{ AB }\\) = \\(\\frac { PA’ }{ PA }\\) …….. (1)
                                  \nThe other set of similar triangles are, ADPF and A BA.’ F. (PD is almost a straight vertical line)
                                  \n\\(\\frac { A’B’ }{ PD }\\) = \\(\\frac { AF’ }{ PF }\\)
                                  \nAs, the distances PD = AB the above equation becomes,
                                  \n\\(\\frac { A’B’ }{ AB }\\) = \\(\\frac { AF’ }{ PF }\\) ……… (2)
                                  \nFrom equations (1) and (2) we can write,
                                  \n\\(\\frac { PA’ }{ PA }\\) = \\(\\frac { AF’ }{ PF }\\)
                                  \nAs, A’ F = PA’ – PF, the above equation becomes,
                                  \n\\(\\frac {PA’ }{ PA }\\) = \\(\\frac { PA’-PF’P }{ PF }\\) ……….. (3)
                                  \nWe can apply the sign conventions for the various distances in the above equation.
                                  \nPA = – u, PA’ = -v, PF = -f
                                  \nAll the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
                                  \n\\(\\frac { -v }{ -u }\\) = \\(\\frac { -v\\left( -f \\right) }{ -f } \\)
                                  \nOn further simplification,
                                  \n\\(\\frac { -v }{ -u }\\) = \\(\\frac { v-f }{ f }\\); \\(\\frac { v }{ u }\\)=\\(\\frac { v }{ f }\\)=1
                                  \nDividing either side with v,
                                  \n\\(\\frac { 1 }{ u }\\)=\\(\\frac { 1 }{ f }\\)=\\(\\frac { 1 }{ v }\\)
                                  \nAfter rearranging,
                                  \n\\(\\frac { 1 }{ v }\\)+\\(\\frac { 1 }{ u }\\)=\\(\\frac { 1 }{ f }\\) ……… (4)
                                  \nThe above equation (4) is called mirror equation.<\/p>\n

                                  Lateral magnification in spherical mirrors:
                                  \nThe lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
                                  \n\"Samacheer
                                  \nm=\\(\\frac { h’ }{ h }\\) ……..(5)
                                  \nApplying proper sign conventions for equation (1),
                                  \n\\(\\frac { A’B’ }{ AB }\\) = \\(\\frac { PA’ }{ PA }\\)
                                  \nA\u2019B’ = -h, AB = h, PA’ = -v, PA = -u
                                  \n\\(\\frac { -h’ }{ h }\\)=\\(\\frac { -v }{ -u }\\)
                                  \nOn simplifying we get,
                                  \nm=\\(\\frac { h’ }{ h }\\)=-\\(\\frac { v }{ u }\\) ………. (6)
                                  \nUsing mirror equation, we can further write the magnification as,
                                  \nm=\\(\\frac { h’ }{ h }\\)–\\(\\frac { f-v }{ f }\\)=\\(\\frac { f }{ f-u }\\) ………… (7)<\/p>\n

                                  Question 2.
                                  \nDescribe the Fizeau\u2019s method to determine speed of light.
                                  \nAnswer:
                                  \nFizeau\u2019s method to determine speed of light:
                                  \nApparatus:
                                  \nThe light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45\u00b0 to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism. The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed w’heel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.<\/p>\n

                                  Working:
                                  \nThe angular speed of rotation of the toothed wheel was increased from zero to a value to until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.<\/p>\n

                                  Expression for speed of light:
                                  \nThe speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
                                  \nv = \\(\\frac { 2d }{ t }\\) c ….. (1)
                                  \nThe distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed co of the toothed wheel.
                                  \nThe angular speed \u03c9 of the toothed wheel when the light disappeared for the first time is,
                                  \n\u03c9 = \\(\\frac { \u03b8 }{ t }\\) ……. (2)
                                  \nHere, 0 is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t.
                                  \n\"Samacheer
                                  \n\"Samacheer
                                  \n\u03b8 = \\(\\frac { 2\u03c0 }{ 2N }\\) = \\(\\frac { \u03c0 }{ N }\\)
                                  \nSubstituting for 0 in the equation (2) for ,
                                  \n\u03c9 = \\(\\frac { \\pi \/n }{ 2 } \\) = \\(\\frac { \u03c0 }{ Nt }\\)
                                  \nRewriting the above equation for t,
                                  \nt = \\(\\frac { \u03c0 }{ N\u03c9 }\\) ……. (3)
                                  \nSubstituting t from equation (3) in equation (1), v = \\(\\frac { 2d }{ \\pi \/N\\omega } \\)
                                  \nAfter rearranging,
                                  \nv = \\(\\frac { 2dN\u03c9 }{ \u03c0 }\\) …….. (4)
                                  \nFizeau had some difficulty to visually estimate the minimum intensity of the light when blocked by the adjacent tooth, and his value for speed of light was very close to the actual value.<\/p>\n

                                  Question 3.
                                  \nObtain the equation for radius of illumination (or) Snell\u2019s window.
                                  \nAnswer:
                                  \nThe radius of Snell\u2019s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell\u2019s law in product form, equation n2 <\/sub>sin i = n2 <\/sub>sin r for the refraction happening at the point B on the boundary between the two media is,
                                  \nn1 <\/sub>sin ic <\/sub>= n2 <\/sub>sin 90\u00b0 ……. (1)
                                  \nn1 <\/sub>sin ic<\/sub>= n2<\/sub> \u2235 sin 90\u00b0 = 1
                                  \nsin ic <\/sub>= \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\) …….. (2)
                                  \n\"Samacheer
                                  \nFrom the right angle triangle \u2206ABC,
                                  \nsin ic <\/sub>= \\(\\frac { R }{ \\sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \\) ……. (3)
                                  \nEquating the above two equation (3) and equation (2).
                                  \n\\(\\frac { R }{ \\sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \\) = \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\)
                                  \n\"Samacheer
                                  \n\"Samacheer
                                  \nIf the rarer medium outside is air, then, n2 <\/sub>= 1, and we can take n1 <\/sub>= n
                                  \n\"Samacheer<\/p>\n

                                  Question 4.
                                  \nDerive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:
                                  \nAnswer:
                                  \nTo ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n1<\/sub>, cladding n2<\/sub> and the outer medium n3<\/sub>. Assume the light is incident at an angle called acceptance angle i at the outer medium and core boundary at A.
                                  \nThe Snell\u2019s law in the product form, equation for this refraction at the point A.
                                  \n\"Samacheer
                                  \nn3<\/sub> sin ia<\/sub> = n1 <\/sub>sin ra<\/sub> …(1)
                                  \nTo have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle ic. Snell\u2019s law in the product form, equation for the refraction at point B is,
                                  \nn1 <\/sub>sin ic <\/sub>= n2 <\/sub>sin 90\u00b0 …(2)
                                  \nn1 <\/sub>sin ic<\/sub>= n2<\/sub> \u2235 sin 90\u00b0 =1
                                  \n\u2234sin ic<\/sub>= \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\) …(3)
                                  \nFrom the right angle triangle \u2206ABC,
                                  \nic <\/sub>= 90\u00b0-ra<\/sub>
                                  \nNow, equation (3) becomes, sin (90\u00b0 – ra<\/sub>) = \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\)
                                  \nUsing trigonometry’, cos ra = ra<\/sub> = \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\) …….. (4)
                                  \nsin ra<\/sub> = \\(\\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \\)
                                  \nSubstituting for cos ra<\/sub>
                                  \n\"Samacheer
                                  \nIf outer medium is air, then n3<\/sub> = 1. The acceptance angle ia<\/sub> becomes,
                                  \nra<\/sub>=sin-1<\/sup>\\(\\left( \\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \\right) \\) …….. (9)
                                  \nLight can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n3 <\/sub>sin ia<\/sub>) is called numerical aperture NA of the optical fibre.
                                  \nNA =n3 <\/sub>sin ia<\/sub> \\(\\left( \\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \\right) \\) …….. (10)
                                  \nIf outer medium is air, then n3<\/sub> = 1. The numerical aperture NA becomes,
                                  \nNA = sin ia<\/sub> = \\(\\left( \\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \\right) \\) …….. (11)<\/p>\n

                                  Question 5.
                                  \nObtain the equation for lateral displacement of light passing through a glass slab.
                                  \nAnswer:
                                  \nConsider a glass slab of thickness t and refractive index n is kept in air medium. The path of the light is ABCD and the refractions occur at two points B and C in the glass slab. The angles of incidence i and refraction r are measured with respect to the normal N1<\/sub>and N2<\/sub> at the two points B and C respectively. The lateral displacement L is the perpendicular distance CE drawn between the path of light and the undeviated path of light at point C. In the right angle triangle \u2206BCE,
                                  \n\"Samacheer
                                  \nsin(i – r) = \\(\\frac { L }{ BC }\\); BC= \\(\\frac { L }{ sin(i-r) } \\) …… (1)
                                  \nIn the right angle triangle \u2206BCF,
                                  \ncos(r) = \\(\\frac { t }{ BC }\\) ; BC= \\(\\frac { t }{ cos(r) } \\) …… (2)
                                  \nEquating equations (1) and (2)
                                  \n\\(\\frac { L }{ sin(i-r) } \\) = \\(\\frac { t }{ cos(r) } \\)
                                  \nAfter rearranging,
                                  \nL = t\\(\\left( \\frac { sin(i-r) }{ cos(r) } \\right) \\)
                                  \nLateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.<\/p>\n

                                  Question 6.
                                  \nDerive the equation for refraction at single spherical surface.
                                  \nAnswer:
                                  \nEquation for refraction at single spherical surface:
                                  \nLet us consider two transparent media having refractive indices n1<\/sub>and n2<\/sub> are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n1<\/sub>. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.
                                  \n\"Samacheer
                                  \nLight from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n2<\/sub> > n1<\/sub>, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
                                  \nSnell\u2019s law in product form for the refraction at the point N could be written as,
                                  \nn1 <\/sub>sin i = n2<\/sub> sin r …(1)
                                  \nAs the angles are small, sin of the angle could be approximated to the angle itself.
                                  \nn1<\/sub> i = n2<\/sub> r …(2)
                                  \nLet the angles,
                                  \n\u2220NOP = \u03b1, \u2220NCP = \u03b2, \u2220NIP = \u03b3
                                  \ntan \u03b1 = \\(\\frac { PN }{ PO }\\);tan \u03b2 = \\(\\frac { PN }{ PC }\\);tan \u03b3 = \\(\\frac { PN }{ PI }\\)
                                  \nAs these angles are small, tan of the angle could be approximated to the angle itself.
                                  \n\u03b1 = \\(\\frac { PN }{ PO }\\); \u03b2 = \\(\\frac { PN }{ PC }\\); \u03b3 = \\(\\frac { PN }{ PI }\\) …….. (3)
                                  \nFor the triangle, \u2206ONC,
                                  \nI = \u03b1 + \u03b2 …….. (4)
                                  \nFor the triangle, \u2206INC,
                                  \n\u03b2 = r + \u03b3 (or) r = \u03b2 – \u03b3 ……… (5)
                                  \nSubstituting for I and r from equations (4) and (5) in the equation (2).
                                  \nn1 <\/sub>(\u03b1 + \u03b2) = n2 <\/sub>(\u03b2 – \u03b3)
                                  \nRearranging,
                                  \nn1<\/sub> \u03b1 + n2<\/sub>\u03b3 = (n2 <\/sub>– n1<\/sub>)\u03b2
                                  \nSubstituting for \u03b1, \u03b2 and \u03b3 from equation (3)
                                  \nn1<\/sub> = (\\(\\frac { PN }{ PO }\\)) + n2<\/sub> = (\\(\\frac { PN }{ PI }\\)) = (n2<\/sub>-n1<\/sub>) (\\(\\frac { PN }{ PC }\\))
                                  \nFurther simplifying by cancelling PN,
                                  \n\\(\\frac { { n }_{ 1 } }{ PO } \\) + \\(\\frac { { n }_{ 2 } }{ PI } \\) = \\(\\frac { { n }_{ 2 }-{ n }_{ 1 } }{ PC } \\)
                                  \nFollowing sign conventions, PO = -u, PI = +v and PC = +R in equation (6),
                                  \n\\(\\frac { { n }_{ 2 } }{ -v } \\) – \\(\\frac { { n }_{ 1 } }{ u } \\) = \\(\\frac { \\left( { n }_{ 2 }-{ n }_{ 1 } \\right) }{ R } \\)
                                  \nAfter rearranging, finally we get,
                                  \n\\(\\frac { { n }_{ 2 } }{ v } \\) – \\(\\frac { { n }_{ 1 } }{ u } \\) = \\(\\frac { \\left( { n }_{ 2 }-{ n }_{ 1 } \\right) }{ R } \\) ……… (7)
                                  \nEquation (7) gives the relation among the object distance n, image distance v, refractive indices of the two media (n1<\/sub> and n2<\/sub>) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
                                  \nIf the first medium is air then, n1<\/sub> = 1 and the second medium is taken just as n2<\/sub> = n, then the equation is reduced to,
                                  \n\\(\\frac { n }{ v}\\)–\\(\\frac { 1 }{ u}\\) = \\(\\frac { \\left( { n }-1 \\right) }{ R } \\) ………(8)<\/p>\n

                                  Question 7.
                                  \nObtain lens maker\u2019s formula and mention its significance.
                                  \nAnswer:
                                  \nLens maker\u2019s formula and lens equation:
                                  \nLet us consider a thin lens made up of a medium of refractive index n2<\/sub>, is placed in a medium of refractive index n1<\/sub>. Let R1<\/sub> and R2<\/sub> be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by
                                  \n\\(\\frac { { n }_{ 2 } }{ v } \\) – \\(\\frac { { n }_{ 1 } }{ u } \\) = \\(\\frac { \\left( { n }_{ 2 }-{ n }_{ 1 } \\right) }{ R } \\)
                                  \nFor the refracting surface (1), the light goes from n1<\/sub>ton2<\/sub>.
                                  \n\\(\\frac { { n }_{ 2 } }{ v’ } \\) – \\(\\frac { { n }_{ 1 } }{ u } \\) = \\(\\frac { \\left( { n }_{ 2 }-{ n }_{ 1 } \\right) }{ R }_{ 1 } \\) …….. (1)
                                  \nFor the refracting surface (2), the light goes from medium n2<\/sub>ton1<\/sub>.
                                  \n\\(\\frac { { n }_{ 1 } }{ v } \\) – \\(\\frac { { n }_{ 2 } }{ v’ } \\) = \\(\\frac { \\left( { n }_{ 1 }-{ n }_{ 2 } \\right) }{ R }_{ 2 } \\) …….. (2)
                                  \nAdding the above two equation (1) and (2)
                                  \n\"Samacheer
                                  \n\\(\\frac { { n }_{ 1 } }{ v } \\) – \\(\\frac { { n }_{ 1 } }{ u } \\) = (n2 <\/sub>– n1<\/sub>) \\(\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\)
                                  \nFurhter simplifying and rearranging,
                                  \n\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\) = \\(\\left(\\frac{n_{2}-n_{1}}{n_{1}}\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\)
                                  \n\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\) = \\(\\left(\\frac{n_{2}}{n_{1}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) ……….. (3)
                                  \nIf the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \u221e, v= f. Then the equation becomes.
                                  \n\\(\\frac { 1 }{ f }\\)–\\(\\frac { 1 }{ \u221e }\\) = \\(\\left(\\frac{n_{2}}{n_{1}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\)
                                  \n\\(\\frac { 1 }{ f }\\)=\\(\\left(\\frac{n_{2}}{n_{1}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) ………… (4)
                                  \nIf the refractive index of the lens is and it is placed in air, then n2<\/sub> = n and n1<\/sub> = 1 equation (4) becomes,
                                  \n\\(\\frac { 1 }{ f }\\) = (n-1) \\(\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) ………. (5)
                                  \nThe above equation is called the lens maker\u2019s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
                                  \n\\(\\frac { 1 }{ v }\\)–\\(\\frac { 1 }{ u }\\) =\\(\\frac { 1 }{ f }\\) …… (6)
                                  \nThis equation is known as lens equation which relates the object distance u and image distance v with the focal length f of the lens. This formula holds good for a any type of lens.<\/p>\n

                                  Question 8.
                                  \nDerive the equation for thin lens and obtain its magnification.
                                  \nAnswer:
                                  \nLateral magnification in thin lens:
                                  \nLet us consider an object 00′ of height h1<\/sub> placed on the principal axis with its height perpendicular to the principal axis. The ray Op passing through the pole of the lens goes undeviated. The inverted real image II’ formed has a height h2<\/sub>.
                                  \n\"Samacheer
                                  \nThe lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
                                  \nm = \\(\\frac { II’ }{ OO’ }\\) …….. (1)
                                  \nFrom the two similar triangles \u2206 POO’ and \u2206 PII’, we can write,
                                  \n\\(\\frac { H’ }{ OO’ }\\) = \\(\\frac { PI }{ PO }\\) ……..(2)
                                  \nApplying sign convention,
                                  \n\\(\\frac{-h_{2}}{h_{1}}\\) = \\(\\frac { v }{ -u}\\)
                                  \nSubstituting this in the equation (1) for magnification,
                                  \nm = \\(\\frac{-h_{2}}{h_{1}}\\) = \\(\\frac { v }{ -u}\\)
                                  \nAfter rearranging,
                                  \nm = \\(\\frac{h_{2}}{h_{1}}\\) = \\(\\frac { v }{ u}\\) …….. (3)
                                  \nThe magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one. We can also have the equations for magnification by combining the lens equation with the formula for magnification
                                  \nas,
                                  \nm = \\(\\frac{h_{2}}{h_{1}}=\\frac{f}{f+u}\\) (or) m = \\(\\frac{h_{2}}{h_{1}}=\\frac{f-v}{f}\\) ……… (4)<\/p>\n

                                  Question 9.
                                  \nDerive the equation for effective focal length for lenses in out of contact.
                                  \nAnswer:
                                  \nConsider a two lenses of focal length f1<\/sub> and f2<\/sub> arranged coaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produce deviations \u03b41<\/sub> and \u03b42 <\/sub>respectively and The net deviation \u03b4 is,
                                  \n\u03b4 = \u03b41<\/sub> + \u03b42<\/sub> ……. (1)
                                  \nFrom Angle of deviation in lens equation, \u03b4 = \\(\\frac { h }{ f }\\)
                                  \n\u03b41<\/sub> = \\(\\frac{h_{1}}{f_{1}}\\); \u03b42<\/sub> = \\(\\frac{h_{2}}{f_{2}}\\) \u03b4 = \\(\\frac{h_{1}}{f}\\) ….. (2)
                                  \nThe equation (1) becomes,
                                  \n\"Samacheer
                                  \n\\(\\frac{h_{1}}{f}\\) = \\(\\frac{h_{1}}{f_{1}}\\) + \\(\\frac{h_{2}}{f_{2}}\\) ……. (3)
                                  \nFrom the geometry,
                                  \nh2 <\/sub>– h1<\/sub> = P2<\/sub>C = CG
                                  \nh2 <\/sub>– h1<\/sub> = BG tan \u03b41<\/sub> \u2248 BG \u03b41<\/sub>
                                  \nh2 <\/sub>– h1<\/sub> = d\\(\\frac{h_{1}}{f_{1}}\\)
                                  \nh2<\/sub> = h1<\/sub> + d\\(\\frac{h_{1}}{f_{1}}\\) …….. (4)
                                  \nSubstituting the above equation in Equation (3)
                                  \n\\(\\frac{h_{1}}{f}=\\frac{h_{1}}{f_{1}}+\\frac{h_{1}}{f_{2}}+\\frac{h_{1} d}{f_{1} f_{2}}\\)
                                  \nOn further simplification,
                                  \n\\(\\frac{1}{f}=\\frac{1}{f_{1}}+\\frac{1}{f_{2}}+\\frac{d}{f_{1} f_{2}}\\) …….. (5)
                                  \nThe above equation could be used to find the equivalent focal length.<\/p>\n

                                  Question 10.
                                  \nDerive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
                                  \nAnswer:
                                  \nAngie of deviation produced by prism:
                                  \nLet light ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face AB are i1<\/sub> and r1<\/sub>. The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is r2<\/sub> and i2<\/sub> respectively. RS is the ray emerging from p the second face. Angle i2<\/sub> is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
                                  \n\"Samacheer
                                  \nThe two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M.
                                  \nThe deviation d1<\/sub> at the surface AB is,
                                  \nangle \u2220RQM = d1<\/sub> = i1 <\/sub>– r1<\/sub> …(1)
                                  \nThe deviation d2<\/sub> at the surface AC is,
                                  \nangle \u2220QRM = d2<\/sub> = i2 <\/sub>– r2<\/sub> …(2)
                                  \nTotal angle of deviation d produced is,
                                  \nd = d1<\/sub> + d2<\/sub> …(3)
                                  \nSubstituting for d1<\/sub> and d2<\/sub>,
                                  \nd=(i1 <\/sub>– r1<\/sub>) + (i2 <\/sub>– r2<\/sub>)
                                  \nAfter rearranging,
                                  \nd = (i1 <\/sub>– r1<\/sub>) + (i2 <\/sub>– r2<\/sub>) …(4)
                                  \nIn the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180\u00b0.
                                  \n\u2220A + \u2220QNR = 180\u00b0 …(5)
                                  \nFrom the triangle \u2206QNR,
                                  \nr1<\/sub>+ r2<\/sub> \u2220QNR = 180\u00b0 …(6)
                                  \nComparing these two equations (5) and (6) we get,
                                  \nr1<\/sub>+ r2<\/sub> = A …(7)
                                  \nSubstituting this in equation (4) for angle of deviation,
                                  \nd= i1<\/sub>+ i2<\/sub> -A …(8)
                                  \nThus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
                                  \nRefractive index of the material of the prism:
                                  \nAt minimum deviation,
                                  \ni1<\/sub> = i2<\/sub> = i and r1<\/sub> = r2<\/sub> = r
                                  \nNow, the equation (8) becomes,
                                  \nD = i1<\/sub> + i2<\/sub>-A (or) i = \\(\\frac { \\left( A+D \\right) }{ 2 } \\)
                                  \nThe equation (7) becomes,
                                  \nr1<\/sub> + r2<\/sub> = A \u21d2 2r = A (or) r = \\(\\frac { A }{ 2 }\\)
                                  \nSubstituting i and r in Snell\u2019s law,
                                  \nn = \\(\\frac { sin i }{ sin r }\\)
                                  \nn = \\(\\frac{\\sin \\left(\\frac{A+D}{2}\\right)}{\\sin \\left(\\frac{A}{2}\\right)}\\)
                                  \nThe above equation is used to find the refractive index of the material of the prism.<\/p>\n

                                  Question 11.
                                  \nWhat is dispersion? Obtain the equation for dispersive power of a medium.
                                  \nAnswer:
                                  \nDispersion. Dispersion is splitting of white light into its constituent colours.
                                  \nDispersive Power:
                                  \nConsider a beam of white light passes through a prism; It gets dispersed into its constituent colours. Let \u03b4V<\/sub>, \u03b4R<\/sub> are the angles of deviation for violet and red light. Let nV<\/sub> and nR<\/sub> are the refractive indices for the violet and red light respectively.
                                  \n\"Samacheer
                                  \nThe refractive index of the material of a prism is given by the equation
                                  \nn = \\(\\frac{\\sin \\left(\\frac{A+D}{2}\\right)}{\\sin \\left(\\frac{A}{2}\\right)}\\)
                                  \nHere A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10\u00b0, the prism is said to be a small angle prism. When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a small angle prism and 5 the angle of deviation then the prism formula becomes.
                                  \nn = \\(\\frac{\\sin \\left(\\frac{A+\\delta}{2}\\right)}{\\sin \\left(\\frac{A}{2}\\right)}\\) ……… (1)
                                  \nFor small angles of A and \u03b4m<\/sub>,
                                  \n\\(\\sin \\left(\\frac{\\mathrm{A}+\\delta}{2}\\right) \\approx\\left(\\frac{\\mathrm{A}+\\delta}{2}\\right)\\) ……. (2)
                                  \n\\(\\sin \\left(\\frac{\\mathrm{A}}{2}\\right) \\approx\\left(\\frac{\\mathrm{A}}{2}\\right)\\) …… (3)
                                  \n\u2234 n = \\(\\frac{\\left(\\frac{A+\\delta}{2}\\right)}{\\left(\\frac{A}{2}\\right)}=\\frac{A+\\delta}{A}=1+\\frac{\\delta}{A}\\)
                                  \nFurther simplifying, \\(\\frac { \u03b4 }{ A }\\) = n – 1
                                  \n\u03b4 = (n – 1) A ……. (4)
                                  \nWhen white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
                                  \nFor Violet light, \u03b4V<\/sub> = (nV <\/sub>– 1)A …(5)
                                  \nFor Red light, \u03b4R<\/sub> = (nR <\/sub>– 1) …(6)
                                  \nAs, angle of deviation for violet colour \u03b4V<\/sub> is greater than angle of deviation for red colour \u03b4R<\/sub>, the refractive index for violet colour nV<\/sub> is greater than the refractive index for red colour nR<\/sub>. Subtracting \u03b4V<\/sub> from \u03b4R<\/sub> we get,
                                  \n\u03b4V <\/sub>– \u03b4R<\/sub> = (nV <\/sub>– nR<\/sub>)A ….. (7)
                                  \nThe term (\u03b4V <\/sub>– \u03b4R<\/sub>) is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.<\/p>\n

                                    \n
                                  1. Angle of the prism<\/li>\n
                                  2. Nature of the material of the prism.<\/li>\n<\/ol>\n

                                    If we take 8 is the angle of deviation for any middly ray (green or yellow) and n the corresponding refractive index. Then,
                                    \n8 = (n – 1) A … (8)
                                    \nDispersive power (\u03c9) is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour. Dispersive power (\u03c9),
                                    \n\"Samacheer
                                    \nSubstituting for (\u03b4V<\/sub>-\u03b4R<\/sub>)and (\u03b4),
                                    \n\u03c9 = \\(\\frac{\\left(n_{\\mathrm{V}}-n_{\\mathrm{R}}\\right)}{(n-1)}\\) ……. (10)
                                    \nDispersive power is a dimensionless quality. It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.<\/p>\n

                                    Question 12.
                                    \nProve laws of reflection using Huygens\u2019 principle.
                                    \nAnswer:
                                    \nProof for laws of reflection using Huygens\u2019 Principle:
                                    \nLet us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY. The incident wavefront is AB and the reflected wavefront is A’B’ in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L’, M’ respectively.<\/p>\n

                                    By the time point A of the incident wavefront touches the reflecting surface, the point B is yet to travel a distance BB’ to touch the reflecting surface a B’. When the point B falls on the reflecting surface at B’ , the point A would have reached A’. This is applicable to all the points on the wavefront.<\/p>\n

                                    Thus, the reflected wavefront A’B’ emanates as a plane wavefront. The two normals N and N’ are considered at the points where the rays L and M fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection. Hence, the time taken for the ray to travel from B to B’ is the same as the time taken for the ray to travel from A to A’. Thus, the distance BB’ is equal to the distance AA’; (AA’= BB’).
                                    \n\"Samacheer
                                    \n(i) The incident rays, the reflected rays and the normal are in the same plane.
                                    \n(ii) Angle of incidence,
                                    \n\u2220i = \u2220NAL = 90\u00b0 – \u2220NAB = \u2220BAB’
                                    \nAngle of reflection,
                                    \n\u2220r = \u2220N’B’M’ = 90\u00b0 – \u2220N’B’A’ = \u2220A’B’A
                                    \nFor the two right angle triangles, A ABB’ and A B’A’A, the right angles, \u2220B and \u2220A’ are equal, (\u2220B and \u2220A’ = 90\u00b0); the two sides, AA’ and BB’ are equal, (AA’ = BB’); the side AB’ is the common. Thus, the two triangles are congruent. As per the property of congruency, the two angles, \u2220BAB’ and \u2220A’B’A must also be equal. i = r
                                    \nHence, the laws of reflection are proved.<\/p>\n

                                    Question 13.
                                    \nProve laws of refraction using Huygens\u2019 principle.
                                    \nAnswer:
                                    \nLet us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface. The incident wavefront AB is in rarer medium (1) and the refracted wavefront A’B’ is in denser medium (2). These wavefronts are perpendicular to the incident rays L, M and refracted rays L’, M’ respectively. By the time the point A of the incident wavefront touches the refracting surface, the point B is yet to travel a distance BB’ to touch the refracting surface at B’
                                    \n\"Samacheer
                                    \nWhen the point B falls on the refracting surface at B’, the point A’ would have reached A in the other medium. This is applicable to all the points on the wavefront. Thus, the refracted wavefront A’B’ emanates as a plane wavefront.
                                    \nThe two normals N and N’ are considered at the points where the rays L and M fall on the refracting surface. As refraction happens from rarer medium (1) to denser medium (2), the speed of light is v1<\/sub> and v2<\/sub> before and after refraction and v1<\/sub> is greater than v2<\/sub> (v1<\/sub> > v2<\/sub>). But, the time taken t for the ray to travel from B to B’ is the same as the time taken for the ray to
                                    \ntravel from A to A’.
                                    \nt = \\(\\frac { BB’ }{ { v }_{ 1 } } \\) = \\(\\frac { AA’ }{ { v }_{ 2 } } \\) (or) \\(\\frac { BB’ }{ AA’ }\\) = \\(\\frac { { v }_{ 1 } }{ { v }_{ 2 } } \\)
                                    \n(i) The incident rays, the refracted rays and the normal are in the same plane.
                                    \n(ii) Angle of incidence,
                                    \ni = \u2220NAL = 90\u00b0 – \u2220NAB = \u2220BAB’
                                    \nAngle of refraction,
                                    \nr =\u2220N’B’M\u2019 = 90\u00b0 – \u2220N’B’A’ = \u2220A’B’A’
                                    \nFor the two right angle triangles \u2206 ABB’ and \u2206 B’A’A,
                                    \n\\(\\frac { sin i }{ sin r }\\) = \\(\\frac{B B^{\\prime} \/ A B^{\\prime}}{A A^{\\prime} \/ A B^{\\prime}}\\) =\\(\\frac { BB’ }{ AA’ }\\) =\\(\\frac{v_{1}}{v_{2}}=\\frac{c \/ v_{2}}{c \/ v_{1}}\\)
                                    \nHere, c is speed of light in vacuum. The ratio c\/v is the constant, called refractive index of the medium. The refractive index of medium (1) is, c\/v1<\/sub> = n1<\/sub> and that of medium (2) is, c\/v2<\/sub> = n2<\/sub>.
                                    \n\\(\\frac { sin i }{ sin r }\\) = \\(\\frac { { n }_{ 2 } }{ { n }_{ 1 } } \\)
                                    \nIn product form,
                                    \nn1<\/sub> sin i = n2<\/sub> sin r
                                    \nHence, the laws of refraction are proved.<\/p>\n

                                    Question 14.
                                    \nObtain the equation for resultant intensify\u2019 due to interference of light.
                                    \nAnswer:
                                    \nLet us consider two light waves from the two sources S1<\/sub> and S2<\/sub> meeting at a point P. The wave from S1<\/sub> at an instant t at P is,
                                    \ny1 <\/sub>= a1<\/sub> sin \u03c9t … (1)
                                    \nThe wave form S2<\/sub> at an instant t at P is,
                                    \n\"Samacheer
                                    \ny2 <\/sub>= a2<\/sub> sin (\u03c9t + \u03a6) … (2)
                                    \nThe two waves have different amplitudes a1<\/sub> and a2<\/sub>, same angular frequency \u03c9, and a phase difference of \u03a6 between them. The resultant displacement will be given by,
                                    \ny = y1<\/sub> + y2<\/sub> = a1<\/sub> sin \u03c9t + a1 <\/sub>sin2<\/sub> (\u03c9t + \u03a6) … (3)
                                    \nThe simplification of the above equation by using trigonometric identities gives the equation,
                                    \ny = A sin (\u03c9t + \u03b8) … (4)
                                    \nWhere, A = \\(\\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \\cos \\phi}\\) …. (5)
                                    \n\u03b8 = tan-1<\/sup> \\(\\frac{a_{2} \\sin \\phi}{a_{1}+a_{2} \\cos \\phi}\\) …. (6)
                                    \nThe resultant amplitude is maximum,
                                    \nAmax<\/sub> = \\(\\sqrt{\\left(a_{1}+a_{2}\\right)^{2}}\\) ; when \u03a6 = 0, \u00b12\u03c0, \u00b14\u03c0 …., …… (7)
                                    \nThe resultant amplitude is minimum,
                                    \nAmax<\/sub> = \\(\\sqrt{\\left(a_{1}-a_{2}\\right)^{2}}\\) ; when \u03a6 = \u00b1\u03c0, \u00b13,\u03c0 \u00b15\u03c0 …., …… (8)
                                    \nThe intensity of light is proportional to square of amplitude,
                                    \nI \u221d A2<\/sup> ….. (9)
                                    \nNow, equation (5) becomes,
                                    \nI \u221d I1<\/sub>+I 2<\/sub>+2 \\(\\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\) cos \u03a6 …… (10)
                                    \nIn equation (10) if the phase difference, \u03a6 = 0, \u00b12\u03c0, \u00b14\u03c0 …., it corresponds to the condition for maximum intensity of light called as constructive interference. The resultant maximum intensity is,
                                    \nImax<\/sub> \u221d ( a1<\/sub> + a2<\/sub>)2<\/sup> \u221d I1<\/sub> + I2 <\/sub>+2 \\(\\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\) …… (11)
                                    \nIn equation (10) if the phase difference, \u03a6 = \u00b1\u03c0, \u00b13,\u03c0 \u00b15\u03c0 …., it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is,
                                    \nImax<\/sub> \u221d( a1<\/sub> – a2<\/sub>)2<\/sup> \u221d I1<\/sub> + I2 <\/sub>-2 \\(\\sqrt { { I }_{ 1 }{ I }_{ 2 } } \\) ……. (12)<\/p>\n

                                    Question 15.
                                    \nExplain the Young\u2019s double slit experimental setup and obtain the equation for path difference.
                                    \nAnswer:
                                    \nI Experimental setup:<\/p>\n

                                    1. Wavefronts from S1<\/sub> and S2<\/sub> spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter s from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
                                    \n\"Samacheer<\/p>\n

                                    2. Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from S1<\/sub> and S2<\/sub> travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.<\/p>\n

                                    3. The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.<\/p>\n

                                    II Equation for path difference:<\/p>\n

                                    1. Let d be the distance between the double slits S1<\/sub> and S2<\/sub> which act as coherent sources of wavelength \u03bb. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S1<\/sub> and S2<\/sub> is C and the midpoint of the screen O is equidistant from S1<\/sub> and S2<\/sub>. P is any point at a distance y from O.
                                    \n\"Samacheer<\/p>\n

                                    2. The waves from S1<\/sub> and S2<\/sub> meet at P either inphase or out-of-phase depending upon the path difference between the two waves.
                                    \n3. The path difference \u03b4 between the light waves from S1<\/sub> and S2<\/sub> to the point P is,
                                    \n\u03b4 = S2 <\/sub>P – S1
                                    \n<\/sub>4. A perpendicular is dropped from the point S1<\/sub> to the line S2 <\/sub>P at M to find the path difference more precisely.
                                    \n\u03b4 = S2 <\/sub>P – MP = S2 <\/sub>M
                                    \nThe angular position of the point P from C is \u03b8. \u2220OCP = \u03b8.
                                    \nFrom the geometry, the angles \u2220OCP and \u2220S2 <\/sub>S1<\/sub>M are equal. \u2220OCP = \u2220S2 <\/sub>S1 <\/sub>M = \u03b8
                                    \nIn right angle triangle \u2206S1 <\/sub>S2 <\/sub>M, the path difference, S2 <\/sub>M = d sin \u03b8
                                    \n\u03b4 = d sin \u03b8
                                    \nIf the angle 0 is small, sin \u03b8 \u2248 tan \u03b8 \u2248 \u03b8. From the right angle triangle \u2206OCP, tan \u03b8 = \\(\\frac { y }{ D }\\)
                                    \nThe path difference, \u03b4 = \\(\\frac { dy }{ D }\\)<\/p>\n

                                    Question 16.
                                    \nObtain the equation for bandwidth in Young\u2019s double slit experiment.
                                    \nAnswer:
                                    \nCondition for bright fringe (or) maxima
                                    \nThe condition for the constructive interference or the point P to be have a bright fringe is,
                                    \nPath difference, \u03b4 = n\u03bb
                                    \nwhere, n = 0, 1, 2,. . .
                                    \n\u2234 \\(\\frac { dy }{ D }\\) = n\u03bb
                                    \ny = n \\(\\frac { \u03bbD }{ d }\\) (or) yn<\/sub> = n \\(\\frac { \u03bbD }{ d }\\)
                                    \nThis is the condition for the point P to be a bright fringe. The distance is the distance of the nth bright fringe from the point O.
                                    \nCondition for dark fringe (or) minima
                                    \nThe condition for the destructive interference or the point P to be have a dark fringe is,
                                    \nPath difference, \u03b4 = (2n – 1) \\(\\frac { \u03bb }{ 2 }\\)
                                    \nwhere, n = 1, 2, 3 …
                                    \n\u2234 \\(\\frac { dy }{ D }\\) = (2n – 1) \\(\\frac { \u03bb }{ 2 }\\)
                                    \ny = \\(\\frac { (2n-1) }{ 2 }\\) \\(\\frac { \u03bbD }{ d }\\) (or) yn<\/sub> = \\(\\frac { (2n-1) }{ 2 }\\) \\(\\frac { \u03bbD }{ d }\\)
                                    \nEquation for bandwidth
                                    \nThe bandwidth (\u03b2) is defined as the distance between any two consecutive bright or dark fringes.
                                    \nThe distance between (n +1)th<\/sup> and nth<\/sup> consecutive bright fringes from O is given by,
                                    \n\u03b2 = y(n +1)<\/sub> – yn<\/sub> = \\(\\left((n+1) \\frac{\\lambda \\mathrm{D}}{d}\\right)-\\left(n \\frac{\\lambda \\mathrm{D}}{d}\\right)\\)
                                    \n\u03b2 = \\(\\frac { \u03bbD }{ d }\\)
                                    \nSimilarly, the distance between (n +1)th<\/sup> and nth<\/sup> consecutive dark fringes from O is given by,
                                    \n\u03b2 = y(n+1)<\/sub> – yn<\/sub> = \\(\\left(\\frac{(2(n+1)-1)}{2} \\frac{\\lambda D}{d}\\right)-\\left(\\frac{(2 n-1)}{2} \\frac{\\lambda D}{d}\\right)\\)
                                    \n\u03b2 = \\(\\frac { \u03bbD }{ d }\\)
                                    \nThe above equation show that the bright and dark fringes are of same width equally spaced on either side of central bright fringe.<\/p>\n

                                    Question 17.
                                    \nObtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
                                    \nAnswer:
                                    \nInterference in thin films:
                                    \nLet us consider a thin film of transparent material of refractive index p (not to confuse with order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted.<\/p>\n

                                    The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back in to the film. Reflected as well as refracted waves are sent by the film as multiple reflections take place inside the film. The interference is produced by both the reflected and transmitted light.
                                    \n\"Samacheer
                                    \nFor transmitted light:
                                    \nThe light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occured. The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD.
                                    \nIf we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is traversed in a medium of refractive index \u00b5, the optical path difference is, \u03b4 = 2\u00b5d.
                                    \nThe condition for constructive interference in transmitted ray is,
                                    \n2\u00b5d = n\u03bb
                                    \nSimilarly, the condition for destructive interference in transmitted ray is,
                                    \n2\u00b5d = (2n – 1) \\(\\frac { \u03bb }{ 2 }\\)
                                    \nFor reflected light:
                                    \nIt is experimentally and theoretically proved that a wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of \u03c0. Hence, an additional path difference of \\(\\frac { \u03bb }{ 2 }\\) should be considered.<\/p>\n

                                    Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film. The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, \u03b4 = 2\u00b5d.
                                    \nThe condition for constructive interference for reflected ray is,
                                    \n2\u00b5d + \\(\\frac { \u03bb }{ 2 }\\) = n\u03bb (or) 2\u00b5d = (2n – 1) \\(\\frac { \u03bb }{ 2 }\\)
                                    \nThe additional path difference \\(\\frac { \u03bb }{ 2 }\\) is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is,
                                    \n2\u00b5d + \\(\\frac { \u03bb }{ 2 }\\) = (2n + 1) \\(\\frac { \u03bb }{ 2 }\\) (or) 2\u00b5d = n\u03bb<\/p>\n

                                    Question 18.
                                    \nDiscuss diffraction at single slit and obtain the condition for nth<\/sup> minimum.
                                    \nAnswer:
                                    \nDiffraction at single slit:
                                    \nLet a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O. We would like to find the intensity at any point P on the screen. The lines joining P to the different points on the slit can be treated as parallel lines, making an angle 9 with the normal CO.<\/p>\n

                                    All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction. We need to give the condition for the point P to be of various minima.<\/p>\n

                                    The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at P with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.
                                    \n\"Samacheer
                                    \nCondition for P to be first minimum:
                                    \nLet us divide the slit AB into two half\u2019s AC and CB. Now the width of AC is (a\/2). We have different points on the slit which are separated by the same width (here a\/2) called corresponding points.
                                    \n\"Samacheer
                                    \nThe path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum. The path difference 8 between waves from these corresponding points
                                    \nis, \u03b4 = \\(\\frac { a }{ 2 }\\) sin \u03b8
                                    \nThe condition for P to be first minimum, \\(\\frac { a }{ 2 }\\) sin \u03b8 = \\(\\frac { \u03bb }{ 2 }\\)
                                    \na sin \u03b8 = \u03bb (first minimum) ….. (1)
                                    \nCondition for P to be first minimum:
                                    \nLet us divide the slit AB into two half\u2019s AC and CB. Now the width of AC is (a\/2). We have different points on the slit which are separated by the same width (here a\/2) called corresponding points.
                                    \nThe path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
                                    \nThe path difference \u03b4 between waves from these corresponding points \u03b4 = \\(\\frac { a }{ 4 }\\) sin \u03b8.
                                    \nThe condition for P to be first minimum, \\(\\frac { a }{ 4 }\\) sin \u03b8 = \\(\\frac { \u03bb }{ 2 }\\)
                                    \na sin \u03b8 = 2\u03bb (second minimum) …(2)
                                    \nCondition for P to be third order minimum:
                                    \nThe same way the slit is divided into six equal parts to explain the condition for P to be third
                                    \nminimum is, \\(\\frac { a }{ 6 }\\) sin \u03b8 = \\(\\frac { \u03bb }{ 2 }\\)
                                    \na sin \u03b8 = 3\u03bb (third minimum) …(3)
                                    \nCondition for P to be nth<\/sup> order minimum:
                                    \nDividing the slit into 2n number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for nth<\/sup>
                                    \norder minimum is, \\(\\frac { a }{ 2n }\\) sin \u03b8 = \\(\\frac { \u03bb }{ 2 }\\)
                                    \na sin \u03b8 = n\u03bb (nth<\/sup> minimum)<\/p>\n

                                    Question 19.
                                    \nDiscuss the diffraction at a grating and obtain the condition for the mth<\/sup> maximum.
                                    \nAnswer:
                                    \nA plane transmission grating is represented by AB. Let a plane wavefront of monochromatic light with wavelength \u03bb be incident normally on the grating. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.
                                    \n\"Samacheer
                                    \nA diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let us consider a point P at an angle 0 with the normal drawn from the center of the grating to the screen. The path difference 5 between the diffracted waves from one pair of corresponding points is,
                                    \n\u03b4 = (a + b) sin \u03b8
                                    \nThis path difference is the same for any pair of corresponding points. The point P will be bright, when
                                    \n\u03b4 = m \u03bb where m = 0, 1 , 2, 3<\/p>\n

                                    Combining the above two equations, we get,
                                    \n(a + b) sin \u03b8 = m \u03bb
                                    \nHere, m is called order of diffraction.<\/p>\n

                                    Condition for zero order maximum, m = 0
                                    \nFor (a + b) sin \u03b8 = 0, the position, \u03b8 = 0. sin \u03b8 = 0 and m = 0. This is called zero order diffraction or central maximum.<\/p>\n

                                    Condition for first order maximum, m = 1
                                    \nIf (a + b) sin \u03b81<\/sub> = \u03bb, the diffracted light meet at an angle \u03b81<\/sub> to the incident direction and the first order maximum is obtained.<\/p>\n

                                    Condition for second order maximum, m = 2
                                    \nSimilarly, (a + b) sin \u03b82<\/sub> = 2\u03bb, forms the second order maximum at the angular position \u03b82<\/sub>.<\/p>\n

                                    Condition for higher order maximum
                                    \nOn either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.
                                    \nIf we take, N = \\(\\frac { 1 }{ a+b }\\)<\/p>\n

                                    Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itself. Now, the equation becomes,
                                    \n\\(\\frac { 1 }{ N }\\) sin \u03b8 = m\u03bb, (or) sin \u03b8 = Nm\u03bb<\/p>\n

                                    Question 20.
                                    \nDiscuss the experiment to determine the wavelength of monochromatic light using diffraction grating.
                                    \nAnswer:
                                    \nExperiment to determine the wavelength of monochromatic light:
                                    \nThe wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. Initially all the preliminary adjustments of the spectrometer are made. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined. The telescope is brought in line with collimator to view the image of the slit. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.<\/p>\n

                                    The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece. The reading of the position of the telescope is noted. Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between two positions gives 2\u03b8. Half of its value gives \u03b8, the diffraction angle for first order maximum. The wavelength of light is calculated from the equation,
                                    \n\u03bb = \\(\\frac { sin \u03b8 }{ N m}\\)
                                    \nHere, N is the number of rulings per metre in the grating and m is the order of the diffraction image.<\/p>\n

                                    Question 21.
                                    \nDiscuss the experiment to determine the wavelength of different colours using diffraction grating.
                                    \nAnswer:
                                    \nDetermination of wavelength of different colours:
                                    \nWhen white light is used, the diffraction pattern consists of a white central maximum and on both sides continuous coloured diffraction patterns are formed. The central maximum is white as all the colours meet here constructively with no path difference.<\/p>\n

                                    As \u03b8 increases, the path difference, (a + b) sin \u03b8, passes through condition for maxima of diffraction of different orders for all colours from violet to red. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula,
                                    \n\u03bb = \\(\\frac { sin \u03b8 }{ N m}\\)
                                    \nHere, N is the number of rulings per metre in the grating and m is the order of the diffraction image.<\/p>\n

                                    Question 22.
                                    \nObtain the equation for resolving power of optical instrument.
                                    \nAnswer:
                                    \nResolution:
                                    \nThe effect of diffraction has an adverse impact in the image formation by the optical instruments such as microscope and telescope. For a single rectangular slit, the half angle \u03b8 subtended by the spread of central maximum (or position of first minimum) is given by the ‘ relation,
                                    \na sin \u03b8 = \u03bb … (1)
                                    \nSimilar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that becomes fainter while moving away from the center. These are known as Airy\u2019s discs. The circle of central maximum has the half angular spread given by the equation,
                                    \na sin \u03b8 = 1.22\u03bb … (2)
                                    \nHere, the numerical value 1.22 comes for central maximum formed by circular apertures. This involves higher level mathematics which is avoided in this discussion.
                                    \n\"Samacheer
                                    \nFor small angles, sin \u03b8 \u2248 \u03b8
                                    \na \u03b8 = 1.22\u03bb … (3)
                                    \nRewriting further, \u03b8 = \\(\\frac { 1.22\u03bb }{ a}\\) and \\(\\frac { { r }_{ 0 } }{ f } \\) = \\(\\frac { 1.22\u03bb }{ a}\\)
                                    \nr0<\/sub> = \\(\\frac { 1.22\u03bbf }{ a}\\) …. (4)<\/p>\n

                                    When two point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.<\/p>\n

                                    According to Rayleigh\u2019s criterion, for two point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh\u2019s criterion is said to be limit of resolution.<\/p>\n

                                    According to Rayleigh\u2019s criterion the two point sources are said to be just resolved when the distance between the two maxima is at least r0<\/sub>. The angular resolution has a unit in radian (rad) and it is given by the equation.
                                    \n\u03b8 = \\(\\frac { 1.22\u03bbf}{ a}\\)
                                    \nIt shows that the first order diffraction angle must be as small as possible for greater resolution. This further shows that for better resolution, the wavelength of light used must be as small as possible and the size of the aperture of the instrument used must be as large as possible. The equation (4) is used to calculate spacial resolution. The inverse of resolution is called resolving power. This implies, smaller the resolution, greater is the resolving power of the instrument. The ability of an optical instrument to \u2022 separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.<\/p>\n

                                    Question 23.
                                    \nDiscuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
                                    \nAnswer:
                                    \nSimple microscope:
                                    \nA simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.<\/p>\n

                                      \n
                                    • Near point focusing :
                                      \nThe image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.<\/li>\n
                                    • Normal focusing :
                                      \nThe image is formed at infinity. In this position the eye is most relaxed to view the image.<\/li>\n<\/ul>\n

                                      Magnification in near point focusing:
                                      \nObject distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
                                      \nm = \\(\\frac { v}{ u}\\)
                                      \nWith the help of lens equation, \\(\\frac { 1}{ v}\\) – \\(\\frac { 1}{ u}\\) \\(\\frac { 1}{ f}\\) the magnification can further be writen as,
                                      \nm = 1- \\(\\frac { v}{ f}\\)
                                      \nSubstituting for v with sign convention, v = -Derivem
                                      \nm = 1 + \\(\\frac { D}{ f}\\)
                                      \nThis is the magnification for near point focusing
                                      \n\"Samacheer
                                      \nMagnification in normal focusing (angular magnification):
                                      \nWe will now find the magnification for the image formed at infinity. If we take the ratio of height of image to height of object \\(\\left(m=\\frac{h^{\\prime}}{h}\\right)\\) to find the magnification, we will not get a practical relation as the image will also be of infinite size when the image is formed at infinity. Hence, we can practically use the angular magnification. The angular magnification is defined as the ratio of angle 0j subtended by the image with aided eye to the angle 90 subtended by the object with unaided eye.
                                      \n\"Samacheer
                                      \nm = \\(\\frac{\\theta_{i}}{\\theta_{0}}\\)
                                      \nFor unaided eye, tan \u03b80 <\/sub>\u2248 \u03b80<\/sub> = \\(\\frac { h }{ D }\\)
                                      \nFor aided eye, tan \u03b8i <\/sub>\u2248 \u03b8i<\/sub> = \\(\\frac { h }{ f }\\)
                                      \nThe angular magnification is, m = \\(\\frac{\\theta_{i}}{\\theta_{0}}\\) = \\(\\frac { h\/f }{ h\/d }\\)
                                      \nm = \\(\\frac { d }{ f }\\)
                                      \nThis is the magnification for normal focusing. The magnification for normal focusing is one less than that for near point focusing.<\/p>\n

                                      Question 24.
                                      \nExplain about compound microscope and obtain the equation for magnification. Compound microscope:
                                      \nAnswer:
                                      \nThe lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.
                                      \nMagnification of compound microscope :
                                      \nFrom the ray diagram, the linear magnification due to the objective is,
                                      \n\"Samacheer
                                      \nm0<\/sub> = \\(\\frac { h’ }{ f}\\)
                                      \nFrom the figure, tan \u03b2 = \\(\\frac { h }{ { f }_{ 0 } } \\) =\\(\\frac { h’ }{ L’}\\),then
                                      \n\\(\\frac { h’ }{ h}\\) = \\(\\frac { L }{ { f }_{ 0 } } \\); m0<\/sub> =\\(\\frac { L }{ { f }_{ 0 } } \\)
                                      \nHere, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length
                                      \nL of the microscope as f0<\/sub> and fe<\/sub> are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification me<\/sub> of the eyepiece is,
                                      \nme<\/sub> = 1 + \\(\\frac { D }{ { f }_{ e } } \\)
                                      \nThe total magnification m in near point focusing is,
                                      \nm = m0<\/sub>me<\/sub> = \\(\\left(\\frac{L}{f_{o}}\\right)\\left(1+\\frac{D}{f_{e}}\\right)\\)
                                      \nIf the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is,
                                      \nme<\/sub> = \\(\\frac { D }{ { f }_{ e } } \\)
                                      \nThe total magnification m in normal focusing is,
                                      \nm = m0<\/sub>me<\/sub> = \\(\\left(\\frac{L}{f_{o}}\\right)\\left(\\frac{D}{f_{e}}\\right)\\)<\/p>\n

                                      Question 25.
                                      \nObtain the equation for resolving power of microscope.
                                      \nAnswer:
                                      \nResolving power of microscope:
                                      \nThe diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance dmin<\/sub>.
                                      \nSmaller the value of dmin<\/sub> better will be the resolving power of the microscope.
                                      \n\"Samacheer
                                      \nThe radius of central maxima is, r0<\/sub> = \\(\\frac { 1.22\u03bbv }{ a }\\)
                                      \nIn the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin<\/sub>, then the magnification m is, m = \\(\\frac{r_{o}}{d_{\\min }}\\)
                                      \ndmin<\/sub> = \\(\\frac{r_{o}}{m}\\) =\\(\\frac{1.22 \\lambda v}{a m}\\) = \\(\\frac{1.22 \\lambda v}{a(v \/ u)}\\) = \\(\\frac{1.22 \\lambda u}{a}\\) [\u2234m = v\/u] [\u2234 [u \u2248 \u0192]
                                      \nOn the object side, 2tan \u03b2 \u2248 2sin \u03b2 = \\(\\frac { a }{ f }\\)
                                      \ndmin<\/sub> = \\(\\frac{1.22 \\lambda}{2 sin \u03b2 }\\) \u2234 [a = \u01922 sin\u03b2]
                                      \nTo further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n.
                                      \ndmin<\/sub> = \\(\\frac{1.22 \\lambda}{2n sin \u03b2 }\\)
                                      \nSuch an objective is called oil immersed objective. The term n sin p is called numerical aperture NA.
                                      \ndmin<\/sub> = \\(\\frac{1.22 \\lambda}{2(NA) }\\)<\/p>\n

                                      Question 26.
                                      \nDiscuss about astronomical telescope.
                                      \nAnswer:
                                      \nAstronomical telescope:
                                      \nAn astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by astronomical telescope will be inverted. It has an objective of long focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image.
                                      \n\"Samacheer
                                      \nMagnification of astronomical telescope:
                                      \nThe magnification m is the ratio of the angle \u03b2 subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye
                                      \nm = \\(\\frac { \u03b2 }{ \u03b1 }\\)
                                      \nFrom the diagram, m = \\(\\frac{h \/ f_{e}}{h \/ f_{0}}\\)
                                      \nm = \\(\\frac { { f }_{ 0 } }{ { f }_{ e } } \\)
                                      \nThe length of the telescope is approximately, L = f0<\/sub> + fe<\/sub><\/p>\n

                                      Question 27.
                                      \nMention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
                                      \nAnswer:
                                      \nThe spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.
                                      \nAdjustments of the spectrometer:
                                      \nThe following adjustments must be made before doing the experiment using spectrometer.<\/p>\n

                                      (a) Adjustment of the eyepiece:
                                      \nThe telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.<\/p>\n

                                      (b) Adjustment of the telescope:
                                      \nThe telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.<\/p>\n

                                      (c) Adjustment of the collimator:
                                      \nThe telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.<\/p>\n

                                      (d) Levelling the prism table:
                                      \nThe prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.<\/p>\n

                                      Question 28.
                                      \nExplain the experimental determination of material of the prism using spectrometer. Determination of refractive index of material of the prism:
                                      \nAnswer:
                                      \nThe preliminary adjustments of the telescope, collimator and the prism table of the spectrometer are made. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum deviation.<\/p>\n

                                      (i) Angle of the prism (A):
                                      \nThe prism is placed on the prism table with its refracting edge facing the collimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the collimator fall on the two faces AB and AC. The telescope is rotated to the position T1<\/sub> until the image of the slit formed by the reflection at the face AB is made to coincide with the vertical cross wire of the telescope.
                                      \n\"Samacheer
                                      \nThe readings of the verniers are noted. The telescope is then rotated to the position T2<\/sub> where the image of the slit formed by the reflection at the face AC coincides with the vertical cross wire. The readings are again noted.
                                      \nThe difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.<\/p>\n

                                      (ii) Angle of minimum deviation (D):
                                      \nThe prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.
                                      \n\"Samacheer
                                      \nThe readings of the verniers are noted. Now, the prism is removed and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula,
                                      \nn = \\(\\frac{\\sin \\left(\\frac{A+D}{2}\\right)}{\\sin \\left(\\frac{A}{2}\\right)}\\)
                                      \nThe refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.<\/p>\n

                                      Samacheer Kalvi 12th Physics Optics <\/span>Conceptual Questions<\/span><\/strong><\/p>\n

                                      Question 1.
                                      \nWhy are dish antennas curved?
                                      \nAnswer:
                                      \nDish antenna is curved so as it can receive parallel signal rays coming from same direction. These parallel signal rays reflect from parabolic dish, and gathered at main antenna part. This increases directivity of antenna, and gives sufficient amplitude signal.<\/p>\n

                                      Question 2.
                                      \nWhat type of lens is formed by a bubble inside water?
                                      \nAnswer:
                                      \nAir bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.<\/p>\n

                                      Question 3.
                                      \nIt is possible for two lenses to produce zero power?
                                      \nAnswer:
                                      \nYes. It is possible for two lenses to produce zero power. Both the surfaces of lenses are equally curved, i.e. R1<\/sub> = R2<\/sub>\u00a0and hence
                                      \nPower (P) = (\u00b5 – 1) \\(\\left(\\frac{1}{R_{1}}-\\frac{1}{R_{2}}\\right)=0\\)<\/p>\n

                                      Question 4.
                                      \nWhy does sky look blue and clouds look white?
                                      \nAnswer:
                                      \nBlue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I \u221d \\(\\frac { 1 }{ { \\lambda }^{ 4 } } \\). So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appear blue. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.<\/p>\n

                                      Question 5.
                                      \nWhy is yellow light preferred to during fog?
                                      \nAnswer:
                                      \nYellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I \u221d \\(\\frac { 1 }{ { \\lambda }^{ 4 } } \\). so yellow colour is least scattered and produces sufficient illumination.<\/p>\n

                                      Question 6.
                                      \nTwo independent monochromatic sources cannot act as coherent sources, why?
                                      \nAnswer:
                                      \nTwo independent sources of light cannot be coherent. This is because light is emitted by individual atoms. When they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.<\/p>\n

                                      Question 7.
                                      \nDoes diffraction take place at the Young\u2019s double slit?
                                      \nAnswer:
                                      \nBoth diffraction and interference in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs.<\/p>\n

                                      Question 8.
                                      \nIs there any difference between colored light obtained from prism and colours of soap bubble?
                                      \nAnswer:
                                      \nYes. there is a difference between colored light obtained from the prism is the phenomenon of\u2018dispersion of light\u2019 and colored light obtained from the soap bubble is the phenomenon of \u2018interference of light\u2019.<\/p>\n

                                      Question 9.
                                      \nA small disc is placed in the path of the light from distance source. Will the center of the shadow be bright or dark?
                                      \nAnswer:
                                      \nWhen a tiny circular small disc is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the disc because, wave diffracted from the edge of the circular disc interface constructively at the centre of the shadow, which produces bright spot.<\/p>\n

                                      Question 10.
                                      \nWhen a wave undergoes reflection at a denser medium, what happens to its phase?
                                      \nAnswer:
                                      \nWhen a wave undergoes a reflection at a denser medium then it’s crest reflected as trough and vice versa. So, its phase changes at 180\u00b0.<\/p>\n

                                      Samacheer Kalvi 12th Physics Optics Numerical Problems<\/span><\/strong><\/p>\n

                                      Question 1.
                                      \nAn object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.
                                      \nSolution:
                                      \n\u0192 = – 20 cm
                                      \nv = – 4u
                                      \nAccording to lens formula
                                      \n\\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ v }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ (-20) }\\) = \\(\\frac { 1 }{ (-4u) }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ (-20) }\\) = \\(\\left[-\\frac{1}{4}+1\\right]\\)
                                      \n= \\(\\frac { 1 }{ u }\\) \\(\\frac { 3 }{ 4 }\\)
                                      \nu = \\(\\frac { 3 \u00d7 20 }{ 4 }\\) = -15 cm.<\/p>\n

                                      Question 2.
                                      \nA compound microscope has a magnification of 30. The focal length of eye piece is 5 cm. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
                                      \nSolution:
                                      \nMagnification of compound microscope, M = 30
                                      \nFocal length, f = 5 cm
                                      \nLeast distance of distinct vision, D = 25 cm
                                      \nNow, M = M0<\/sub> x Me<\/sub>
                                      \n= M0<\/sub> x \\(\\left[1+\\frac{\\mathrm{D}}{f_{e}}\\right]\\)
                                      \n30 = M0<\/sub> x \\(\\left[1+\\frac{25}{5}\\right]\\)
                                      \nM0<\/sub> = \\(\\frac { 30 }{ 6 }\\) = 5<\/p>\n

                                      Question 3.
                                      \nAn object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
                                      \nSolution:
                                      \nm = + 3 and m = – 3; f = – 20 cm
                                      \nWhen, m = 3
                                      \nm = \\(\\frac { v }{ u }\\) = 3
                                      \nv = – 3 u
                                      \nFrom mirror equation,
                                      \n\\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ v }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ -20 }\\) = \\(\\frac { 1 }{ -3u }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ u }\\) \\(\\left[-\\frac{1}{3}+1\\right]\\)
                                      \n\\(\\frac { 1 }{ -20 }\\) = \\(\\frac { 2 }{ 3u }\\)
                                      \nu = –\\(\\frac { 40 }{ 3 }\\) cm
                                      \nWhen, m = – 3 \u21d2 v = 3u
                                      \n\\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ v }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ -20 }\\) = \\(\\frac { 1 }{ 3u }\\) + \\(\\frac { 1 }{ u }\\)
                                      \n\\(\\frac { 1 }{ -20 }\\) = \\(\\frac { 4 }{ 3u }\\)
                                      \nu = \\(\\frac { 4×20 }{ 3 }\\)
                                      \nu = –\\(\\frac { 80 }{ 3 }\\) cm<\/p>\n

                                      Question 4.
                                      \nA small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
                                      \nSolution:
                                      \nActual depth of the bulb in water
                                      \nd1<\/sub> = 80 cm = 0.8 m
                                      \nRefractive index of water, p = 1.33
                                      \nThe given situation is shown in the figure.
                                      \n\"Samacheer
                                      \nWhere, i = Angle of incidence
                                      \nr = Angle of refraction = 90\u00b0
                                      \nSince the bulb is a point source, the emergent light can be considered as a circle AT
                                      \nR= \\(\\frac { AC }{ 2 }\\) = AO = OB
                                      \nUsing snell\u2019s law, the relation for the refractive index of water is sin
                                      \n\u00b5 = \\(\\frac { sin r }{ sin i }\\)
                                      \ni = sin-1<\/sup>\u00a0(0.75) = 48.75\u00b0
                                      \nUsing the given figure, we have the relation
                                      \ntan i = \\(\\frac { OC }{ BC }\\) = \\(\\frac { R }{ { d }_{ 1 } } \\)
                                      \nR = tan i x d1<\/sub> = tan 48.75\u00b0 x 0.8
                                      \nR = 0.91 m
                                      \nArea of the surface of water = \u03c0R\u00b2
                                      \n= 3.14 x (0.91)\u00b2 = 2.61 m\u00b2
                                      \nHence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m\u00b2.<\/p>\n

                                      Question 5.
                                      \nA thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
                                      \nSolution:
                                      \nAccording to Lens maker\u2019s formula, the focal length for a convex lens placed in air can be obtained as
                                      \n\"Samacheer
                                      \n\\(\\frac { 0.5 }{ 5 }\\) = \\(\\frac { 1.5 }{ n }\\) -1
                                      \n0.9 = \\(\\frac { 1.5 }{ n }\\)
                                      \nn = \\(\\frac { 1.5 }{ 0.9 }\\) = \\(\\frac { 5 }{ 3 }\\) -1<\/p>\n

                                      Question 6.
                                      \nIf the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
                                      \nSolution:
                                      \nLet us fix the position of object and place the screen to get the enlarged image first. Also, let us fix the position of screen where we get the enlarged image.
                                      \nLet D be the distance between object and screen. Let us mark the position of lens dv Then let us move the lens away from the object to get a diminished image. Let this position of lens be d2<\/sub>. Let d be the distance between the lens position d1<\/sub> and d2<\/sub>. Let V be the distance b\/w image and lens. Let ‘u’ be the distance between object and lens.
                                      \nFrom mirror equation,
                                      \n\"Samacheer
                                      \n\\(\\frac { 1 }{ v }\\) + \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ f }\\)
                                      \nLet us replace v by substituting v = D – u
                                      \n\\(\\frac { 1 }{ D-u }\\) + \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ f }\\)
                                      \nwe get the equation, u\u00b2 -Du + fD = 0
                                      \nthe quadratic equation for above equation,
                                      \nu = \\(\\frac{\\mathrm{D} \\pm \\sqrt{\\mathrm{D}^{2}-4 f \\mathrm{D}}}{2}\\)
                                      \nWhen D = 4f, we get only position of lens to get image.
                                      \nThis corresponds to placing the object at 2f and getting the image at 2f on the otherside. Hence, for displacement method we need D > 4 f. When this condition is satisfied we get
                                      \nu1<\/sub> = \\(\\frac{\\mathrm{D}-\\sqrt{\\mathrm{D}^{2}-4 f \\mathrm{D}}}{2}\\) ; corresponding v1<\/sub> – D – u2<\/sub> = \\(\\frac{\\mathrm{D}+\\sqrt{\\mathrm{D}^{2}-4 f \\mathrm{D}}}{2}\\) after changing the location
                                      \nu1<\/sub> =\\(\\frac{\\mathrm{D}+\\sqrt{\\mathrm{D}^{2}-4f \\mathrm{D}}}{2}\\) ; corresponding v2<\/sub> – D – u2<\/sub> = \\(\\frac{\\mathrm{D}-\\sqrt{\\mathrm{D}^{2}-4 f \\mathrm{D}}}{2}\\)
                                      \nnow the displacement d = v1<\/sub> – u1<\/sub> = u1<\/sub> – v1<\/sub> = \\(\\sqrt{D^{2}-4 f D}\\)
                                      \nHence we get focal length, \u0192 = \\(\\frac{D^{2}-d^{2}}{4 D}\\)<\/p>\n

                                      Question 7.
                                      \nA beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?
                                      \nSolution:
                                      \nFor first minimum (n = 1) on either side of central maximum.
                                      \nsin \u03b8 = \\(\\frac { \u03bb }{ a }\\)
                                      \nWhere a is width of the slit Since 0 is very small, (sin \u03b8 \u2248 \u03b8)
                                      \n\u03b8 = \\(\\frac { \u03bb }{ a }\\) …… (1)
                                      \nsin \u03b8 \u2248 \u03b8 = \\(\\frac { x }{ 2D }\\) …… (2)
                                      \nWhere, x is distance of first dark fringe from central maximum.
                                      \nD is distance between slit and screen.
                                      \nFrom equation (1) and (2)
                                      \n\\(\\frac { x }{ 2D }\\) = \\(\\frac { \u03bb }{ a }\\)
                                      \nx = \\(\\frac { 2D\u03bb }{ a }\\) = \\(\\frac { 2\\times 2\\times 600\\times { 10 }^{ -9 } }{ 1\\times { 10 }^{ -3 } } \\)
                                      \n= 2400 x 10-6<\/sup> = 2.4 x 10-3<\/sup>m
                                      \nx = 2.4 mm<\/p>\n

                                      Question 8.
                                      \nIn Young\u2019s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength \u03bb0<\/sub> = 750 nm and \u03bb = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
                                      \nSolution:
                                      \nNow from the question we can infer that
                                      \nD = 2; d = 2
                                      \nn1<\/sub> \u03bb1<\/sub> = n2<\/sub> \u03bb2<\/sub>
                                      \n\\(\\frac{n_{1}}{n_{2}}=\\frac{\\lambda_{2}}{\\lambda_{1}}\\) =\\(\\frac { 900 }{ 750 }\\) \\(\\frac { 1 }{ 2 }\\)
                                      \nThus, we have
                                      \n\\(\\frac{n_{1}}{n_{1}}=\\frac{\\lambda_{2}}{\\lambda_{1}}\\) =\\(\\frac { 6 }{ 5 }\\)
                                      \n5th<\/sup> and 6th<\/sup> fringes will coincide respectively. The minimum distance is given as
                                      \nXmin<\/sub> = \\(\\frac{n_{2} \\lambda_{2} \\mathrm{D}}{d}\\) = \\(\\frac{5 \\times 900 \\times 10^{-9} \\times 2}{2 \\times 10^{-3}}\\)
                                      \n= 4500 x 10-6<\/sup> = 4.5 x 10-3<\/sup>m
                                      \nXmin<\/sub> = 4.5 mm<\/p>\n

                                      Question 9.
                                      \nIn Young\u2019s double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength 5893 \u00c5. If violet light of wavelength 4359 \u00c5 is used in place of sodium light, then what is the number of fringes seen?
                                      \nSolution:
                                      \nFrom young\u2019s double slit experiment,
                                      \n\u03bb1<\/sub> = 5893 \u00c5 ; \u03bb2<\/sub> = 4359 \u00c5
                                      \n\\(\\frac{n_{1} \\lambda_{1} D}{d}\\) = \\(\\frac{n_{2} \\lambda_{2} \\mathrm{D}}{d}\\)
                                      \nThe above condition is total extent of fringes is constant for both wavelengths.
                                      \n\\(\\frac{62 \\times 5893 \\times 10^{-10} \\times D}{d}\\) = \\(\\frac{n_{2} \\times 4359 \\times 10^{-10} \\times D}{d}\\)
                                      \nn2<\/sub> = \\(\\frac{62 \\times 5893}{4359}\\) = \\(\\frac{365366}{4359}\\) = 83.8
                                      \nn2<\/sub> = 84<\/p>\n

                                      Question 10.
                                      \nA compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.
                                      \nSolution:
                                      \nMagnifying Power, m = 100
                                      \nFocal length of the objective,f0<\/sub> = 0.5 cm
                                      \nTube length, l = 6.5 cm
                                      \nSince the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
                                      \nv0<\/sub> +\u0192e<\/sub>= 6.5 cm ….. (1)
                                      \nThe magnifying power for normal adjustment is given by
                                      \nm = \\(\\left(\\frac{v_{o}}{u_{o}}\\right) \\times \\frac{\\mathrm{D}}{f_{e}}\\)
                                      \n= – \\(\\left[1-\\frac{v_{o}}{f_{\\mathrm{o}}}\\right] \\frac{\\mathrm{D}}{f_{e}}\\)
                                      \n100 = – \\(\\left[1-\\frac{v_{o}}{0.5}\\right] \\times \\frac{25}{f_{e}}\\)
                                      \n2v0<\/sub> -4\u0192e<\/sub>= 1
                                      \nOn solving equations (1) and (2), we get
                                      \nv0<\/sub> = 4.5 cm and \u0192e<\/sub> = 2 cm
                                      \nThus, the focal length of the eyepiece is 2 cm.<\/p>\n

                                      Samacheer Kalvi 12th Physics Optics Additional Questions<\/h3>\n

                                      Samacheer Kalvi 12th Physics Optics Multiple Choice Questions<\/span><\/strong><\/p>\n

                                      Question 1.
                                      \nWhen light is refracted, which of the following does not change?
                                      \na) Wavelength
                                      \nb) Frequency
                                      \nc) Velocity
                                      \nd) Amplitude
                                      \nAnswer:
                                      \nb) Frequency<\/p>\n

                                      Question 2.
                                      \nA source emits sound of frequency 600 Hz inside water. The frequency heard in air (velocity of sound in water = 1500 m\/s, the velocity of sound in air = 300 m\/s) will be
                                      \n(a) 300 Hz
                                      \n(b) 120 Hz
                                      \n(c) 600 Hz
                                      \n(d) 6000 Hz
                                      \nAnswer:
                                      \n(c) 600 Hz
                                      \nHint:
                                      \nFrequency does not change when sound travels from one medium to another
                                      \n\u2234 Frequency of sound in air = Frequency of sound in water = 600 Hz<\/p>\n

                                      Question 3.
                                      \nTwo beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60\u00b0). In the position of minimum deviation, the angle of refraction will be
                                      \n(a) 30\u00b0 for both the colours
                                      \n(b) greater for the violet colour
                                      \n(c) greater for the violet colour
                                      \n(d) equal but not 30\u00b0 for both the colours
                                      \nAnswer:
                                      \n(a) 30\u00b0 for both the colours
                                      \nHint:
                                      \nFor any prism, r1<\/sub> = r2<\/sub> = A
                                      \nIn the position of minimum deviation for any wavelength,
                                      \nr1<\/sub> = r2<\/sub> = \\(\\frac { A }{ 2 }\\) = \\(\\frac { 60\u00b0 }{ \u03b8 }\\) = 30\u00b0<\/p>\n

                                      Question 4.
                                      \nTo get three images of a single object, one should have two plane mirrors at an angle of
                                      \n(a) 60\u00b0
                                      \n(b) 90\u00b0
                                      \n(c) 120\u00b0
                                      \n(d) 30\u00b0
                                      \nAnswer:
                                      \n(b) 90\u00b0
                                      \nHint:
                                      \nThe number of images formed,
                                      \nn = \\(\\frac { 360\u00b0 }{ \u03b8 }\\) – 1 or 3 = \\(\\frac { 360\u00b0 }{ \u03b8 }\\) -1 or \u03b8 = 90\u00b0<\/p>\n

                                      Question 5.
                                      \nWhich of the following phenomena is used in optical fibres?
                                      \na) Total internal reflection
                                      \nb) Scattering
                                      \nc) Diffraction
                                      \nd) Refraction
                                      \nAnswer:
                                      \na) Total internal reflection<\/p>\n

                                      Question 6.
                                      \nTwo lenses of power – 15 D and +15D are in contact with each other. The focal length of the combination is
                                      \n(a) + 10 cm
                                      \n(b) -20 cm
                                      \n(c) – 10 cm
                                      \n(d) + 20cm
                                      \nAnswer:
                                      \n(c) – 10 cm
                                      \nHint P = P1<\/sub> + P2<\/sub> = -15 + 5= -10 D
                                      \nF = \\(\\frac { 1 }{ P }\\) = \\(\\frac { 1 }{ -10 }\\) m = -10 cm<\/p>\n

                                      Question 7.
                                      \nThe refractive index of glass is 1.520 for red light and 1.525 for blue light. Let \u03b41<\/sub> and \u03b42<\/sub> be angles of minimum deviation for red and blue light respectively in a prism of this glass, then
                                      \n(a) \u03b41<\/sub>, can be less than or greater than \u03b42<\/sub> depending upon the values of \u03b41<\/sub> and \u03b42<\/sub>
                                      \n(b) \u03b41<\/sub> > \u03b42<\/sub>
                                      \n(c) \u03b41<\/sub> < \u03b42<\/sub>
                                      \n(d) \u03b41<\/sub> = \u03b42<\/sub>
                                      \nAnswer:
                                      \n(c) \u03b41<\/sub> < \u03b42<\/sub>
                                      \nHint:
                                      \n\u03b41<\/sub> = (\u03bcR<\/sub> – 1)A, \u03b42<\/sub> = (\u03bcB<\/sub> – 1)A
                                      \nAs, \u03bcR<\/sub> < \u03bcB<\/sub> \u2234 \u03b41<\/sub> < \u03b42<\/sub><\/p>\n

                                      Question 8.
                                      \nA parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light.
                                      \na) remains constant
                                      \nb) continuously increases
                                      \nc) continuously decreases
                                      \nd) first increases and then decreases
                                      \nAnswer:
                                      \na) remains constant<\/p>\n

                                      Question 9.
                                      \nAn astronomical telescope has a large aperture to,
                                      \n(a) reduce spherical aberration
                                      \n(b) have high resolution
                                      \n(c) increase span of observation
                                      \n(d) have low dispersion
                                      \nAnswer:
                                      \n(b) have high resolution<\/p>\n

                                      Question 10.
                                      \nTwo plane mirrors are inclined to each other at an angle of 60\u00b0. A point object is placed in between them. The total number of images produced by both the mirror is
                                      \n(a) 2
                                      \n(b) 4
                                      \n(c) 5
                                      \n( d) 6
                                      \nAnswer:
                                      \n(c) 5
                                      \nHint:
                                      \nNumber of images formed, \u03b8 = \\(\\frac { 360\u00b0 }{ \u03b8 }\\) – 1 = \\(\\frac { 360 }{ 60 }\\) = 1 = 5.<\/p>\n

                                      Question 11.
                                      \nA boy 1.5 m tall with his eye level at 1.38 m stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is
                                      \n(a) 0.75 m
                                      \n(b) 0.06 m
                                      \n(c) 0.69 m
                                      \n(d) 0.12 m
                                      \nAnswer:
                                      \n(a) 0.75 m
                                      \nHint
                                      \nMinimum length of mirror required \\(\\frac { 1 }{ 2 }\\) x Height of boy = \\(\\frac { 1 }{ 2 }\\) x 1.5 = 0.75 m.<\/p>\n

                                      Question 12.
                                      \nThe equi-convex lens has a focal length f. The focal length of each half is the lens cut along.
                                      \na) f\/2
                                      \nb) f
                                      \nc) \\(\\frac{3 \\mathrm{f}}{2}\\)
                                      \nd) 2f
                                      \nAnswer:
                                      \nd) 2f<\/p>\n

                                      Question 13.
                                      \nFor a real object, which of the following can produce a real image?
                                      \n(a) plane mirror
                                      \n(b) concave lens
                                      \n(c) convex lens
                                      \n(d) concave mirror
                                      \nAnswer:
                                      \n(d) concave mirror
                                      \nHint:
                                      \nOnly a concave mirror produces real image provided the object is not placed between its focus and pole.<\/p>\n

                                      Question 14.
                                      \nWhich mirror is to be used to obtain a parallel beam of light from a small lamp?
                                      \n(a) Plane mirror
                                      \n(b) Convex mirror
                                      \n(c) Concave mirror
                                      \n(d) None of the above
                                      \nAnswer:
                                      \n(c) Concave mirror
                                      \nHint:
                                      \nWhen the small lamp is placed at the focus of the concave mirror, the reflected light is a parallel beam.<\/p>\n

                                      Question 15.
                                      \nWhen a plane electromagnetic wave enters a glass slab, then which of the following will not change?
                                      \n(a) Wavelength
                                      \n(b) Frequency
                                      \n(c) Speed
                                      \n(d) Amplitude
                                      \nAnswer:
                                      \n(b) Frequency
                                      \nOnly the frequency of the electromagnetic wave remains unchanged.<\/p>\n

                                      Question 16.
                                      \nIf the wavelength of light in air is 2400 x 10-10<\/sup> m, then what will be the wavelength of light in glass (\u03bc = 1.5)?
                                      \n(a) 1600 \u00c5
                                      \n(b) 7200 \u00c5
                                      \n(c) 1080 \u00c5
                                      \n(d) None of these
                                      \nAnswer:
                                      \n(a) 1600 \u00c5
                                      \nHint:
                                      \n\u03bc = \\(\\frac { { \\lambda }_{ a } }{ { \\lambda }_{ g } } \\) \u21d2\u03bb\u03bb<\/sub>g = \\(\\frac { { \\lambda }_{ a } }{ \u03bc } \\) = \\(\\frac{2400 \\times 10^{-10} \\mathrm{m}}{1.5}\\) = 1600 \u00c5<\/p>\n

                                      Question 17.
                                      \nWhy is refractive index in a transparent medium greater than one?
                                      \n(a) Because the speed of light in vacuum is always less than speed in a transparent medium.
                                      \n(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
                                      \n(c) Frequency of wave changes when it Gasses medium.
                                      \n(d) None of the above.
                                      \nAnswer:
                                      \n(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
                                      \nHint:
                                      \n\"Samacheer
                                      \nAs c > v, \u03bc > 1<\/p>\n

                                      Question 18.
                                      \nBy properly combining two prisms made of different materials, it is not possible to have
                                      \na) dispersion without average deviation
                                      \nb) deviation without dispersion
                                      \nc) both dispersion and average deviation
                                      \nd) neither dispersion nor average deviation
                                      \nAnswer:
                                      \nd) neither dispersion nor average deviation<\/p>\n

                                      Question 19.
                                      \nA glass slab (\u03bc = 1.5) of thickness 6 cm is placed over a paper. Qhat is the shift in the letters?
                                      \n(a) 4 cm
                                      \n(b) 2 cm
                                      \n(c) 1 cm
                                      \n(d) None of these
                                      \nAnswer:
                                      \n(b) 2 cm
                                      \nHint:
                                      \nNormal shift, x = \\(t\\left(1-\\frac{1}{\\mu}\\right)=6\\left(1-\\frac{1}{1.5}\\right)\\) cm = 2cm<\/p>\n

                                      Question 20.
                                      \nLight traveling from a transparent medium to air undergoes total internal reflection at an angle of incident of 45\u00b0. Then refractive index of the medium may be
                                      \n(a) 1.5
                                      \n(b) 1.3
                                      \n(c) 1.1
                                      \n(d) \\(\\frac { 1 }{ \u221a2 }\\)
                                      \nAnswer:
                                      \n(a) 1.5
                                      \nHint:
                                      \n\u03bc = \\(\\frac{1}{\\sin i_{c}}\\) = \\(\\frac{1}{\\sin 45^{\\circ}}\\) = \u221a2 = 1.414 \u2248 1.5.<\/p>\n

                                      Question 21.
                                      \nWhich light rays undergo two internal reflections inside a raindrop, which of the rainbow is formed?
                                      \na) Primary rainbow
                                      \nb) Secondary rainbow
                                      \nc) Both (a) and (b)
                                      \nd) Can’t say
                                      \nAnswer:
                                      \nb) Secondary rainbow<\/p>\n

                                      Question 22.
                                      \nIn optical fibres, propagation of light is due to
                                      \n(a) diffraction
                                      \n(b) total internal reflection
                                      \n(c) reflection
                                      \n(d) refraction
                                      \nAnswer:
                                      \n(b) total internal reflection
                                      \nHint:
                                      \nIn optical fiberes, light propagates due to total internal reflection.<\/p>\n

                                      Question 23.
                                      \nSparkling of diamond is due to
                                      \n(a) reflection
                                      \n(b) dispersion
                                      \n(c) total internal reflection
                                      \n(d) high refractive index of diamond
                                      \nAnswer:
                                      \n(c) total internal reflection<\/p>\n

                                      Question 24.
                                      \nFor a given lens, the magnification was found to be twice as large as when the object was 0.15m distant from it as when the distance was 0.2 m. The focal length of the lens is
                                      \n(a) 1.5 m
                                      \n(b) 0.20 m
                                      \n(c) 0.10 m
                                      \n(d) 0.05 m
                                      \nAsnwer:
                                      \n(c) 0.10 m
                                      \nHint:
                                      \nHere m1<\/sub> = 2m2<\/sub>
                                      \n\\(\\frac { f }{ f-0.15 }\\) = 2 \\(\\frac { f }{ f-0.20 }\\)
                                      \n2f – 0.30 = f- 0.20 ; f = 0.10 m<\/p>\n

                                      Question 25.
                                      \nTwo lenses of focal lengths f1<\/sub> and f2<\/sub> are kept in contact coaxially. The resultant power of combination will be
                                      \n(a) \\(\\frac { { f }_{ 1 }{ f }_{ 2 } }{ { { f }_{ 1 }-f }_{ 2 } } \\)
                                      \n(b) \\(\\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \\)
                                      \n(c) \\({ f }_{ 1 }{ +f }_{ 2 }\\)
                                      \n(d) \\(\\frac { { f }_{ 1 } }{ { f }_{ 2 } } +\\frac { { f }_{ 2 } }{ { f }_{ 1 } } \\)
                                      \nAnswer:
                                      \n(b) \\(\\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \\)
                                      \nHint:
                                      \nP = \\(\\frac { 1 }{ { f }_{ 1 } } +\\frac { 1 }{ { f }_{ 2 } } \\) = \\(\\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \\).<\/p>\n

                                      Question 26.
                                      \nAn astronomical telescope has a large aperture to
                                      \na) reduce spherical aberration
                                      \nb) have high resolution
                                      \nc) increases span of observation
                                      \nd) have low dispersion
                                      \nAnswer:
                                      \nb) have high resolution<\/p>\n

                                      Question 27.
                                      \nIf two thin lenses are kept coaxially together, then their power is proportional (R1<\/sub>,R2<\/sub>) being the radii of curved surfaces) to
                                      \n(a) R1<\/sub> + R2<\/sub>
                                      \n(b) \\(\\left[\\frac{\\mathbf{R}_{1}+\\mathbf{R}_{2}}{\\mathbf{R}_{1} \\mathbf{R}_{2}}\\right]\\)
                                      \n(c) \\(\\left[\\frac{\\mathrm{R}_{1} \\mathrm{R}_{2}}{\\mathrm{R}_{1} \\mathrm{R}_{2}}\\right]\\)
                                      \n(d) None of these
                                      \nAnswer:
                                      \n(b) \\(\\left[\\frac{\\mathbf{R}_{1}+\\mathbf{R}_{2}}{\\mathbf{R}_{1} \\mathbf{R}_{2}}\\right]\\)
                                      \nHint:
                                      \nP = \\(\\frac{1}{f_{1}}+\\frac{1}{f_{2}}=\\frac{2}{\\mathrm{R}_{1}}+\\frac{2}{\\mathrm{R}_{2}}\\) = 2\\(\\left[\\frac{\\mathbf{R}_{1}+\\mathbf{R}_{2}}{\\mathbf{R}_{1} \\mathbf{R}_{2}}\\right]\\)
                                      \nor p \u221d \\(\\left[\\frac{\\mathbf{R}_{1}+\\mathbf{R}_{2}}{\\mathbf{R}_{1} \\mathbf{R}_{2}}\\right]\\)<\/p>\n

                                      Question 28.
                                      \nA ray incident at 15\u00b0 on one refracting surface of a prism of angle 60\u00b0, suffers a deviation of 55\u00b0. What is the angle of emergence?
                                      \n(a) 95\u00b0
                                      \n(b) 45\u00b0
                                      \n(c) 30\u00b0
                                      \n(d) none of these
                                      \nAnswer:
                                      \n(d) none of these
                                      \nHint:
                                      \nA + \u03b4 = i + each
                                      \n60\u00b0 + 55\u00b0 = 15\u00b0 + each e = 115 -15 =100\u00b0<\/p>\n

                                      Question 29.
                                      \nWhen monochromatic light is replaced by white light in Fresnel’s bi-prism arrangement, the central fringe is
                                      \na) coloured
                                      \nb) white
                                      \nc) dark
                                      \nd) None of these
                                      \nAnswer:
                                      \nb) white<\/p>\n

                                      Question 30.
                                      \nWhite light is incident on one of the refracting surfaces of a prism of angle 50\u00b0. If the refractive indices for red and blue colours are 1.641 and 1.659 respectively, the angular separation between these two colours when they emerge out of the prism is
                                      \n(a) 0.9\u00b0
                                      \n(b) 0.09\u00b0
                                      \n(c) 1.8\u00b0
                                      \n(d) 1.2\u00b0
                                      \nAnswer:
                                      \n(b) 0.09\u00b0
                                      \nHint:
                                      \nAngular dispersion,
                                      \n\u03b4B<\/sub> – \u03b4R<\/sub> = (\u00b5B<\/sub> – \u00b5R<\/sub>) A
                                      \n= (1.659 – 1.641) x 5\u00b0 = 0.09\u00b0<\/p>\n

                                      Question 31.
                                      \nThe sky would appear red instead of blue if
                                      \n(a) atmospheric particles scatter blue light more than red light
                                      \n(b) atmospheric particles scatter all colours equally
                                      \n(c) atmospheric particle scatter red light more than blue light
                                      \n(d) the sun was much hotter
                                      \nAnswer:
                                      \n(c) atmospheric particle scatter red light more than blue light<\/p>\n

                                      Question 32.
                                      \nA setting sun appears to be at an altitude higher than it really is. This is because of
                                      \n(a) absorption of light
                                      \n(b) reflection of light
                                      \n(c) refraction of light
                                      \n(d) dispersion of light
                                      \nAnswer:
                                      \n(c) refraction of light
                                      \nHint:
                                      \nThis is due to the refraction of light by the earth\u2019s atmosphere.<\/p>\n

                                      Question 33.
                                      \nThe reddish appearance of the rising and setting sun is due to
                                      \n(a) reflection of light
                                      \n(b) diffraction of light
                                      \n(c) scattering of light
                                      \n(d) interference of light
                                      \nAnswer:
                                      \n(c) scattering of light
                                      \nHint:
                                      \nThe reddish appearance of the rising and the setting sun is due to the scattering of light.<\/p>\n

                                      Question 34.
                                      \nIn the formation of a rainbow, the light from the sun on water droplets undergoes
                                      \n(a) dispersion only
                                      \n(b) only total internal reflection
                                      \n(c) dispersion and total internal reflection
                                      \n(d) none of the above
                                      \nAnswer:
                                      \n(c) dispersion and total internal reflection
                                      \nHint:
                                      \nRainow is formed due to the dispersion of sunlight by raindrops which also deviate the colours by total internal reflection.<\/p>\n

                                      Question 35.
                                      \nIn the case of linearly polarized light, the magnitude of the electric field vector
                                      \na) is parallel to the direction of propagation
                                      \nb) does not change with time
                                      \nc) increases linearly with time
                                      \nd) varies periodically with time
                                      \nAnswer:
                                      \nd) varies periodically with time<\/p>\n

                                      Question 36.
                                      \nFor compound microscope f0<\/sub> = 1 cm, fe<\/sub> = 2.5 cm. An object is placed at distance 1.2 cm from objective lens. What should be length of microscope for normal adjustment?
                                      \n(a) 8.5 cm
                                      \n(b) 8.3 cm
                                      \n(c) 6.5 cm
                                      \n(d) 6.3 cm
                                      \nAnswer:
                                      \n(a) 8.5 cm
                                      \nHint:
                                      \nIn the normal adjustment of a compound microscope, L = \u03c50<\/sub> + \u03c5e<\/sub> = \\(\\frac{\u03c5_{0} f_{e}}{\u03c5_{0}+f_{e}}\\) + fe<\/sub> = \\(\\frac { 1.2\u00d71 }{ -1.2+1 }\\) + 2.5 = 6 + 25 = 8.5 cm<\/p>\n

                                      Question 37.
                                      \nMagnifying power of an astronomical telescope for normal vision with usual notation is
                                      \n(a) -f0 <\/sub>\/ fe<\/sub>
                                      \n(b) -f0<\/sub> x fe<\/sub>
                                      \n(c) -f0 <\/sub>\/ f0<\/sub>
                                      \n(d) -f0<\/sub> + fe<\/sub>
                                      \nAnswer:
                                      \n(a) -f0 <\/sub>\/ fe<\/sub>
                                      \nHint:
                                      \nIn normal adjustment of the telescope, m = -f0 <\/sub>\/ fe<\/sub><\/p>\n

                                      Question 38.
                                      \nOptically active substances are those substances which
                                      \na) produces polarised light
                                      \nb) produces double refraction
                                      \nc) rotate the plane of polarisation of polarised light
                                      \nd) converts a plane polarised light into
                                      \ncircularly polarised light
                                      \nAnswer:
                                      \nc) rotate the plane of polarisation of polarised light<\/p>\n

                                      Question 39.
                                      \nFocal length of objective and eyepiece of telescope are 200 cm and 4 cm respectively. What is length of telescope for normal adjustment?
                                      \n(a) 196 cm
                                      \n(b) 204 cm
                                      \n(c) 250 cm
                                      \n(d) 225 cm
                                      \nAnswer:
                                      \n(b) 204 cm
                                      \nHint L = \\(-{ f }_{ 0 }+{ f }_{ e }\\) = 200 + 4 = 204 cm<\/p>\n

                                      Question 40.
                                      \nFor normal vision, what is minimum distance of object from eye?
                                      \n(a) 30 cm
                                      \n(b) 25 cm
                                      \n(c) Infinite
                                      \n(d) 40 cm
                                      \nAnswer:
                                      \n(b) 25 cm
                                      \nHint:
                                      \nFor normal eye, the least distance of distinct vision is 25 cm.<\/p>\n

                                      Question 41.
                                      \nThe focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5\u00b0 ; the angle subtended by the moon\u2019s image will be
                                      \n(a) 10\u00b0
                                      \n(b) 250
                                      \n(e) 100\u00b0
                                      \n(d) 75\u00b0
                                      \nAnswer:
                                      \nb) 250
                                      \nHint ;
                                      \nm = \\(\\frac { \u03b2 }{ \u03b1 }\\) \u03b2 = \\(\\frac { { f }_{ 0 } }{ { f }_{ e } } \\) ; \u03b1 = \\(\\frac { 100 }{ 2 }\\) x 0.5\u00b0 = 25\u00b0<\/p>\n

                                      Question 42.
                                      \nA person cannot clearly see distance more than 40 cm. He is advised to use lens of power,
                                      \n(a) – 2.5 D
                                      \n(b) 2.5 D
                                      \n(c) – 6.25 D
                                      \n(d) 1.5 D
                                      \nAnswer:
                                      \n(a) – 2.5 D
                                      \nHint;
                                      \nFor the remedial lens, u = \u221e,
                                      \nv = – 40 cm = – 0.40 m
                                      \n\u2234 \\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ v }\\) – \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ -0.40 }\\) – \\(\\frac { 1 }{ \u221e }\\) = -2.5 \u21d2 P = 2.5D<\/p>\n

                                      Question 43.
                                      \nThe number of images produced by two parallel plane mirrors is.
                                      \na) infinity
                                      \nb) zero
                                      \nc) 3
                                      \nd) 8
                                      \nAnswer:
                                      \na) infinity<\/p>\n

                                      Question 44.
                                      \nThe amount of light entering into the camera depends upon
                                      \n(a) focal length of objective lens
                                      \n(b) product of focal length and diameter of the objective lens
                                      \n(c) distance of object from camera
                                      \n(d) aperture setting of the camera
                                      \nAnswer:
                                      \n(d) aperture setting of the camera
                                      \nHint:
                                      \nThe amount of light entering into the camera depends upon the aperture setting of the camera.<\/p>\n

                                      Question 45.
                                      \nLine spectrum can be obtained from
                                      \n(a) sun
                                      \n(b) candle
                                      \n(c) mercury vapour lamp
                                      \n(d) electric bulb
                                      \nAnswer:
                                      \n(c) mercury vapour lamp<\/p>\n

                                      Question 46.
                                      \nThe Production of band spectra is caused by
                                      \n(a) atomic nuclei
                                      \n(b) hot metals
                                      \n(c) molecules
                                      \n(d) electrons
                                      \nAnswer:
                                      \n(c) molecules<\/p>\n

                                      Question 47.
                                      \nTwo mirrors are kept at 60\u00b0 to each other and a body is placed at middle. The total number of images formed is
                                      \n(a) six
                                      \n(b) four
                                      \n(c) five
                                      \n(d) three
                                      \nAnswer:
                                      \n(a) six
                                      \nHint:
                                      \nNumber of images formed, n = \\(\\frac { 360 }{ \u03b8 }\\) -1 = \\(\\frac { 360 }{ 60 }\\) -1 = 5<\/p>\n

                                      Question 48.
                                      \nA point source kept at a distance of 1000 m has an illumination I. To change the illumination to 161, the new distance should become
                                      \n(a) 250 m
                                      \n(b) 500 m
                                      \n(c) 750 m
                                      \n(d) 800 m
                                      \nAnswer:
                                      \n(a) 250 m
                                      \nHint:
                                      \n\\(\\frac{I_{2}}{I_{1}}=\\left(\\frac{r_{1}}{r_{2}}\\right)^{2}\\)
                                      \n\\(\\frac{16 \\mathrm{I}_{1}}{\\mathrm{I}_{1}}=\\frac{(1000)^{2}}{r_{2}^{2}}\\) ; r2<\/sub> \\(\\frac { 1000 }{ 4 }\\) = 250 cm<\/p>\n

                                      Question 49.
                                      \nA concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual will be
                                      \n(a) 22.5 cm
                                      \n(b) 7.5 cm
                                      \n(c) 30 cm
                                      \n(d) 45 cm
                                      \nAnswer:
                                      \n(b) 7.5 cm
                                      \nHint:
                                      \nFor virtual images, m = \\(\\frac { -v }{ u }\\) = + 2 or v = – 2u
                                      \nAs \\(\\frac { 1 }{ u }\\) + \\(\\frac { 1 }{ v }\\) = \\(\\frac { 1 }{ f }\\)
                                      \n\u2234 \\(\\frac { 1 }{ u }\\) – \\(\\frac { 1 }{ 2u }\\) = \\(\\frac { 1 }{ -15 }\\) or \\(\\frac { 1 }{ 2u }\\) = \\(\\frac { 1 }{ -15 }\\) \u21d2 u = -7.5 cm<\/p>\n

                                      Question 50.
                                      \nThe magnification of a telescope is given by
                                      \na) \\(\\frac{f_{e}}{f_{o}}\\)<\/p>\n

                                      b) \\(\\frac{f_{o}}{f_{e}}\\)<\/p>\n

                                      c) \\(\\frac{2 f_{e}}{f_{o}}\\)<\/p>\n

                                      d) \\(\\frac{f_{o}+f_{e}}{2}\\)
                                      \nAnswer:
                                      \nb) \\(\\frac{f_{o}}{f_{e}}\\)<\/p>\n

                                      Question 51.
                                      \nA light wave of frequency u and wavelength A travels from air to glass. Then,
                                      \n(a) y changes
                                      \n(b) y does not change, A changes
                                      \n(c) A does not change
                                      \n(d) y and A change
                                      \nAnswer:
                                      \n(b) y does not change, A changes
                                      \nHint:
                                      \nSame reasoning as in the above question.<\/p>\n

                                      Question 52.
                                      \nIn refraction, light waves are bent on passing from one medium to the second medium, because in the second medium.
                                      \n(a) the frequency is different
                                      \n(b) the coefficient of elasticity is different
                                      \n(c) the speed is different
                                      \n(d) the amplitude is smaller
                                      \nAnswer:
                                      \n(c) the speed is different
                                      \nHint:
                                      \nSpeed of light in second medium is different than that in first medium.<\/p>\n

                                      Question 53.
                                      \nA ray of light having a wavelength 720 nm enters in a glass of refractive index 1.5. The wavelength of the ray within the glass will be
                                      \n(a) 360 nm
                                      \n(b) 480 nm
                                      \n(e) 720 nm
                                      \n(d) 1080 nm
                                      \nAnswer:
                                      \n(b) 480 nm
                                      \nHint:
                                      \n\\({ \\lambda }_{ g }\\) = \\(\\frac { { \\lambda }_{ 0 } }{ \\mu } \\) = \\(\\frac { 790nm }{ 1.5 }\\) = 480nm<\/p>\n

                                      Question 54.
                                      \nBrilliance of a diamond is due to
                                      \n(a) shape
                                      \n(b) cutting
                                      \n(c) reflection
                                      \n(d) total internal reflection
                                      \nAnswer:
                                      \n(d) total internal reflection
                                      \nHint:
                                      \nBrilliance of a diamond is due to total internal reflection of light.<\/p>\n

                                      Question 55.
                                      \nAn endoscope is employed by a physician to view the internal parts of a body organ. If is based on the principle of
                                      \n(a) refraction
                                      \n(b) reflection
                                      \n(c) total internal reflection
                                      \n(d) dispersion
                                      \nAnswer:
                                      \n(c) total internal reflection
                                      \nHint:
                                      \nAn endoscope is made of optical fibres which work on the principle of total internal reflection.<\/p>\n

                                      Question 56.
                                      \n\u2018Mirage\u2019 is a phenomenon due to
                                      \n(a) reflection of light
                                      \n(b) refraction of light
                                      \n(c) total internal reflection of light
                                      \n(d) diffraction of light Hint Mirage occurs due to total internal reflection of light.
                                      \nAnswer:
                                      \n(c) total internal reflection of light<\/p>\n

                                      Question 57.
                                      \nTwo lenses of power +12D and -2D are combined together. What is their equivalent focal length?
                                      \n(a) 10 cm
                                      \n(b) 12.5 cm
                                      \n(c) 16.6 cm
                                      \n(d) 8.33 cm
                                      \nAnswer:
                                      \n(a) 10 cm
                                      \nHint:
                                      \nP = P1<\/sub> + P2<\/sub>\u00a0= + 12 – 2 = 10 D
                                      \nF = \\(\\frac { 1 }{ P }\\) = \\(\\frac { 1 }{ 10 }\\)m = 10 cm<\/p>\n

                                      Question 58.
                                      \nIf two lenses of power + 1.5 D and + 1.0 D are placed in contact, then the effective power of combination will be
                                      \n(a) 2.5 D
                                      \n(b) 1.5 D
                                      \n(c) 0.5 D
                                      \n(d) 3.25 D
                                      \nAnswer:
                                      \n(a) 2.5 D
                                      \nHint:
                                      \nP = P1<\/sub> + P2<\/sub>\u00a0= + 1.5 + 1.0 = + 2.5D<\/p>\n

                                      Question 59.
                                      \nThe angle of a prism is 6\u00b0 and its refractive index for green light is 1.5. If a green-ray passes through it, the deviation will be
                                      \n(a) 30\u00b0
                                      \n(b) 15\u00b0
                                      \n(c) 3\u00b0
                                      \n(d) 0\u00b0
                                      \nAnswer:
                                      \n(c) 3\u00b0
                                      \nHint:
                                      \n5 = (p – 1) A = (1.5 – 1) x 6 = 3\u00b0<\/p>\n

                                      Question 60.
                                      \nSky appears to be blue in a clear atmosphere due to light\u2019s
                                      \n(a) diffraction
                                      \n(b) dispersion
                                      \n(c) scattering
                                      \n(d) polarisation
                                      \nAnswer:
                                      \n(c) scattering
                                      \nHint:
                                      \nSky appears blue due to the scattering of light by atmospheric molecules.<\/p>\n

                                      Question 61.
                                      \nOne can not see through fog, because
                                      \n(a) fog absorbs the light
                                      \n(b) light suffers total reflection at droplets
                                      \n(c) the refractive index of the fog is infinity
                                      \n(d) light is scattered by the droplets
                                      \nAnswer:
                                      \n(d) light is scattered by the droplets<\/p>\n

                                      Question 62.
                                      \nFraunhofer lines of the solar system is an example of
                                      \n(a) emission lines spectrum
                                      \n(b) emission band spectrum
                                      \n(c) continuous emission spectrum
                                      \n(d) line absorption spectrum
                                      \nAnswer:
                                      \n(d) line absorption spectrum
                                      \nHint:
                                      \nFraunhofer lines are an example of a line absorption spectrum.<\/p>\n

                                      Question 63.
                                      \nA person using a lens as a sample microscope sees an
                                      \n(a) inverted virtual image
                                      \n(b) inverted real magnified image
                                      \n(c) upright virtual image
                                      \n(d) upright real magnified image
                                      \nAnswer:
                                      \n(d) upright real magnified image
                                      \nHint:
                                      \nA person sees an upright virtual image in a simple microscope.<\/p>\n

                                      Question 64.
                                      \nFour lenses of focal length + 10 cm, + 50 cm, + 100 cm and + 200 cm are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be
                                      \n(a) + 10 cm
                                      \n(b) + 50 cm
                                      \n(c) + 100 cm
                                      \n(d) + 200 cm
                                      \nAnswer:
                                      \n(a) + 10 cm
                                      \nHint:
                                      \nTo produce the largest magnification, the eyepiece should have a minimum focal length.<\/p>\n

                                      Question 65.
                                      \nThe camera lens has an aperture of f and the exposure time is 1\/60 s. What will be the new exposure time if the aperture becomes 1.4f?
                                      \n(a) \\(\\frac { 1 }{ 42 }\\) s
                                      \n(b) \\(\\frac { 1 }{ 56 }\\) s
                                      \n(c) \\(\\frac { 1 }{ 72 }\\) s
                                      \n(d) \\(\\frac { 1 }{ 31 }\\) s
                                      \nAnswer:
                                      \n(d) \\(\\frac { 1 }{ 31 }\\) s
                                      \nHint:
                                      \nTime of exposure \u221d (f – number)2
                                      \n<\/sup>\\(\\frac{t}{\\left(\\frac{1}{60}\\right)}=\\left(\\frac{1.4}{1}\\right)^{2}\\) \u21d2 t = \\(\\frac { 1.4\u00d71.4 }{ 60 }\\) \u2248 \\(\\frac { 1 }{ 31 }\\) s.<\/p>\n

                                      Question 66.
                                      \nFor a person near point of vision is 100 cm. Then the power of lens he must wear so as have normal vision, should be
                                      \n(a) + ID
                                      \n(b) – ID
                                      \n(c) + 3D
                                      \n(d) – 3D
                                      \nAnswer:
                                      \n(c) + 3D
                                      \nHint:
                                      \nf = \\(\\frac { yD }{ y-D }\\) = \\(\\frac { 100\u00d725 }{ 100-25 }\\) = \\(\\frac { 100 }{ 3 }\\) cm = \\(\\frac { 1 }{ 3 }\\) cm ; P = \\(\\frac { 1 }{ f }\\) = +3d<\/p>\n

                                      Question 67.
                                      \nRay optics is valid, when characteristic dimension ions are
                                      \n(a) much smaller than the wavelength of light
                                      \n(b) much larger than the wavelength of light
                                      \n(c) of the same order as the wavelength of light
                                      \n(d) of the order of one millimetre
                                      \nHint:
                                      \nRay optics is valid, when characteristic dimensions are much larger than the wavelength of
                                      \nlight.<\/p>\n

                                      Question 68.
                                      \nA tall man of height 6 feet, wants to see his full image. Then required minimum length of the mirror will be
                                      \n(a) 12 feet
                                      \n(b) 3 feet
                                      \n(e) 6 feet
                                      \n(d) any length
                                      \nHint:
                                      \nminimum height of mirror required for seeing full image Height of the man = 3 feet.<\/p>\n

                                      Question 69.
                                      \nThe refractive index of water is 1.33. What will be the speed of light in water?
                                      \n(a) 3 x 108<\/sup> ms-1<\/sup>
                                      \n(b) 2.26 x 108<\/sup> ms-1<\/sup>
                                      \n(c) 4 x 108<\/sup> ms-1<\/sup>
                                      \n(d) 1.33 x 108<\/sup> ms-1<\/sup>
                                      \nAnswer:
                                      \n(b) 2.26 x 108<\/sup> ms-1<\/sup>
                                      \nHint:
                                      \nAs, \u03bc = \\(\\frac { c }{ v }\\) \u21d2 v = \\(\\frac { c }{ \u03bc }\\) = \\(\\frac { 3\\times { 10 }^{ 8 } }{ 1.33 } \\) = 2.26 x 108<\/sup> ms-1<\/sup><\/p>\n

                                      Question 70.
                                      \nA beam of monochromatic light is refracted from vacuum into a medium of refractive index
                                      \n1.5. The wavelength of refracted light will be
                                      \n(a) same
                                      \n(b) dependent on intensity of refracted light
                                      \n(c) larger
                                      \n(d) smaller
                                      \nAnswer:
                                      \n(d) smaller
                                      \nHint:
                                      \nAs light enters the medium, its wavelength decreases and becomes equal to \\(\\frac { \u03bb }{ \u03bc }\\).<\/p>\n

                                      Question 71.
                                      \nOptical fibers are based on
                                      \n(a) total internal reflection
                                      \n(b) less scattering
                                      \n(c) refraction
                                      \n(d) less absorption coefficient
                                      \nAnswer:
                                      \n(a) total internal reflection<\/p>\n

                                      Question 72.
                                      \nA convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the lens. Then, its focal length will
                                      \n(a) become zero
                                      \n(b) becomes infinite
                                      \n(e) remain unchanged
                                      \n(d) become small, but non-zero
                                      \nAnswer:
                                      \n(b) becomes infinite
                                      \nHint:
                                      \n\\(\\frac { 1 }{ { f }_{ 1 } } \\) = \\(\\left(\\frac{\\mu_{g}}{\\mu_{e}}-1\\right)\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) = (1 – 1) \\(\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) = 0 ; f1<\/sub> = \u221e<\/p>\n

                                      Question 73.
                                      \nA convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power of the combination (in diopter) is
                                      \n(a) zero
                                      \n(b) 25
                                      \n(c) 50
                                      \n(d) infinite
                                      \nAnswer:
                                      \n(a) zero
                                      \nHint:
                                      \n\\(\\frac { 1 }{ F }\\) = \\(\\frac { 1 }{ { f }_{ 1 } } \\) + \\(\\frac { 1 }{ { f }_{ 2 } } \\) = \\(\\frac { 1 }{ +25 }\\) + \\(\\frac { 1 }{ -25 }\\) = 0 ; P = \\(\\frac { 1 }{ F }\\) = 0<\/p>\n

                                      Question 74.
                                      \nThe focal length of a converging lens is measured for violet, green and red colours. If is fV,<\/sub> fG<\/sub> and fR<\/sub> respectively. We will get
                                      \n(a) fV<\/sub> = fG<\/sub>
                                      \n(b) fG<\/sub> = fR<\/sub>
                                      \n(c) fV<\/sub> < fR<\/sub>
                                      \n(d) fV<\/sub> > fR<\/sub>
                                      \nAnswer:
                                      \n(c) fV<\/sub> < fR<\/sub>
                                      \nHint:
                                      \n\\(\\frac { 1 }{ f }\\) = (\u03bc – 1) \\(\\left(\\frac{1}{\\mathrm{R}_{1}}-\\frac{1}{\\mathrm{R}_{2}}\\right)\\) i.e., f \u221d \\(\\frac { 1 }{ \u03bc – 1 }\\)
                                      \nAs \u03bcV<\/sub> > \u03bcR<\/sub>, so, fV<\/sub> < fR<\/sub><\/p>\n

                                      Question 75.
                                      \nRainbow is formed due to a combination of
                                      \n(a) refraction and scattering
                                      \n(b) refraction and absorption
                                      \n(c) dispersion and total internal reflection
                                      \n(d dispersion and focusing
                                      \nAnswer:
                                      \n(c) dispersion and total internal reflection
                                      \nHint:
                                      \nRainbow is formed due to dispersion and total internal reflection of sunlight by raindrops.<\/p>\n

                                      Question 76.
                                      \nThe blue colour of the sky is due to the phenomenon of
                                      \n(a) scattering
                                      \n(b) dispersion
                                      \n(c) reflection
                                      \n(d) refraction
                                      \nAnswer:
                                      \n(a) scattering
                                      \nHint:
                                      \nThe blue colour of the sky is due to the scattering of sunlight by atmospheric molecules.<\/p>\n

                                      Question 77.
                                      \nAn astronomical telescope often fold angular magnification has a length of 44 cm. The focal length of the object is
                                      \n(a) 4 cm
                                      \n(b) 40 cm
                                      \n(c) 44 cm
                                      \n(d) 440 cm
                                      \nAnswer:
                                      \n(b) 40 cm
                                      \nHint:
                                      \nHere, f0<\/sub> + fe<\/sub> = 44 cm
                                      \nm =\\(\\frac { { f }_{ 0 } }{ { f }_{ e } } \\) (or) f0<\/sub> = 10 fe<\/sub>
                                      \n\u2234 10 fe<\/sub> + 10 fe<\/sub> = 44 cm (or) fe<\/sub> = 4 cm
                                      \nHence, f0<\/sub> = 10 x 4 (or) f0<\/sub> = 40 cm<\/p>\n

                                      Question 78.
                                      \nExposure time of a camera lens at the \\(\\frac { f }{ 2.8 }\\) Setting is \\(\\frac { 1 }{ 200 }\\) second. The correct time of exposure at \\(\\frac { f }{ 5.6 }\\) is
                                      \n(a) 0.20 second
                                      \n(b) 0.40 second
                                      \n(c) 0.02 second
                                      \n(d) 0.04 second
                                      \nAnswer:
                                      \n(c) 0.02 second
                                      \nHint:
                                      \nTime of exposure, t \u221d (f- number)2<\/sup>
                                      \n\u2234 \\(\\frac{t}{\\left(\\frac{1}{200}\\right)}=\\left(\\frac{5.6}{2.8}\\right)^{2}\\) = 4 \u21d2 t = 0.02s<\/p>\n

                                      Question 79.
                                      \nWhich of the following is not due to total internal reflection?
                                      \n(a) Working of optical fibre
                                      \n(b) Difference between apparent and real depth of a pond
                                      \n(c) Mirage on hot summer day
                                      \n(d) Brilliance of diamond
                                      \nAnswer:
                                      \n(b) Difference between apparent and real depth of a pond
                                      \nHint:
                                      \nDifference between apparent and real depth of a pond is due to refraction of light and the other three phenomena involve total internal reflection.<\/p>\n

                                      Question 80.
                                      \nAn object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be
                                      \n(a) 0.25 m
                                      \n(b) 0.5 m
                                      \n(c) 1.0 m
                                      \n(d) 2.0 m
                                      \nAnswer:
                                      \n(c) 1.0 m
                                      \nHint:
                                      \nDistance between object and image = 0.5 + 0.5 = 1.0 m.<\/p>\n

                                      Question 81.
                                      \nAn observer moves towards a stationary plane mirror at a speed of 4 ms-1<\/sup> with what speed – will his image move towards him?
                                      \n(a) 2 ms-1<\/sup>
                                      \n(b) 4 ms-1<\/sup>
                                      \n(c) 8 ms-1<\/sup>
                                      \n(d) the image will stay at rest
                                      \nAnswer:
                                      \n(c) 8 ms-1<\/sup>
                                      \nHint:
                                      \nSpeed of the image towards the observer = 2 x 4 = 8 ms-1<\/sup><\/p>\n

                                      Question 82.
                                      \nIf two mirrors are kept at 60\u00b0 to each other and a body is placed at the middle, then total number of images formed is
                                      \n(a) six
                                      \n(b) four
                                      \n(c) five
                                      \n(d) three
                                      \nAnswer:
                                      \n(c) five
                                      \nHint:
                                      \nNumber of images formed = \\(\\frac { 360 }{ \u03b8 }\\) -1 = \\(\\frac { 360 }{ 60 }\\) -1 = 5<\/p>\n

                                      Question 83.
                                      \nIf an object is placed at 10 cm infront of a concave mirror of focal length 15 cm. The magnification of image is
                                      \n(a) -1.5
                                      \n(b) 1.5
                                      \n(c) -3
                                      \n(d) 3
                                      \nAnswer:
                                      \nd) 3
                                      \nHint:
                                      \nm = \\(\\frac { f }{ f-u }\\) = \\(\\frac { -15 }{ -15(-10) }\\) = +3<\/p>\n

                                      Question 84.
                                      \nAn object of length 2.5 cm is placed at the principal axis of a concave mirror at a distance 1.5f. The image height is
                                      \n(a) + 5 m
                                      \n(b) -5 cm
                                      \n(c) – 10 cm
                                      \n(d) + 1 cm
                                      \nAnswer:
                                      \n(b) -5 cm
                                      \nHint:
                                      \nm = \\(\\frac { f }{ f-u }\\) = \\(\\frac { f }{ f-1.5f}\\) = -2cm
                                      \nHeight image = m x height of object -2 x 2.5 = – 5 cm<\/p>\n

                                      Question 85.
                                      \nWhich of the following mirror is used by a dentist to examine a small cavity?
                                      \n(a) Concave mirror
                                      \n(b) Convex mirror
                                      \n(c) Combination of (a) and (b)
                                      \n(d) None of these
                                      \nAnswer:
                                      \n(a) Concave mirror
                                      \nHint:
                                      \nA concave mirror, because it forms erect and enlarged image when held closer to the cavity.<\/p>\n

                                      Question 86.
                                      \nWhen a ray of light enters from one medium to another, then which of the following does not change?
                                      \n(a) Frequency
                                      \n(b) Wavelength
                                      \n(c) Speed
                                      \n(d) Amplitude
                                      \nAnswer:
                                      \n(a) Frequency
                                      \nHint:
                                      \nOnly frequency remains unchanged.<\/p>\n

                                      Question 87.
                                      \nWhen light travels from one medium to the other medium of which the refractive index is different, then which of the following will change?
                                      \n(a) Frequency, wavelength and velocity
                                      \n(6) Frequency and wavelength
                                      \n(c) Frequency and velocity
                                      \n(d) Wavelength and velocity
                                      \nAnswer:
                                      \n(d) Wavelength and velocity
                                      \nHint:
                                      \nWavelength and velocity will change while frequency remains unchanged.<\/p>\n

                                      Question 88.
                                      \nThe time taken by the light to cross a glass of thickness 4 mm and refractive index (\u03bc = 3), will be
                                      \n(a) 4 x 10-11<\/sup> sec
                                      \n(b) 16 x 10-11<\/sup> sec
                                      \n(c) 8 x 10-11<\/sup> sec
                                      \n(d) 24 x 10-10<\/sup> sec
                                      \nAnswer:
                                      \n(a) 4 x 10-11<\/sup> sec
                                      \nHint:
                                      \n\"Samacheer
                                      \n\\(\\frac { d }{ c\/a }\\) = \\(\\frac { \u03bcd }{ c }\\) = \\(\\frac { 3\u00d74\u00d710^{-3} }{ 3\u00d710^{8} }\\) =\u00a0 4 \u00d7 10-11 <\/sup>s<\/p>\n

                                      Question 89.
                                      \nThe critical angle of a medium with respect to air is 45\u00b0. The refractive index of a medium is
                                      \n(a) 1.41
                                      \n(b) 1.2
                                      \n(c) 1.5
                                      \n(d) 2
                                      \nAnswer:
                                      \n(a) 1.41
                                      \nHint:
                                      \n\u03bc = \\(\\frac{1}{\\sin i_{c}}\\) = \\(\\frac { 1 }{ sin 45\u00b0 }\\) = \\(\\frac { 1 }{ 1\/\u221a2 }\\) \u2248 1.41<\/p>\n

                                      Question 90.
                                      \nIf the critical angle for total internal reflection from a medium to vacuum is 30\u00b0, then velocity of light in the medium is
                                      \n(a) 6 x 108<\/sup> m\/sec
                                      \n(b) 2 x 108<\/sup> m\/sec
                                      \n(c) 3 x 108<\/sup> m\/sec
                                      \n(d) 1.5 x 108<\/sup> m\/sec
                                      \nAnswer:
                                      \n(c) 3 x 108<\/sup> m\/sec
                                      \nHint:
                                      \n\u03bc = \\(\\frac{1}{\\sin i_{c}}\\) = \\(\\frac { c }{ v }\\)<\/p>\n

                                      Question 91.
                                      \nWhen a ray of light enters from one medium to another, its velocity is doubled. The critical angle for the ray for two internal reflections will be
                                      \n(a) 30\u00b0
                                      \n(b) 60\u00b0
                                      \n(c) 90\u00b0
                                      \n(d) Information is incomplete
                                      \nAnswer:
                                      \n(a) 30\u00b0
                                      \nHint:
                                      \n\u03bc = \\(\\frac{1}{\\sin i_{c}}\\) = \\(\\frac { { v }_{ 1 } }{ { v }_{ 2 } } \\) =\\(\\frac { { 2v }_{ 2 } }{ { v }_{ 2 } } \\) = 2 ; sin ic<\/sub> = \\(\\frac { 1 }{ 2 }\\) \u2234ic<\/sub> = 30\u00b0<\/p>\n

                                      Question 92.
                                      \nA driver at a depth 12 m inside water (\u03bc = 4\/3) see the sky in a cone of semi-vertical angle is
                                      \n(a) sin-1<\/sup>\\(\\left( \\frac { 4 }{ 3 } \\right) \\)
                                      \n(b) tan-1<\/sup>\\(\\left( \\frac { 4 }{ 3 } \\right) \\)
                                      \n(c) sin-1<\/sup>\\(\\left( \\frac { 3 }{ 4 } \\right) \\)
                                      \n(d) 90\u00b0
                                      \nAnswer:
                                      \n(c) sin-1<\/sup>\\(\\left( \\frac { 3 }{ 4 } \\right) \\)
                                      \nHint:
                                      \nRequired semi vertical angle = Critical angle ic<\/sub> = sin-1<\/sup> \\(\\frac { 1 }{ \u03bc }\\) = sin-1<\/sup>\\(\\left( \\frac { 3 }{ 4 } \\right) \\)<\/p>\n

                                      Question 93.
                                      \nThe principle behind optical fibres is
                                      \n(a) total internal reflection
                                      \n(b) total external reflection
                                      \n(c) both (a) and (b)
                                      \n(d) diffraction
                                      \nAnswer:
                                      \n(a) total internal reflection
                                      \nHint:
                                      \nOptical fibres work on the principle of total internal reflection.<\/p>\n

                                      Question 94.
                                      \nAir bubble in water behaves as
                                      \n(\u0430) some times concave, sometimes convex lens
                                      \n(b) concave lens
                                      \n(c) convex lens
                                      \n(d) always refracting surface
                                      \nAnswer:
                                      \n(b) concave lens<\/p>\n

                                      Question 95.
                                      \nA convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is
                                      \n(a) -1.5 D
                                      \n(b) – 6.5 D
                                      \n(c) + 6.6 D
                                      \n(d) + 6.5 D
                                      \nAnswer:
                                      \n(a) -1.5 D
                                      \nHint:
                                      \nP = P1<\/sub> + P2<\/sub> = \\(\\frac { 1 }{ { f }_{ 1 } } \\) + \\(\\frac { 1 }{ { f }_{ 2 } } \\) = \\(\\frac { 100 }{ 40 }\\) + \\(\\frac { 100 }{ -25 }\\) = -1.5d.<\/p>\n

                                      Question 96.
                                      \nTwo thin lenses, one of focal length + 60 cm and the other of focal length – 20 cm are put in contact, the combined focal length is,
                                      \n(a) 15 cm
                                      \n(b) – 15 cm
                                      \n(c) – 30 cm
                                      \n(d) 30 cm
                                      \nAnswer:
                                      \n(c) – 30 cm
                                      \nHint:
                                      \nF = \\(\\frac { { f }_{ 1 }{ f }_{ 2 } }{ { f }_{ 1 }{ +f }_{ 2 } } \\) = \\(\\frac { 60 \u00d7(-20) }{ 60-20 }\\) = 30 cm<\/p>\n

                                      Question 97.
                                      \nHow does refractive index (\u03bc) of a material vary with respect to wavelength (\u03bb). (A and B are constants).
                                      \n(a) \u03bc = A + \\(\\frac { B }{ { \\lambda }^{ 2 } } \\)
                                      \n(b) \u03bc = A + B\u03bb2<\/sup>
                                      \n(c) \u03bc = A + \\(\\frac { B }{ \u03bb }\\)
                                      \n(d) \u03bc = A + B\u03bb
                                      \nAnswer:
                                      \n(a) \u03bc = A + \\(\\frac { B }{ { \\lambda }^{ 2 } } \\)
                                      \nHint:
                                      \nAccording to cauchy’s relation, \u03bc = A + \\(\\frac { B }{ { \\lambda }^{ 2 } } \\)<\/p>\n

                                      Question 98.
                                      \na prism of a refracting angle 60\u00b0 is made with a material of refractive index p. For a certain wavelength of light, the angle of minimum deviation is 30\u00b0. For this wavelength, the value of p of material is
                                      \n(a) 1.820
                                      \n(b) 1.414
                                      \n(c) 1.503
                                      \n(d) 1.231
                                      \nAnswer:
                                      \n(b) 1.414
                                      \nHint:
                                      \n\u03bc = \\(\\frac{\\sin \\left(\\frac{60^{\\circ}+30^{\\circ}}{2}\\right)}{\\sin 30^{\\circ}}\\) = \\(\\frac { sin 45\u00b0 }{ sin 30\u00b0 }\\) = \\(\\frac { 1 }{ \u221a2 }\\) x \\(\\frac { 2 }{ 1 }\\) \u221a2 = 1.414<\/p>\n

                                      Question 99.
                                      \nRefractive index of red and violet light are 1.52 and 1.54 respectively. If the angle of prism is 10\u00b0, the angular dispersion will be
                                      \n(a) 0.02\u00b0
                                      \n(b) 0.20\u00b0
                                      \n(c) 3.06\u00b0
                                      \n(d) 30.6\u00b0
                                      \nAnswer:
                                      \n(b) 0.20\u00b0
                                      \nHint:
                                      \nAngular dispersion = \u03b4V<\/sub> – \u03b4R<\/sub> = A(\u03bcV<\/sub> – \u03bcV<\/sub>) = 10\u00b0 (1.54 – 1.52) = 0.20\u00b0<\/p>\n

                                      Question 100.
                                      \nIn a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then magnifying power is
                                      \n(a) \\(\\frac { 25 }{ f }\\)
                                      \n(b) 1 + \\(\\frac { 25 }{ f }\\)
                                      \n(c) \\(\\frac { f }{ 25 }\\)
                                      \n(d) \\(\\frac { f }{ 25 }\\) + 1
                                      \nAnswer:
                                      \n(b) 1 + \\(\\frac { 25 }{ f }\\)
                                      \nHint:
                                      \nWhen the final image is formed at the least distance of distinct vision in a simple microscope,
                                      \nm = 1 + \\(\\frac { 25 }{ f }\\)<\/p>\n

                                      Question 101.
                                      \nMagnification at least distance of distinct vision of a simple microscope of focal length 5 cm is
                                      \n(a) 2
                                      \n(b) 5
                                      \n(c) 4
                                      \n(d) 6
                                      \nAnswer:
                                      \n(d) 6
                                      \nHint:
                                      \nm = 1 + \\(\\frac { 25 }{ f }\\) = 1 + \\(\\frac { 25 }{ 5 }\\) = 6<\/p>\n

                                      Question 102.
                                      \nMagnification of a compound microscope is 30. Focal length of eyepiece is 5 cm and the image is formed at a distance of distinct vision of 25 cm. The magnification of the objective
                                      \nlens is
                                      \n(a) 6
                                      \n(b) 5
                                      \n(c) 7.5
                                      \n(d) 10
                                      \nAnswer:
                                      \n(b) 5
                                      \nHint:
                                      \nme<\/sub> = \\(\\frac { v }{ u}\\) = \\(\\frac { D }{ { u }_{ c } } \\) = 1 + \\(\\frac { D }{ { f }_{ e } } \\) = 1 + \\(\\frac { 25 }{ 5 }\\) = 6
                                      \nFor the compound microscope, m = m0<\/sub> x me<\/sub> \u21d2 30 = m0<\/sub>\u00a0x 6 (or) m0<\/sub> = 5<\/p>\n

                                      Question 103.
                                      \nThe astronomical microscope consists of objective and eyepiece. The focal length of the objective is
                                      \n(a) equal to that of the eyepiece
                                      \n(b) shorter than that of the eyepiece
                                      \n(c) greater than that of the eyepiece
                                      \n(d) five times shorter than that of eyepiece
                                      \nAnswer:
                                      \n(c) greater than that of the eyepiece
                                      \nHint:
                                      \nFor producing large magnification,f0 <\/sub>> fe<\/sub><\/p>\n

                                      Question 104.
                                      \nThe number of lenses in terrestrial telescope is
                                      \n(a) 2
                                      \n(b) 4
                                      \n(c) 3
                                      \n(d) 6
                                      \nAnswer:
                                      \n(c) 3
                                      \nHint:
                                      \nA terrestrial telescope consists of three lenses: objective, erecting lens and eyepiece.<\/p>\n

                                      Question 105.
                                      \nAn achromatic combination of lenses is formed by joining
                                      \n(a) 2 convex lens
                                      \n(b) 1 convex, 1 concave lens
                                      \n(c) 2 concave lenses
                                      \n(d) 1 convex and 1 plane mirror
                                      \nAnswer:
                                      \n(b) 1 convex, 1 concave lens
                                      \nHint:
                                      \nAn achromatic doublet should satisfy the condition \\(\\frac { { w }_{ 1 } }{ { f }_{ 1 } } \\) + \\(\\frac { { w }_{ 2 } }{ { f }_{ 2 } } \\) = 0.<\/p>\n

                                      Question 106.
                                      \nThe amount of light received by a camera depends upon
                                      \n(a) diameter only
                                      \n(b) ratio of focal length and diameter
                                      \n(c) product of focal length and diameter
                                      \n(d) only one of the focal length
                                      \nAnswer:
                                      \n(b) ratio of focal length and diameter
                                      \nHint:
                                      \nThe amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture.<\/p>\n

                                      Question 107.
                                      \nMyopia is corrected by using a
                                      \n(a) cylindrical lens
                                      \n(b) bifocal lens
                                      \n(c) convex lens
                                      \n(d) concave lens
                                      \nAnswer:
                                      \n(d) concave lens<\/p>\n

                                      Question 108.
                                      \nThe critical angle for total internal reflection in diamond is 24.5\u00b0. The angle refractive index of diamond is
                                      \n(a) 2.41
                                      \n(b) 1.41
                                      \n(c) 2.59
                                      \n(d) 1.59
                                      \nAnswer:
                                      \n(a) 2.41
                                      \nHint:
                                      \n\u03bc = \\(\\frac { 1 }{ { sin i }_{ c } } \\) = \\(\\frac { 1 }{ sin 24.5\u00b0 }\\) = 2.41<\/p>\n

                                      Question 109.
                                      \nWhen a glass lens with \u03bc = 1.47 is immersed in a trough of liquid, it looks to be disappeared. The liquid in the trough could be
                                      \n(a) water
                                      \n(b) kerosene
                                      \n(c) glycerine
                                      \n(d) alcohol
                                      \nAnswer:
                                      \n(c) glycerine
                                      \nHint:
                                      \nGlass lens will disappear, if \u00b5l<\/sub> = \u00b5g<\/sub>.<\/p>\n

                                      Question 110.
                                      \nIn optical fibres, the refractive index of the core is
                                      \n(a) greater than that of the cladding
                                      \n(b) equal to that of the cladding
                                      \n(c) smaller than that of the cladding
                                      \n(d) independent of that of the cladding
                                      \nAnswer:
                                      \n(a) greater than that of the cladding
                                      \nHint:
                                      \nIn optical fibres, refractive index of core material > refractive index of the cladding.<\/p>\n

                                      Question 111.
                                      \nFor a wavelength of light \u2018\u03bb\u2019 and scattering object of size \u2018a\u2019, all wavelength are scattered nearly equally, if
                                      \n(a) a = \u03bb
                                      \n(b) a >> \u03bb
                                      \n(c) a << \u03bb
                                      \n(d) a \u2265 \u03bb
                                      \nAnswer: (b) a >> \u03bb
                                      \nHint:
                                      \nFor a >> \u03bb, the scattering power is not selective.<\/p>\n

                                      Question 112.
                                      \nTwo coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beams are
                                      \n(a) 5I and I
                                      \n(b) 9I and I
                                      \n(c) 5I and 3I
                                      \n(d) 9I and 3I
                                      \nAnswer:
                                      \n(b) 9I and I
                                      \nHint:
                                      \nImax<\/sub> = \\((\\sqrt{\\mathrm{I}_{1}}+\\sqrt{\\mathrm{I}_{2}})^{2}\\) = \\((\\sqrt{41}+\\sqrt{1})^{2}\\) = 9I
                                      \nImax<\/sub> = \\((\\sqrt{\\mathrm{I}_{1}}-\\sqrt{\\mathrm{I}_{2}})^{2}\\) = \\((\\sqrt{41}-\\sqrt{1})^{2}\\) = I<\/p>\n

                                      Question 113.
                                      \nscreen is doubled. The fringe width is
                                      \n(a) unchanged
                                      \n(b) halved
                                      \n(c) doubled
                                      \n(d) quadrupled
                                      \nAnswer:
                                      \n(d) quadrupled
                                      \nHint:
                                      \n\u03b2 = \\(\\frac { \u03bbD }{ d }\\) ; \u03b2 = \\(\\frac { \u03bb \u00d72D }{ D\/2 }\\) = 4\u03b2<\/p>\n

                                      Question 114.
                                      \nIn a young\u2019s double-slit experiment, 12 fringes are observed to be formed in a certain segment of the screen, when the light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by
                                      \n(a) 12
                                      \n(b) 18
                                      \n(c) 24
                                      \n(d) 30
                                      \nAnswer:
                                      \n(b) 18
                                      \nHint:
                                      \nn1<\/sub>\u03bb1<\/sub>= n2<\/sub>\u03bb2<\/sub> \u21d212 x 600 = n2<\/sub> x 400 or n2<\/sub> = 18<\/p>\n

                                      Question 115.
                                      \nConsider Fraunhofer diffraction pattern obtained with a single slit illuminated at the normal incidence. At the angular position of the first diffraction minimum, the phase difference between the wavelets from the opposite edges of the slits is
                                      \n(a) \\(\\frac { \u03c0 }{ 4 }\\)
                                      \n(b) \\(\\frac { \u03c0 }{ 2 }\\)
                                      \n(c) 2\u03c0
                                      \n(d) \u03c0
                                      \nAnswer:
                                      \n(c) 2\u03c0
                                      \nHint:
                                      \nAt the angular position of first minimum wavelets from opposite edges of the slits have a path difference of X and a phase difference of 2\u03c0 radian.<\/p>\n

                                      Question 116.
                                      \nA beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is
                                      \n(a) 1.2 cm
                                      \n(b) 1.2 mm
                                      \n(c) 2.4 cm
                                      \n(d) 2.4 mm
                                      \nAnswer:
                                      \n(c) 2.4 cm
                                      \nHint:
                                      \nDistance between the first dark fringes on either side = width of central maximum
                                      \n= \\(\\frac { 2D\u03bb }{ d }\\) = \\(\\frac{2 \\times 2 \\times 600 \\times 10^{-9}}{1.00 \\times 10^{-3}}\\) m = 2.4 x 10-3<\/sup> m = 2.4mm<\/p>\n

                                      Question 117.
                                      \nA young\u2019s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.
                                      \n(a) hyperbola
                                      \n(b) circle
                                      \n(c) straight line
                                      \n(d) parabola
                                      \nAnswer:
                                      \n(a) hyperbola
                                      \nHint:
                                      \nIn young\u2019s double slit experiment, the fringes obtained are hyperbolic in shape. But in a small interference pattern, the fringes appear straight.<\/p>\n

                                      Question 118.
                                      \nThe initial shape of the wavefront of the beam is
                                      \n(a) planar
                                      \n(b) convex
                                      \n(c) concave
                                      \n(d) convex near the axis and concave near the periphery
                                      \nAnswer:
                                      \n(a) planar
                                      \nHint:
                                      \nAs the beam is initially parallel, the shape of wavefront is planar.<\/p>\n

                                      Question 119.
                                      \nThe angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index \u03bc) is
                                      \n(a) sin-1<\/sup> (\u03bc)
                                      \n(b) sin-1<\/sup> \\(\\left( \\frac { 1 }{ \u03bc } \\right) \\)
                                      \n(c) tan-1<\/sup> \\(\\left( \\frac { 1 }{ \u03bc } \\right) \\)
                                      \n(d) tan-1<\/sup> (\u03bc)
                                      \nAnswer:
                                      \n(d) tan-1<\/sup> (\u03bc)
                                      \nAccording to Brewster\u2019s law,
                                      \n\u03bc = tan ip<\/sub> \u2234 ip<\/sub> = tan-1<\/sup> (\u03bc)<\/p>\n

                                      Question 120.
                                      \nAccording to Huygen\u2019s principle, light is a form of
                                      \n(a) particle
                                      \n(b) rays
                                      \n(c) wave
                                      \n(d) none of the above
                                      \nAnswer:
                                      \n(c) wave
                                      \nHint:
                                      \nAccording to Huygen\u2019s principle, light travels in the form of a longitudinal wave.<\/p>\n

                                      Question 121.
                                      \nWhich one of the following phenomena is not explained by Huygen\u2019s construction of wavefront?
                                      \n(a) refraction
                                      \n(b) reflection
                                      \n(c) diffraction
                                      \n(d) origin of spectra
                                      \nAnswer:
                                      \n(d) origin of spectra
                                      \nHint:
                                      \nHuygen\u2019s construction of wavefront cannot explain origin of spectra which can be explained on the basis of quantum theory.<\/p>\n

                                      Samacheer Kalvi 12th Physics Optics Additional Problems<\/span><\/strong><\/p>\n

                                      Question 1.
                                      \nLight from a point source in air falls on a convex spherical glass surface (n = 1.5, radius of curvature = 20 cm). The distance of light source from the glass surface is 100 cm. At
                                      \nWhat position is the image formed?
                                      \nSolution:
                                      \nn1<\/sub> = 1; n2<\/sub> = 1.5
                                      \nu = 100cm ; R = + 20 cm
                                      \n(R is + Ve for a convex refracting surfce)
                                      \nAs \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { 1 }{ 2 }\\)
                                      \n\\(\\frac{n_{2}}{v}-\\frac{n_{1}}{u}\\) = \\(\\frac{n_{2}-n_{1}}{R}\\)
                                      \n\\(\\frac{1.5}{v}+\\frac{1}{100}\\) = \\(\\frac{1.5-1}{20}=\\frac{1}{40}\\)
                                      \n\\(\\frac { 3 }{ 2v }\\) = \\(\\frac{1}{40}-\\frac{1}{100}=\\frac{5-2}{200}=\\frac{3}{200}\\)
                                      \n\\(\\frac { 1 }{ v }\\) = \\(\\frac { 1 }{ 100 }\\)
                                      \nv = 100 cm
                                      \nThus the image is formed at a distance of 100 cm from the glass surface in the direction of incident light.<\/p>\n

                                      Question 2.
                                      \nFind the value of critical angle for a material of refractive index \u221a3 .
                                      \nSolution:
                                      \nHere, n = \u221a3
                                      \nsin ic<\/sub> = \\(\\frac { 1 }{ n }\\) = \\(\\frac { 1 }{ \u221a3 }\\) = \\(\\frac { \u221a3 }{ 3 }\\)
                                      \n\u2234 critical angle, ic<\/sub> = 35.3\u00b0<\/p>\n

                                      Question 3.
                                      \nThe radius of curvature of each face of biconcave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air.
                                      \nSolution:
                                      \nHere n = 1.5 ; R1<\/sub> = – 30 cm ; R2<\/sub> = 30 cm
                                      \nUsing len’s maker’s formula,
                                      \n\\(\\frac { 1 }{ f }\\) = (n – 1) \\(\\left(\\frac{1}{R_{1}}-\\frac{1}{R_{2}}\\right)\\)
                                      \n= (1.5-1)\\(\\left[\\frac{1}{-30}-\\frac{1}{30}\\right]\\) = 0.5 \u00d7 \\(\\left(\\frac{-2}{30}\\right)\\)
                                      \n\\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ 30 }\\)
                                      \nf = -30<\/p>\n

                                      Question 4.
                                      \nThe radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length is 12 cm. What is the refractive index of glass?
                                      \nSolution:
                                      \nf = +12 cm ; R1<\/sub> = 10 cm ; R2<\/sub> = – 15 cm ; n = ?
                                      \nAs, \\(\\frac { 1 }{ f }\\) = (n – 1) \\(\\left(\\frac{1}{R_{1}}-\\frac{1}{R_{2}}\\right)\\)
                                      \n\\(\\frac { 1 }{ 12 }\\) = (n – 1) \\(\\left(\\frac{1}{10}+\\frac{1}{15}\\right)\\) = (n – 1) x latex]\\frac { 5 }{ 30 }[\/latex]
                                      \n(n – 1) = \\(\\frac { 6 }{ 12 }\\) = 0.5
                                      \nn = 0.5 +1
                                      \nn = 1.5<\/p>\n

                                      Question 5.
                                      \nA double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height is placed at a distance of 10 cm from the lens. Find the position, nature and size of the image.
                                      \nSolution:
                                      \nHere n = 1.5 ; R1<\/sub>= +20 cm ; R2<\/sub> = – 20 cm
                                      \nUsing lens maker’s formula,
                                      \n\\(\\frac { 1 }{ f }\\) = (n – 1) \\(\\left(\\frac{1}{R_{1}}-\\frac{1}{R_{2}}\\right)\\)
                                      \n(1.5-1) = \\(\\left[\\frac{1}{20}-\\frac{1}{-20}\\right]\\) 0.5 x \\(\\left(\\frac{2}{20}\\right)\\)
                                      \n\\(\\frac { 1 }{ f }\\) = \\(\\frac { 1 }{ 20 }\\)
                                      \nf = 20 cm
                                      \nNow , u = -10 cm andf = + 20 cm
                                      \nFrom thin lens formula, \\(\\frac { 1 }{ v }\\) = \\(\\frac { 1 }{ f }\\) + \\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ 20 }\\) – \\(\\frac { 1 }{ 10 }\\) = \\(\\frac { 1 }{ 20 }\\)
                                      \n\u2234 v = -20 cm
                                      \nMagnification, m = \\(\\frac{h_{2}}{h_{1}}\\)= \\(\\frac { v }{ u }\\)
                                      \n\\(\\frac{h_{2}}{5cm}\\) = \\(\\frac { -20 }{ -10 }\\)
                                      \nh2<\/sub>
                                      \nHence a virtual and erect image of height 10 cm is formed at a distance of 20 cm from the lens on the same side as the the object.<\/p>\n

                                      Question 6.
                                      \nThe image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.
                                      \nSolution.
                                      \nHeref= 20 cm, m = + 4 for a virtual image.
                                      \nTo calculate u, we have
                                      \nm = \\(\\frac { f }{ u+f }\\)<\/p>\n

                                      Question 7.
                                      \nThe radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power.
                                      \nSolution:
                                      \nHere, n = 105 ; R1<\/sub> = + 40cm = 0.40 m
                                      \nR2<\/sub> = – 40cm = – 0.40 m
                                      \nPower (p) = \\(\\frac { 1 }{ f }\\) = (n – 1) \\(\\left(\\frac{1}{R_{1}}-\\frac{1}{R_{2}}\\right)\\)
                                      \n= (1.5 – 1)\\(\\left[\\frac{1}{0.40}-\\frac{1}{(-0.40)}\\right]\\) = 0.5 x \\(\\frac { 2 }{ 0.40 }\\)
                                      \np = 2.5 D<\/p>\n

                                      Question 8.
                                      \nA ray of light incident on an equilateral glass prism shows minimum deviation of 30\u00b0. Calculate the speed of light through the prism.
                                      \nSolution:
                                      \nHere, A = 60\u00b0 ; D = 30\u00b0
                                      \nRefractive index, n = \\(\\frac{\\sin \\left(\\frac{A+D}{2}\\right)}{\\sin \\left(\\frac{A}{2}\\right)}\\) = \\(\\frac{\\sin \\left(\\frac{60+30}{2}\\right)}{\\sin \\left(\\frac{60}{2}\\right)}\\)
                                      \nn = \\(\\frac{\\sin 45^{\\circ}}{\\sin 30^{\\circ}}\\) = \\(\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{1}{2}}\\) = \u221a2
                                      \nn = 1.414
                                      \nVelocity or light in glass, v = \\(\\frac { c }{ n }\\) = \\(\\frac{3 \\times 10^{8}}{1.414}\\)
                                      \nv = 2.12 x 108<\/sup> ms-1<\/sup><\/p>\n

                                      Question 9.
                                      \nTwo sources of intensity I and 41 are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase
                                      \n(i) Zero
                                      \n(ii) \\(\\frac { \u03c0 }{ 2 }\\)
                                      \n(iii) \u03c0
                                      \nSolution:
                                      \nThe resultant intensity at a point where phase difference is \u03a6 is
                                      \nIR<\/sub>+I1<\/sub>+ I2<\/sub> 2\\(+\\sqrt{\\mathrm{I}_{1} \\mathrm{I}_{2}}\\) cos \u03a6
                                      \nAs I1<\/sub> + I and I2<\/sub> = 4I, therefore,
                                      \nIR<\/sub> = I + 4I + 2\\(\\sqrt{I.4I}\\) cos \u03a6
                                      \nIR<\/sub> = 5I + 4I cos \u03a6
                                      \n(i) When \u03a6 = 0 ; IR<\/sub> = 5I + 4I cos 0 = 9I
                                      \n(ii) When \u03a6 = \\(\\frac { \u03c0 }{ 2 }\\) ; IR<\/sub> = 5I + 4I cos \\(\\frac { \u03c0 }{ 2 }\\) = 5I
                                      \n(iii) When \u03a6 = \u03c0 ; IR<\/sub> = 5I + 4I cos \u03c0 = 51 – 41 = I
                                      \n\u03a6 = 0 ; IR<\/sub> = 9I
                                      \n\u03a6 = \\(\\frac { \u03c0 }{ 2 }\\) ; IR<\/sub> = 5I
                                      \n\u03a6 = \u03c0 ; IR<\/sub> = I<\/p>\n

                                      Question 10.
                                      \nAssume that light of wavelength 600 A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
                                      \nSolution:.
                                      \nThe limit of resolution of a telescope, d\u03b8 = \\(\\frac { 1.22\u03bb }{ D }\\)
                                      \nD = 100 inch = 254 cm
                                      \n[\u2234 1 inch = 2.54 cm]
                                      \n\u03bb = 6000 \u00c5 = 6000 x 10-10<\/sup> m\/s
                                      \nd\u03b8 = \\(\\frac{1.22 \\times 6000 \\times 10^{-10}}{254 \\times 10^{-2}}\\) = 2.9 x 10-7<\/sup>
                                      \nd\u03b8 = 2.9 x 10-7<\/sup> rad.<\/p>\n

                                      Question 11.
                                      \nTwo polarising sheet have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?
                                      \nSolution:
                                      \nHere, I = \\(\\frac{I_{o}}{2}\\)
                                      \nUsing Malus law,,
                                      \nI = Io<\/sub> cos2<\/sup> \u03b8
                                      \n\\(\\frac{I_{o}}{2}\\) = Io<\/sub> cos2<\/sup> \u03b8
                                      \ncos \u03b8 \u00b1 \\(\\frac { 1 }{ \u221a2 }\\)
                                      \n\u03b8 = \u00b1 45\u00b0, \u00b1 135\u00b0<\/p>\n

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