PU–Question Bank Wave Optics

Multiple Choice Questions

1. Wavefront is the locus of all points, where the particles of the medium vibrate with the same _________ .
(a) phase (b) amplitude (c) frequency (d) period
Sol: Wavefront is the locus of all points, where the particles of the medium vibrate with the same phase.
Ans: (a)
2. Huygens concept of secondary wave
(a) allows us to find the focal length of a thick lens (b) is a geometrical method to find a wavefront
(c) is used to determine the velocity of light (d) is used to explain polarisation
Sol: Huygen’s principle gives us a geometrical method of tracing a wavefront.
Ans: (b)
3. Huygens wave theory allows us to know
(a) The wavelength of the wave (b) The velocity of the wave
(c) The amplitude of the wave (d) The propagation of wavefronts
Sol: Huygen’s theory explains propagation of wavefronts.
Ans: (d)
4. The fringe width in a Young’s double slit experiment can be increased if we decrease
(a) width of slits (b) separation of slits
(c) wavelength of light used (d) distance between slits and screen
D 1
Sol: Fringe width,  =  
d d
where d = distance between slits
Ans: (b)
5. Consider a sunlight incident on a slit of width 104 Å. The image seen through the slit shall
(a) be a fine sharp slit white in colour at the centre
(b) a bright slit white at the centre diffusing to zero intensities at the edges
(c) a bright slit white at the centre diffusing to regions of different colours
(d) only be a diffused slit white in colour
Sol: As given that the width of the slit = 104 Å = 10000Å = 104  10−10 m = 10−6 m = 1 μm
Wavelength of visible sunlight varies from 4000Å to 8000 Å.
Thus the width of slit is 10000Å comparable to that of wavelength visible light i.e., 8000Å. So diffraction occurs
with maxima at centre. Hence at the centre all colours appear i.e., mixing of colour form white patch at the centre.
Ans: (a)
6. In a Young’s double-slit experiment, the source is white light. One of the holes is covered by a red filter and another
by a blue filter. In this case,
(a) there shall be alternate interference patterns of red and blue
(b) there shall be an interference pattern of red distinct from that for blue
(c) there shall be no interference fringes
(d) there shall be an interference pattern for red mixing with one for blue
Sol: For sustained interference pattern to be formed on the screen, the sources must be coherent and emits lights of
same frequency and wavelength. In a Young’s double-slit experiment, when one of the holes is covered by a red
filter and another by a blue filter. In this case due to filtration only red and blue lights are present which has
different frequency. So, in that case there shall be no interference fringes.
Ans: (c)

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PU–Question Bank Wave Optics

7. The condition for obtaining secondary maxima in the diffraction pattern due to signal slit is
 n
(a) a sin = n (b) a sin  = ( 2n − 1) (c) a sin  = ( 2n − 1)  (d) a sin  =
2 2
n
Sol: The condition for obtaining secondary maxima in the diffraction pattern due to signal slit is a sin  =
2
Ans: (d)
8. According to Huygen’s construction which of the following wavefront does not exists?
(a) forward wavefront (b) backward wavefront (c) cylindrical wavefront (d) can not be predicted
Sol: According to Huygen’s construction backward wavefront does not exists.
Ans: (b)
9. In interference pattern, the energy is:
(a) created at the maximum (b) destroyed at the minimum
(c) conserved but redisturbed (d) None of the above
Sol: In interference pattern, the energy is conserved but redisturbed.
Ans: (c)
10. Two identical light source S1 and S 2 emit light of same wavelength  . These light rays will exhibit interference if
(a) their phase difference remain constant (b) their phases are distributed randomly
(c) their light intensities remain constant (d) their light intensities change randomly
Sol: For interference phase difference must be constant.
Ans: (a)
11. Which one of the following statements is correct?
(a) Monochromatic light is never coherent
(b) Monochromatic light is always coherent
(c) Two independent monochromatic sources are coherent
(d) Coherent light is always monochromatic
Ans: (d)
12. If yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by a monochromatic blue
light of the same intensity
(a) fringe width will decrease (b) fringe width will increase
(c) fringe width will remain unchanged (d) fringes will become less intense
D
Sol: As  = and b   y ,
d
 fringe width  will decrease.
Ans: (a)
13. If we observe the single slit Fraunhofer diffraction with wavelength  and slit width b , the width of the central
maxima is 2 . On decreasing slit width for the same  .
(a)  increases (b)  remains unchanged
(c)  decreases (d)  increases or decreases depending on the intensity of light
 2n + 1   
Sol: We know that for maxima b sin  = ( 2n + 1) or sin  =  
2 2 b
So on decreasing the slit width, ' b ', keeping  same, sin  and hence  increases.
Ans: (a)

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PU–Question Bank Wave Optics

14. In a single slit diffraction experiment, the width of the slit is made double its original width. Then the central
maxima of the diffraction pattern will become
(a) narrower and fainter (b) narrower and brighter
(c) broader and fainter (d) broder and brighter
Sol: The width of the central maximum is given by
2 D
=  If d → 2d , then  decreases.
d
 I increases as d increases
 The central maximum will become narrower and brighter.
Ans: (b)
15. When unpolarised light is incident on a plane glass plate at Brewster’s angle, then which of the following
statements is correct?
(a) Reflected and refracted rays are completely polarised with their planes of polarization parallel to each other
(b) Reflected and refracted rays are completely polarised with their planes of polarization perpendicular to
each other
(c) Reflected light is plane polarised but transmitted light is partially polarised
(d) Reflected light is partially polarised but refracted light is plane polarised
Sol: At Brewster’s angle, only the reflected light is plane polarised, but transmitted light is partially polarised.
Ans: (c)
16. In the propagation of light waves, the angle between the plane of vibration and plane of polarisation is
(a) 0 (b) 90 (c) 45 (d) 80
Sol: Angle between plane of vibration and plane of polarisation in 90 .
Ans: (b)
17. Polaroid glass is used in sun glasses because
(a) It reduces the light intensity to half on account of polarisation
(b) It is fashionable
(c) It has good colour
(d) It is cheaper.
Sol: Polaroid glass polarises light reducing the light intensity to half its original value.
Ans: (a)
18. Spherical wavefronts, emanating from a point source, strike a plane reflecting surface. What will happen to these
wave fronts, immediately after reflection?
(a) The will remain spherical with the same curvature, both in magnitude and sign.
(b) They will become plane wave fronts.
(c) They will remain spherical, with the same curvature, but sign of curvature reversed.
(d) They will remain spherical, but with different curvature, both in magnitude and sign.
Sol: They will remain spherical, with the same curvature, but sign of curvature reversed.
Ans: (c)
19. Two plane wavefronts of light, one incident on a thin convex lens and another on the refracting face of a thin prism.
After refraction at them, the emerging wavefronts respectively become
(a) Plane wavefront and plane wavefront (b) Plane wavefront and spherical wavefront
(c) Spherical wavefront and plane wavefront (d) Spherical wavefront and spherical wavefront

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PU–Question Bank Wave Optics

Sol: Emerging wavefront will be spherical from convex lens and plane wavefront from the prism.
Ans: (c)
20. In Young’s double slit experiment, the minimum amplitude is obtained when the phase difference of super-
imposing waves is (where n = 1,2,3,... )
(a) zero (b) ( 2n − 1)  (c) n (d) ( n + 1) 
Sol: For minima, phase difference = odd integral multiple of  = ( 2n − 1)  .
Ans: (b)

Fill in the Blanks

21. The locus of all particles in a medium, vibrating in the same phase is called wavefront.
22. Wavefront is locus of all points, where the particles of the medium vibrate with the same phase.
23. The idea of secondary wavelets was given by huygen.
24. Spherical is the shape of a wave front obtained from a point source
25. Plane type of wave front is obtained by a point source at a very large distance
26. According to wave theory, intensity of light is directly proportional to square of amplitude.
27. The physicist who experimentally studied the interference of light for the first time is Thomas Young.
28. For constructive interference (formation of bright fringes) the path difference of interfering waves must be an
Integral multiple of wavelength.
29. For destructive interference (formation of bright fringes) the path difference of interfering waves must be an
Odd integral multiple of  (wavelength ) /2 .
30. The condition for observing Fraunhoffer diffraction from a single slit is that the light wavefront incident on the slit
should be plane
31. The fringe width in a Young’s double slit experiment can be increased if we decrease the width of slits.
32. The phenomenon of polarization confirmed that light is a transverse wave.
33. A particular angle for which the reflected ray is fully polarised is called the Brewster’s angle .
34. Large aperture of objective lens in an astronomical telescope increases the resolving power of telescope.
35. The resolving power of an optical instrument decreases when the wavelength of light used is increased.

One Mark Questions

36. Who gave the corpuscular model for light before Newton? (P)
Answer: Descartes
37. What was the drawback of corpuscular theory of light? (D)
Answer: Speed of light is more in denser medium than in rarer medium which was incorrect.
38. Who proposed wave theory of light? (P)
Answer: Christian Huygens.
39. Who proved experimentally that the speed of light in denser medium is less than in rarer medium? (D)
Answer: Foucault
40. Which model of light predicted that the speed of light in denser medium is less than in rarer medium? (D)
Answer: Wave theory of light (Huygens model)
41. Who experimentally proved the wave nature of light? (D)
Answer: Thomas Young
42. Which experiment proved the wave theory of light? (P)
Answer: Interference experiment of young.
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PU–Question Bank Wave Optics

43. Who proposed electromagnetic theory of light? (P)

Answer: Maxwell
44. What is the nature of light waves according to Maxwell’s electromagnetic theory of light? (P)
Answer: Transverse nature
45. Define a wave front. (P)
Answer: A wave front is defined as the locus of all points in a medium which are vibrating in the same phase.
46. What is the shape of a wave front obtained from a point source? (P)
Answer: Spherical wave front.
47. Which type of wave front is obtained by a point source at a very large distance? (P)
Answer: Plane wave front.
48. To get a cylindrical wave front what should be the shape of the light source? (P)
Answer: A narrow slit
49. What is the shape of the wave front obtained from a distant star? (P)
Answer: A plane wave front
50. Which type of wave front is obtained when a plane wave is reflected by a concave mirror? (D)
Answer: Spherical wave front.
51. State Huygen’s principle. (D)
Answer: Huygen’s principle states that every point on primary wavefront acts as a source of secondary
disturbance.
52. Name the physicist who experimentally studied the interference of light for the first time. (P)
Answer: Thomas Young.
53. State the principle of superposition of waves. (P)
Answer: At a particular point in medium, the resultant displacement produced by an number of waves is the vector
sum of the displacements produced by each of the waves.
54. What is interference of light? (P)
Answer: Interference is a phenomenon in which there is a modification in the distribution of light energy (intensity)
due to superposition of light waves emitted from coherent sources.
55. What is constructive interference of light? (P)
Answer: It is a phenomenon in which there is a maximum intensity of light due to the superposition of coherent
waves of phase difference 2n .
56. What is destructive interference of light? (P)
Answer: It is phenomenon in which there is a minimum intensity of light due to the superposition of coherent
waves of phase difference ( 2n − 1)  .
57. Define fringe width of interference pattern in Young’s double slit experiment. (P)
Answer: Fringe width is defined as distance between two successive bright points or dark points in interference
pattern.
58. Write the expression for fringe width of interference pattern in Yong’s double slit experiment. (P)
D
Answer:  =
d
59. What is the effect on the interference fringes in Young’s double-slit experiment when the monochromatic
source is replaced by a source of white light? (F)
Answer: When monochromatic light in YDSE is replaced by white light, the central fringe is white. This is because
the path difference is zero for all wave lengths.

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PU–Question Bank Wave Optics

60. How does the fringe width of interference pattern vary with the wavelength of incident ray? (P)
D
Answer:  = , when  is decreased,  decreases.
d
61. How does the fringe width of interference pattern vary with the intensity of incident light? (D)
Answer: It does not depend on intensity of light.
62. Draw the graph of the intensity distribution of light in Young’s double-slit experiment. (D)
Answer:

63. Instead of using two slits as in Young’s experiment, if two separate but identical sodium lamps are used, what is
the result on interference pattern? (D)
Answer: There will be no interference pattern.
64. Does longitudinal wave exhibit the phenomenon of interference? (D)
Answer: Yes, they exhibit.
65. If Young’s double slit experiment is performed in water, what will be the effect on the fringe width? (D)
Answer: Fringe width decreases (    ;   1 /  )
66. In which direction the energy of the light wave travels with respect to the wave front? (D)
Answer: In a perpendicular direction to wave front.
67. Do the backward moving wave fronts exist in Huygens wave model for light? (D)
Answer: No.
68. Young’s double slit experiment is performed using red light. If red light is replaced by blue light, what is the
effect on the fringe width of interference pattern? (D)
Answer: Fringe width decreases (    )
69. What happens to the fringe width of interference pattern when the distance of separation between two slits is
doubled in Young’s experiment? (P)
Answer. When the slit separation ( d ) is increased by two times, fringe width becomes half.
70. The distance between the slits and the screen in Young’s double slit experiment is doubled. What happens to
the fringe width? (P)
D
Answer:  = , when D is decreased,  decreases
d
71. Let the fringe width in Young’s double slit experiment be  . What is the fringe width if the distance between
the slits and the screen is doubled and slit separation is halved? (D)
D
Answer:  = ; increases by four times.
d
72. What is the intensity of light due to constructive interference in Young's double slit experiment if the intensity
of light emerging from each slit is I 0 ? (D)
Answer: I Res = 4 I 0
73. In a single slit diffraction experiment if the width of the slit is doubled what happens to the width of the central
diffraction band? (D)
Answer: Becomes half

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74. Name a phenomenon which confirms the wave nature of light. (P)
Answer: Interference of light
75. What is diffraction of light? (P)
Answer: The phenomenon of bending of light around the corners (or edges) of very small obstacles or apertures
and its spreading into geometrical shadow region is called diffraction of light.
76. Which colour of light undergoes diffraction to the maximum extent? (D)
Answer: Blue and violet colour
77. How will the diffraction pattern due to a single slit change if violet light replaces green light? (D)
Answer: Diffraction bands width increases but diffraction amount decreases.
78. Do all types of waves exhibit diffraction or only light waves? (D)
Answer: All types of waves except waves on strings.
79. What happens to the resolving power of an optical instrument when the wavelength of light used is increased?
(D)
Answer: Resolving power decreases ( R.P  1 /  )
80. Define the resolving power of an optical instrument. (P)
Answer: Resolving power of an optical instrument is defined as an ability of an optical instrument to see clearly
two objects which are placed very closely.
81. Write the expression for limit of resolution of telescope. (P)
1.22
Answer: Limit of resolution = d =
a
82. How can resolving power of telescope be increased? (P)
Answer: The resolving power of a telescope can be increased by increasing the diameter of the objective.
83. Name a factor which affects the resolving power of a microscope. (D)
 2n sin  
Answer: Refractive index  R.P =
  
84. Mention the expression for limit of resolution of microscope. (P)

Answer: Limit of resolution a microscope dx =
2n sin 
85. Express Doppler shift in terms of wavelength of light used. (D)
−v 
Answer: Doppler shift  = radial
c
86. Write the formula for the Doppler shift in terms of frequency of light used. (D)
−v
Answer: Doppler shift f = radial f
c
87. Give one application of the study of Doppler effect in light. (D)
Answer: In astronomy it is used to measure the radial velocities of distant galaxies.
88. Which phenomenon confirms the transverse wave nature of light? (P)
Answer: Polarization of light
89. What is polarization of light? (P)
Answer: Polarisation of light is a phenomenon in which the vibrations of light occur only in one plane.
90. What is pass axis of the Polaroid? (P)
Answer: When un-polarised light is incident on a Polaroid, then the light wave will get linearly polarised with the
electric vector oscillating along a direction perpendicular to the aligned molecules. This direction is called pass axis
of the Polaroid.

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PU–Question Bank Wave Optics

91. By what percentage the intensity of light decreases when an ordinary un-polarised (like from sodium lamp)
light is passed through a Polaroid sheet? (D)
Answer: 50%
92. The intensity of incident light on a Polaroid P1 is I . What is the intensity of light crossing another Polaroid P2
when the pass-axis of P2 makes an angle 90° with the pass-axis of P1 ? (D)
Answer: Zero
93. What should be the angle between the pass axes of two Polaroids to get the maximum intensity of transmitted
light form the second Polaroid? (D)
Answer: Zero
94. State Brewster's Law. (P)
Answer: Brewster’s law states that at certain angle of incidence reflected light is completely polarised and  = tan i p
95. Define Brewster’s angle (OR polarising angle). (P)
Answer: A particular angle for which the reflected ray is fully polarised is called the angle of polarisation or
Brewster’s angle.
96. Write the relation between refractive index of a medium (reflector) and polarising angle. (P)
Answer:  = tan i p
97. The intensity of light incident on a polariser is I and that of the light emerging from it is also I . Is the incident
light polarised or un-polarised? (D)
Answer: Polarised light.
98. A ray of light is incident at polarising angle on a glass plate. What is the angle between the reflected ray and
refracted ray? (D)
Answer: 90°
99. What is partially polarised light? (P)
Answer: When polarised light is viewed through a rotating analyser, one sees a maximum and minimum of
intensity, but not complete darkness this kind of light is called partially polarised.
100. When can we have total transmission of light through a prism? (D)
Answer: If A  2C ( A → angle of prism, C → critical angle)

101. Does the polarising angle depend on the refractive index of that medium? (D)
Answer: Yes it depends.

Two Marks Questions

102. Name the two theories of light in support of its wave nature. (P)
Answer: Huygen’s wave theory and Maxwell’s electromagnetic wave theory. (2 marks)
103. What are coherent sources? Give an example. (P)
Answer: Two sources are said to be coherent when the waves emanated from them have the same frequency and a
constant phase difference. (1 mark)
Ex: Young’s double slit (1 mark)
104. Can two sodium vapour lamps be considered as coherent sources? Justify your answer. (P)
Answer: Each source of light consists of a large number of atoms and light is emitted by these atoms independently
and randomly. There will be no constant phase relation between the two sources. The phase difference between the
two sources is rapidly changing with time. Thus two independent sources cannot be considered as coherent
sources. (2 marks)

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105. Which of the two, red and yellow produces wider interference fringes? Why? (D)
Answer: Red produces wider interference fringes than yellow because fringe width is directly proportional to
wavelength. (2 marks)
106. Name any two factors affecting fringe width of interference pattern in Young's double slit experiment. (P)
Answer: Wavelength, Distance of screen from plane slits and slit separation. (2 marks)
107. Is it possible to conclude that light is either transverse or longitudinal wave from interference phenomenon?
Justify your answer. (D)
Answer: No because longitudinal waves also give interference. (2 marks)
108. Write the conditions for constructive and destructive interference in terms of path difference of interfering
waves. (P)
Answer: Condition for constructive interference (formation of bright fringes): Path difference must be an integral
multiple of  . (1 mark)

Condition or ( x = n ) for destructive interference (formation of dark fringes): Path difference must be

 
an odd integral multiple of  /2 or  x = ( 2n − 1)  (1 mark)
 2

109. Write the conditions for constructive and destructive interference in terms of phase difference of interfering
waves. (P)
Answer: Conditions for constructive interference (Formation of bright fringes): Phase difference must be an even
multiple of  . (1 mark)
Conditions for destructive interference (Formation of dark fringes): Phase difference must be an odd multiple of  .
(1 mark)
110. We do not encounter diffraction effects of light in everyday observations. Explain why? (D)
Answer: We do not observe distraction effects in everyday life because wavelength of light is much smaller than
the size of interacting objects. (2 marks)
111. Why diffraction effects due to sound waves are more noticeable than due to light waves? (P)
Answer: Diffraction effects are observable only when the size of the obstacle is comparable to the wavelength of the
waves. In the case of sound waves, the wavelength of sound waves is comparable to the size of the obstacle or
aperture. So diffraction effects are easily observable. But in the case of light waves, the wavelength of light is
extremely small compared to the size of the obstacle or object. So diffraction effects in the case of light waves are
not easily observed. (1+1 mark)
112. Explain how the principle of conservation of energy is consistent with interference and diffraction phenomena
of light. (D)
Answer: In interference and diffraction, light energy is re distributed, if it reduces in one region producing dark
fringe, it increases in another region producing a bright fringe, there is no gain or loss of energy which is consistent
with the principle of conservation of energy. (2 marks)
113. Mention the conditions for diffraction minima and maxima in diffraction due to single slit. (F)
Answer: Condition for diffraction maxima:

 1 
The angular position of bright fringe is  =  n + 
 2 d (1 mark)
Where,  − wavelength of light used, d − width of the slit and n = 1, 2,3 . . . . .

(1 mark)
Condition for diffraction minima:
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n
The angular position of dark fringe is  =
d (1 mark)

Where,  − wavelength of light used, d − width of the slit and n = 1, 2,3 . . . . .

114. Represent graphically the variation of intensity of light due to diffraction at single slit. (P)
Answer:

(2 marks)
115. Give any two methods of increasing the resolving power of a microscope. (F)
Answer:
(i) By increasing the R.I. of the medium (1 mark)
(ii) By decreasing the wavelength of light used (1 mark)
116. What is Fresnel distance? Write its expression. (F)
Answer: It is defined as the distance from an obstacle or aperture causing diffraction at which the diffraction spread
of the beam becomes comparable with the size of the aperture.
a2
Expression for Fresnel distance = Z f =

Where a = size of the aperture,  = wave length of light. (2 marks)
117. What is red shift? What is its significance? (D)
Answer: Increase in wave length due to Doppler effect is red shift. From red shift, it can be said that the distant
astronomical object is moving away from you. (2 marks)
118. What is blue shift? When does it occur? (D)
Answer: Decrease in wavelength due to Doppler effect is blue shift, it happens when distant source moves towards
observer. (2 marks)
119. Write the mathematical expression for Malus law and explain the terms. (P)
Answer: I = I 0 cos 
2

I 0 = Intensity from polariser,  = angle between axis of polariser analyser. (2 marks)

120. Diagrammatically represent polarised light and unpolarised light. (P)

Answer:

(2 marks)

121. Mention any two methods of producing plane polarised light. (D)
Answer: Reflection of light, scattering of light. (2 marks)

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122. Write any two uses of Polaroids? (P)

Answer:
(a) Polaroid films are used in photography. (1 mark)
(b) Polaroid glasses are worn by pilots, race-car drivers…etc for glare – free vision. (1 mark)
123. Find the value of Brewster angle for air to glass transmission of light if the refractive index of glass is 1.5 .
(D)
Answer:  = tan i p
1.5 = tan i p
i p = tan −1 (1.5 )
 i p = 56° 18' (2 marks)
124. The polarising angle for a medium is 520. Find its critical angle? (P)
Answer:  = tan i p ,  = tan 52
1
= tan 52
sin C
 1 
C = sin −1    C = sin −1 ( 0.769 ) = 51.4 (2 marks)
 1.2799 
125. Brewster's angle for a certain medium is 52 . Find the refractive index of the medium. (P)
Answer:  = tan i p (1 mark)

 = tan 52   = 1.23 (1 mark)

126. The refractive index of certain glass is 1.5 for light whose wavelength in vacuum is 600 nm . Find the
wavelength of this light in glass?
(D)
c  
Answer:  =  air = m
v m air
1 
= m
 600
m = 400nm (2 marks)
127. Why interference pattern cannot be seen when pin hole of young's double slit experiment is illuminated by two
identical but separate sodium sources? Explain (D)
Answer: Two identical separate sodium sources cannot be coherent sources. Because due to internal atomic
phenomena the phase difference between two waves of identical sources cannot be same. (2 marks)
128. How to get two coherent sources of light? Why they are said to be coherent? (P)
Answer: When two slits are illuminated by monochromatic source of light we get two coherent sources. Two
sources are said to be coherent in nature if they emit light of same frequency and of a stable path difference.
(2 marks)
129. A monochromatic light of wavelength 700 nm is incident on a 3.5 mm wide aperture. Find the distance up to
which the ray of light can travel so that its spread is less than the size of the aperture (D)
Answer: Given  = 700nm   = 7  10−7 m , d = 3.5  10−3 m
d2
Distance upto which light travels without much spread is Fresnel distance =


=
(3.5 10 ) −3 2
=
12.25  10−6
= 17.5m (2 marks)
7  10−7 7  10−7
130. What are Polaroids? Mention any two uses of Polaroids. (P)
Answer: Polaroids are thin glass sheets or plastic sheets containing long chain molecules which are used to produce
polarised light. (1 mark)
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(a) Polaroid films are used in photography.

(b) Polaroid glasses are worn by pilots, race-car drivers…etc., for glare – free vision. (1 mark)
131. In Young's double slit experiment, fringes of certain width are produced on the screen kept at a certain distance
from the slits. When the screen is moved away from the slits by 0.1m , fringe width increases by 6  10−5 m . The
separation between the slits is 1mm . Calculate the wavelength of the light used. (D)
D D '
Answer: Given d = 10−3 m ,  = , '= and D ' = D + 0.1
d d
 '−  = 6  10−5
 ( D + 0.1)  D
− = 6  10−5
d d
 D 0.1  D −5
+ − = 6  10
d d d
0.1
= 6  10−5
10−3
 = 6  10−7 m = 600nm (2 marks)

Three Marks Questions

132. Who solved the major drawback of Huygens's wave theory? Explain how it was solved. (D)
Answer: Michelson and Morley solved the drawback of Huygens’s wave theory that is existence of ether medium.
It was solved by finding no significant difference in speed of light in direction of movement through pre assumed
ether medium and at right angles from Michelson and Morley experiment. (3 marks)
133. Using Huygens’ wave theory of light, show that the angle of incidence is equal to angle of reflection in case of
reflection of a plane wave by a plane surface. (P)
Answer: Consider a plane wave front AB striking a plane reflecting surface MN at an angle of incidence i . Let
v be the speed of light in the medium. In a time t , the wave travels from B to C a distance BC = vt . In order to

construct the reflected wave front we draw a sphere of radius vt from the point A as shown in the figure.
Let CE represent the tangent plane drawn from the point C to this sphere.
AE = BC = vt (1 mark)
Compare triangles BAC and EAC
ABC = AEC = 90 Side AC is common

N Incident
wavefront

E
B
Reflected
i wavefront

A i r C
M N
(Fig1 mark)
So triangles are congruent
 BAC = ECA  i = r

Also, NAE = 90 − EAC

= 90 − ( 90 − r ) = r

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Hence the laws of reflection are proved (1 mark)

134. Using Huygens’ principle show that the frequency of light wave remains the same when the light travels from
one optical medium to another. (P)
Answer: Consider a plane surface XY that separates a denser medium of Q
refractive index n from a rarer medium. If v1 is the velocity of light in the
rarer medium and v2 is the velocity of light in the denser medium, then by P B
definition, i
v X i A'
n= 1 … (1) A r Y
v2
Consider two incident rays PA and QA which are incident on this surface. Draw B'

AB perpendicular to QA . AB represents the incident wave front. According Q'

P'
to Huygens’ principle, every point on the primary wavefront AB acts as a
secondary source of disturbance. (fig 1 mark)
Let the secondary wavelet strike the surface XY at A ' . The secondary wavefront from B travels a distance v1t in
time t seconds.
 BA = v1 t … (2) (1 mark)
The secondary wavelet from A travels a distance v2t in the same interval of time. So with A as centre draw an arc
of radius v2t . From A’ draw a tangent plane which touches this spherical arc at B ' . Then A' B ' will be
perpendicular to AB ' . So A’B’ is the refracted wavefront. So the ray AP ' perpendicular to this refracted wavefront
AP represents the refracted ray at A and A ' Q ' represents the refracted ray at A ' .
BA
In ABA ', BAA = i , sin i = … (1)
AA
AB
In AA ' B ' , BAA = r , sin r = … (2)
AA
Dividing (1) by (2) we get,
 BA 
sin i  AA  BA v1t v1
= = = =
sin r  AB  AB v2t v2
 
 AA 
sin i v1
n= = (1 mark)
sin r v2
Thus the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair
of media for a given colour of light.
This is the Snell’s law of refraction. Further, if 1 and 2 denote the wavelengths of light in medium 1 and
medium 2 , respectively and if the distance BA ' is equal to 1 then the distance AB ' will be equal to 2 (because if
the crest from B has reached A ' in time t , then the crest from A should have also reached B ' in time t ); thus,
1 BA '
= … (4)
2 AB '
From (3) and (4),
v1 1 v v
=  1 = 2 (1 mark)
v2 2 1 2
The above equation implies that when a wave gets refracted into a denser medium ( v1  v2 ) , the wavelength and
the speed of propagation decrease but the frequency ( v /  ) remains the same.

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135. Draw diagram representing refraction of a plane wave incident on a rarer medium from a denser medium, and
explain critical angle and total internal reflection. (P)
Answer:

(fig 1 mark)

Consider refraction of a plane wave at a rarer medium, i.e., v2  v1 . Proceeding in an exactly similar manner we can
construct a refracted wavefront as shown in figure. The angle of refraction will now be greater than angle of
incidence; however, we will still have n1 sin i = n2 sin r . We define an angle ic by the following equation.
n
sin ic = 2
n1
Thus, if i = ic , then sin r = 1 and r = 90 . Obviously, for i  ic , there cannot be any refracted wave. The angle ic is
known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any
refracted wave and the wave will undergo what is known as total internal reflection. (2 marks)
136. Illustrate with the help of suitable diagram, the refraction of a plane wave by (i) a thin prism (ii) a convex lens
and reflection by a concave mirror. (D)
Answer:

fig ( a ) fig ( b ) fig ( c )

(i) Action of the prism when a plane wavefront incident on it: In adjacent fig (a), consider a plane wave passing
through a thin prism. Since the speed of light waves is less in glass, the lower portion of the incoming
wavefront which travels through the greatest thickness of glass will get delayed resulting in a tilt in the
emerging plane wavefront.
(ii) Action of the convex lens when a plane wavefront incident on it: In the adjacent fig (b), a plane wave incident
on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and
is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront
becomes spherical (radius = f , focal length) and converges to the point focus F .
(iii) Action of the concave mirror when a plane wavefront incident on it: In adjacent fig (c), a plane wave is incident
on a concave mirror and on reflection we have a spherical wave converging to the focus F . (3 marks)

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137. Briefly describe Young's double slit experiment with the help of a schematic diagram. (P)

(fig. 1 mark)

Answer: Young’s double slit apparatus consist of a narrow rectangular slit S . It is illuminated by a monochromatic
light. The light emerging from S illuminates the two identical close slits S1 and S 2 which acts as coherent sources.
The waves emerging from S1 and S 2 get superposed and interference fringes are produced as shown on the
observation screen. Bright fringes are formed on the screen due to the superposition of crests of two waves or
trough of two waves. Dark fringes are formed due to the superposition of crest and trough of two waves. At the
centre of the screen always bright fringe is formed. This is because, all the waves meeting at the centre of the screen
have the zero path difference and zero phase difference. (2 marks)
138. What is Doppler's effect in light? Write the formula for the Doppler shift. (D)
Answer: It is the apparent change in the frequency (or wavelength) of light wave due to the relative motion
between the source of light and the observer.
f v
=  radial
f c

Where, f is the Doppler shift, f is the true frequency, c is speed of light and vradial is the component of the
source velocity along the line joining the observer and source. (2 marks)
139. Give the theory of interference of light. (D)
Answer: If the displacement produced by source S1 is given by y1 = a cos (t ) , then the displacement produced by
S 2 would be y2 = a cos (t +  ) where  is the phase difference between the two waves.
The resultant displacement: y = y1 + y2 =  a cos (t ) + a cos (t +  ) 
   
After solving, we will get y = 2a cos   cos  t + 
2  2
 
The amplitude of the resultant displacement is 2a cos  
2
Conditions for constructive interference: The phase difference:
 = 0, 2 ,  4 ....
 = 2n
[OR path difference:  = n , (where n = 0,1, 2,3.. )] constructive interference takes place leading to maximum
intensity = 4I 0 and resultant amplitude = 2a .
Conditions for destructive Interference:
 =  ,  3 , 5 ....
 = ( 2n + 1) 
[OR path difference: (where n = 0,1, 2,3.... )], destructive interference takes place leading to zero amplitude and zero
intensity. (3 marks)
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140. Arrive at the condition for constructive and destructive interference in terms of phase difference between the
two waves. (P)
Answer: Conditions for constructive Interference: If the two coherent sources S1 and S 2 vibrating in phase, then at
an arbitrary point P ,
The phase difference:  = 0,  2 ,  4 ....
 = 2n
[OR path difference:  = n , (where n = 0,1, 2,3.. )] constructive interference takes place leading to maximum
intensity = 4I 0 and resultant amplitude = 2a .
Conditions for destructive Interference: If the point P is such that the phase difference:
 =   ,  3 ,  5 ....
 = ( 2n + 1) 
[OR path difference: (where n = 0,1, 2,3... )] , destructive interference takes place, leading to zero amplitude and
zero intensity. (3 marks)
141. What is the effect on (i) the angular fringe width (ii) the linear fringe width in Young's double-slit experiment
due to each of the following operations:
(a) The screen is moved away from the plane of the slits.
(b) One monochromatic source is replaced by another monochromatic source of shorter wavelength and
(c) The separation between the two slits is increased? (D)
Answer:
(a) Angular fringe width remains same but linear fringe width increases. (1 mark)
(b) Both angular fringe width and linear fringe width decreases. (1 mark)
(c) Both angular fringe width and linear fringe width decreases. (1 mark)
142. Compare the interference pattern of light obtained by Young’s double slit experiment with diffraction pattern
due to single slit. (D)
Answer: (Each carry 1 mark)
Intensity distribution curve for interference fringe pattern

Intensity

−2 −
3 −  O   3 2 Path difference
−
2 2 2 2

−2 − O  2 Phase difference

Intensity distribution curve for diffraction fringe system
Intensity
For secondary maxima:
Central maximum
I secondary maximum
Second secondary maximum
Third secondary maximum

O  3 2 5 3
2 2
III minimum
I minimum II minimum (3 marks)
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143. Obtain the expression for limit of resolution of microscope. (D)

D
Answer: It can be seen from fig  2 tan  where 2 is the angle subtended by diameter of objective lens at the
f
focus of the microscope.
When the separation between two points in a microscopic specimen is comparable to the wavelength  of the
light, the diffraction effects become important. The image of a point object will again be a diffraction pattern whose
size in the image plane will be
 1.22 
v = v  
 D 
Two objects whose images are closer than this distance will not be resolved, they will be seen as one. The
corresponding minimum separation, d min , in the object plane is given by
  1.22  
v  D  
 
d min = 
m
1.22 v 1.22 f 
=  =
D m D
Now, from above equations
1.22 1.22
d min = =
2 tan  2sin 
If the medium between the object and the objective lens is not air but a medium of refractive index n , gets modified
1.22
to d min =
2n sin 
144. Briefly explain Polarization by reflection with the help of a diagram. (P)

Answer: Polarization by reflection: It is found that when a beam of ordinary light is reflected by the surface of a
transparent medium like glass or water, the reflected light is partially polarized. The degree of polarization
depends on the angle of incidence. As the angle of incidence is gradually increased from a small value, the degree
of polarization also increases. At a particular angle of incidence the reflected light is completely plane polarized.
This angle of incidence is called Brewster’s angle or polarizing angle. If the angle of incidence is further increased,
the degree of polarization decreases. (3 marks)

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145. With the help of a diagram explain how polarised sun light is produced by scattering. (D)

Answer: Dots are the vibrations of electric field vector perpendicular to plane of paper and Double arrows are the
vibrations of electric field vector parallel to plane of paper. Sun light is un-polarised light. When sun light falls on
the air molecule, the electrons in the molecule begin to vibrate in these planes. The electrons vibrating parallel to
the double arrows cannot send energy towards the observer looking perpendicular to the direction of incident
light.
This is because light is a transverse wave. But the electrons vibrating parallel to the dots can send energy towards
the observer. It is plane polarized light which contains only dots. (3 marks)
146. Show that the refractive index of a reflector is equal to tangent of the polarising angle.
(OR) Show that n = tan iB or Arrive at Brewster’s law. (P)
Answer: Consider a ray of light which is made to fall on a slab of a transparent material. A part of this ray is
reflected. Another part is refracted. For a particular angle of incidence, it is found that the reflected ray is
perpendicular to the refracted ray.
In this case it is found that the reflected ray is fully polarised. This particular angle for which the reflected ray
is fully polarised is called the angle of polarisation. ( P ) .The refracted ray is partially polarised.

MOQ + QOR + RON =180°

  P + 90° + rP =180°

  P + rP = 90° … (1) (1 mark)

 rP = 90° −  P
M
P Q

P P
A 90
B
O
rP
N (Fig.1 mark)
R
D C

Let n be the refractive index of the material of the medium.

sin i
n=
sin r

sin  p sin  p
n= = = tan  p
sin ( 90 − P ) cos  p

n = tan  p

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is called Brewster’s law of polarisation.

“The refractive index of a medium is equal to the tangent of the angle of polarisation”.
n = tan  p ,  p is called Brewster’s angle. (1 mark)

147. State and explain law of Malus in polarisation. (P)

Answer: “When a beam of polarised light is incident on an analyser, the intensity of light transmitted from the
analyser varies directly as the square of the cosine of the angle between the planes of polariser and analyser.”
Plane of
I  cos 2  polariser (1 mark)
I = I o cos 
2 Plane of
analyser
Ao sin  Ao


Ao cos  (1 mark)
where I o = Intensity of plane polarised light from the polariser and Ao its corresponding amplitude. I is the
intensity of the transmitted beam. (1 mark)
148. Assume that light of wavelength 5000Å is coming from a star. What is the limit of resolution and resolving
power of a telescope whose objective has a diameter of 200 inch? (D)
−7
Answer: Given  = 5000 Å ,  = 5  10 m
Diameter d = 200inch = 200  2.54cm
d = 508  10−2 m
1.22 1.22  5  10−7
Limit of resolution = = = 0.012  10−5 rad
d 508  10−2
 = 1.2  10−7 rad
1 1
Resolving power = =
 1.27  10−7
R.P = 0.82  107 rad −1 (3 marks)
149. If Young's double slit experiment, distances between 2nd and 10 th bright fringes for a light of wavelength
486nm is same as that of the distance between 3rd and 9 th bright fringes for the light of different wavelength
is used. Find the wavelength of light. (D)
Answer: Given 1 = 486nm
( y10 − y2 ) = ( y9 − y3 )
101D 21D 92 D 32 D
− = −  81 = 62
d d d d
4
2 =  486
3
2 = 648nm (3 marks)
150. If Young's double slit experiment with monochromatic light and slit separation of 1.2mm , the fringes is
obtained on a screen placed at some distance from the slits. If the screen is moved by 5 cm towards the slits, the
change in fringe width is 20μm . Calculate the wavelength of the light used. (D)
Answer: Given d =1.2mm
D '
d = 1.2  10−3 m  ' =
d
Where D ' = D − 5  10−2
 −  ' = 20  10−6

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D D '   5  10−2
− = 2  10−5  = 2  10−5
d d d
2d
=  10−3 = 480nm (3 marks)
5

Five Marks Questions

151. Using Huygens’ wave theory of light, derive Snell’s law of refraction. (P)
Answer: Consider a plane surface XY that separates a denser medium of refractive index n from a rarer medium.
If v1 is the velocity of light in the rarer medium and v2 is the velocity of light in the denser medium, then by
definition,
Q

P
B
i
i A'
X A r Y

B'
Q'
P' (fig 1 mark)

v
n= 1 … (1) (1 mark)
v2
Consider two incident rays PA and QA which are incident on this surface. Draw AB perpendicular to QA . AB
represents the incident wave front. According to Huygens’ principle, every point on the primary wavefront AB
acts as a secondary source of disturbance. Let the secondary wavelet strike the surface XY at A ' . The secondary
wavefront from B travels a distance v1t in time t seconds.
 BA = v1 t … (2) (1 mark)
The secondary wavelet from A travels a distance v2t in the same interval of time. So with A as centre draw an arc
of radius v2t . From A’ draw a tangent plane which touches this spherical arc at B ' . Then A' B ' will be
perpendicular to AB ' . So A’B’ is the refracted wavefront. So the ray AP ' perpendicular to this refracted wavefront
AP represents the refracted ray at A and A ' Q ' represents the refracted ray at A ' .
In ABA ', BAA = i
BA
sin i = … (1)
AA
In AA ' B ' , BAA = r
AB
sin r = … (2)
AA
Dividing (1) by (2), we get
 BA 
sin i  AA  BA v1t v1
= = = = (1 mark)
sin r  AB  AB v2t v2
 
 AA 
sin i v1
n= = (1 mark)
sin r v2
Thus the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair
of media for a given colour of light.

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152. Obtain the expressions for resultant displacement and amplitude when two light waves having same amplitude
and a phase difference  superpose. Hence give the conditions for constructive and destructive interference in
terms of path difference/phase difference. (OR)
Give the theory of interference. Hence arrive at the conditions for constructive and destructive interferences in
terms of path difference/phase difference. (P)
Answer: Consider two waves having amplitude a1 and a2 , coming out from a source of light of wavelength  .
They have a constant phase difference  between them. They are represented by
y1 = a1 sin t … (1) and
y2 = a2 sin (t +  ) ... (2) respectively
y1 and y2 are the particle displacements produced by the respective waves at any instant of time t.
These two waves are made to overlap. The resultant displacement produced by the two waves on superposition is
given by
y = y1 + y2 (by the principle of superposition of waves) (1 mark)
y = a1 sin t + a2 sin (t +  )
= a1 sin t + a2 sin t cos  + cos t sin   ... (3)
= ( a1 + a2 cos  ) sin  t + ( a2 sin  ) cos  t
Put ( a1 + a2 cos  ) = A cos ... (4)
a2 sin  = A sin  ... (5)
Substituting these in (3), we get y = sin t  A cos  + cos t  A sin  
y = sin t  A cos  + cos t  A sin  
y = A sin (t +  ) … (6) (1 mark)
This is the equation of the combined wave formed by the superposition of the waves represented by (1) and (2).
This equation is similar to the equation of the component waves. Hence A represents the resultant amplitude of
the combined wave. We know that a wave carries energy. The intensity of a wave is the energy transmitted by a
wave per unit area per unit time.
Let I1 and I 2 represent the intensity of the first and second waves respectively. Let I represent the intensity of the
combined wave. The intensity of a wave is directly proportional to the square of the amplitude.
Then I1  a12 , I 2  a22 and I  A2 .
Square and add (4) and (5)
( A cos )2 + ( A sin  )2 = ( a1 + a2 cos )2 + ( a2 sin  )2
A2 cos 2  + A2 sin 2  = a12 + 2a1a2 cos  + a22 cos2  + a22 sin 2 

( ) (
A2 cos 2  + sin 2  = a12 + 2a1a2 cos  + a22 cos 2  + sin 2  )
A2 = a12 + 2a1a2 cos  + a22 ... (7)
A = a12 + 2a1a2 cos  + a22
2
... (8)
Hence the resultant intensity of the combined wave is directly proportional to cos  . (1 mark)
I is maximum when cos  = +1 .
Case (1)
cos  becomes+1 when  = 0, 2 , 4 ,6 .........2n = even multiple of  where n is an integer. n = 0,1, 2,3.... ...

( )
2
I max = I1 + 2 I1 I 2 + I 2 = I1 + I 2

= ( a1 + a2 )
2
A2 = a12 + 2a1a2 + a22  Amax
2

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In this case the two waves are said to undergo constructive interference. Since the resultant intensity is maximum,
bright fringes are produced. The crest of one wave falls on the crest of the other. The amplitudes get added up.
We know that a phase difference of 2 corresponds to a path difference of  .
 2n = n ( 2 )  n where n = 0,1, 2,3...
Hence the condition for constructive interference is that the path difference between the two waves must be n  .
Case (2)
I is minimum when cos  = −1 . (1 mark)
cos  becomes -1 when  =  , 3 , 5 ....... ( 2 n − 1)  = odd multiple of  where n is an integer. n = 1, 2,3....

( )
2
I max = I1 − 2 I1 I 2 + I 2 = I1 − I 2
2
A2 = a12 − 2a1a2 + a22  Amin = ( a1 − a2 )
2

In this case the two waves are said to undergo destructive interference. Since the resultant intensity is minimum,
dark fringes are produced. The crest of one wave falls on the trough of the other.

We know that a phase difference of  corresponds to a path difference of .
2
Hence the condition for destructive interference is that the path difference between the two waves must be

( 2n − 1) where n = 1, 2,3... (1 mark)
2

( )
2
I min = I1 − 2 I1 I 2 + I 2 = I1 − I 2 (5)  ( 4)

( 1 2)
2 = a −a 2
A2 = a12 − 2a1a2 + a22  Amin
a2 sin 
tan  =
a1 + a2 cos 

( )
2
I (a + a )
Note: max = 1 2 =
2 I1 + I 2
I min ( a − a )2
( I2 )
2
1 2 I1 −

When the two light waves have the same amplitude, then a1 = a2 = a

In this case, A2 = a12 + a22 + 2a1a2 cos  = a 2 + a 2 + 2a 2 cos  = 2a 2 + 2a 2 cos  = 2a 2 (1 + cos  )

( )    
= 2a 2  2cos 2  = 4a 2 cos 2  
 2 2
 
 I = 4 I 0 cos2  
2
Now I o is the maximum intensity at centre of the fringe systems. I is the intensity at any point where the phase
difference between the two superposing waves is  .
153. Obtain the expressions for resultant displacement when two light waves having same amplitude and a
phase difference  superpose and write the expression for intensity at that point.
Answer: For an arbitrary point G let the phase difference between the two displacements be  . Thus, if the

displacement produced by S1 is given by

y1 = a cos t (1 mark)

Then, the displacement produced by S 2 would be

y2 = a cos (t +  ) (1 mark)

And the resultant displacement will be given by

y = y1 + y2 (1 mark)

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PU–Question Bank Wave Optics

= a cos t + cos (t +  ) 

= 2a cos ( / 2 ) cos (t +  / 2 ) (1 mark)

  A+ B   A − B 
 cos A + cos B = 2cos  2  cos  2  
    

The amplitude of the resultant displacement is 2a cos ( / 2 ) and therefore the intensity at that point will be

I = 4 I 0 cos 2 ( / 2 ) (1 mark)

154. Derive the expression for the width of interference fringes in Young’s double slit experiment. (P)
Answer: S is a monochromatic source of light of wavelength  . S1 and S 2 are two slits separated by a distance d.
Two sets of wave fronts are produced when light passes through the two slits. The paths of these waves are
S1P and S2 P respectively. The path difference between them is  = S2 P − S1P . (1 mark)
2 2 2
S1P = S1Q + PQ
2 P
 d
S2 P 2 = D 2 +  xn −  ... (1) S1 Q
 2 d xn
2 2 2
S2 P = S2 R + PR 2
O (fig. 1 mark)
2 d
 d S
S2 P 2 = D 2 +  xn +  … (2) 2
 2 S2 R
( 2 ) − (1) gives
D Screen
S2 P 2 − S1P 2
  d  
2
 d 
2
=  D2 +  xn +   −  D 2 +  xn −  
  2    2 
   ( n + 1)th bright fringe
( S2P2 − S1P2 ) = 4xn  d2   
( S2 P + S1P )( S2 P − S1P ) = 4 xn 
d
 n th bright fringe
2
xn xn+1
S2 P  S1P = D
d
 ( D + D )  = 4 xn
2
d D
2 D = 4 xn ; xn =
2 d
This is an expression for the distance of the nth fringe from the centre of the fringe system.
For the nth bright fringe to be formed at P, the path difference  must be equal to n  . (1 mark)
D n
xn = (1 mark)
d
Similarly, the (n+1)th bright fringe will be formed at distance xn +1 from the centre of the fringe system. Fringe
width (  ) is defined as the distance between two consecutive bright fringes.
D ( n + 1) 
xn +1 =
d
D ( n + 1)  D n  D
 = xn +1 − xn = − =
d d d
D
= (1 mark)
d

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155. Explain the phenomenon of diffraction of light due to a single slit and mention the conditions for diffraction
minima and maxima. (D)

Answer: DIFFRACTION OF LIGHT AT SINGLE SLIT:

• When single narrow slit illuminated by a monochromatic light source, a broad pattern with a central bright
region is seen. On both sides, there are alternate dark and bright regions; the intensity becomes weaker away
from the centre.
• A parallel beam of light falling normally on a single slit LN of width a. The diffracted light goes on to meet a
screen. The midpoint of the slit is M. A straight line through M perpendicular to the slit plane meets the screen
at C.
• The straight lines joining P to the different points L, M, N, etc., can be treated as parallel, making an angle θ
with the normal MC. This is to divide the slit into smaller parts, and add their contributions at P with the
proper phase differences.
• Different parts of the wavefront at the slit are treated as secondary sources. Because the incoming wavefront is
parallel to the plane of the slit, these sources are in phase.
• The path difference between the two edges of the slit N and P is NP − LP = NQ = a sin   a
• At the central point C on the screen, the angle θ is zero. The path difference is zero and hence all the parts of
the slit contribute in phase. This gives maximum intensity at C, the central maximum.
• Secondary maxima is formed at
 1
   n +  where n = 1,  2,  3,....
 2 a 
• Minima (zero intensity) is formed
 n 
   where n = 1,  2,  3,.... (5 marks)
 a 

Numerical Problems
156. Light is incident on a glass plate at an angle of 60 . The reflected and refracted rays are mutually perpendicular
to each other. Calculate the refractive index of the material of the plate? (P)
Answer: Given i = 60°
r + r ' = 90°
r → angle of refraction
r ' → angle of reflection
r + i = 90° (1 mark)
sin i
From  = (1 mark)
sin r
sin 60
= (1 mark)
sin ( 90 − i )

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PU–Question Bank Wave Optics

sin 60
= (1 mark)
cosi
sin 60
=
cos 60
= 3 (1 mark)
157. In a Young’s double-slit experiment, the slits are separated by 2.8 mm and the screen is placed 1.4 m away. The
distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 mm. Determine the
wavelength of light used in the experiment. Also find the distance of fifth dark fringe from the central bright
fringe. (D)
Answer: Given d = 2.8mm
d = 2.8  10−3 m
D = 1.4 m
Y4 = 1.2mm for fourth bright fringe
n D
We know yn = (1 mark)
d
4 D
y4 =
d
4  1.4
1.2  10−3 = (1 mark)
2.8  10−3
 = 6000Å (1 mark)
Distance of fifth dark fringe is

y5 =
( 2n − 1)  D = 2 ( 5) − 1 6 10−7 1.4
2d 2  2.8  10−3
y5 = 1.35mm
158. A beam of un-polarised is incident on an arrangement of two Polaroids successively. If the angle between the
pass axes of the two Polaroids is 60° , then what percentage of light intensity emerges out of the second
Polaroid sheet?
(D)
Answer: Given  = 60°
I
I = 0 cos2  (1 mark)
2
I
I = 0 cos 2 60° (1 mark)
2
I0 I 1
I=  = (1 mark)
8 I0 8
= 0.125 or 12.5% (2 marks)
159. An optical instrument resolves two points at a distance from it using light of wavelengths 450nm and 600nm,
find the ratio of their respective resolving powers. (D)
Answer: Given 1 = 450nm and 2 = 600nm
1
Resolving power  (1 mark)

R1 2
= (1 mark)
R2 1
R1 600
= (1 mark)
R2 450
R1 60
= (1 mark)
R2 45

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PU–Question Bank Wave Optics

R 4
 1= (1 mark)
R2 3
160. In Young’s double slit experiment the two slits are 0.3 mm apart and are illuminated by a light of wavelength
650nm . Calculate the distance of (i) the 3rd dark and (ii) 5th bright fringes from the midpoint in the
interference pattern obtained on a screen 1.2 m away from the slits. (D)
−4 −9
Answer: Given d = 0.3mm and D = 1.2 m , d = 3  10 m ,  = 650nm ,  = 650  10 m

 1  D
(i) yn =  n +  , n = 0,  1,  2 for dark fringe (1 mark)
 2 d

 1  65  10−8  1.2
y3 =  2 + 
 2  3  10−4 (1 mark)

y3 = 65  10−4 m

y3 = 6.5mm
(1 mark)
n D
(ii) yn = for bright fringe n = 0,  1,  2 (1 mark)
d

5  65  10−8  1.2
y5 =
3  10−4

y5 = 130  10−4

y5 = 13mm (1 mark)

161. A parallel beam of light of wavelength 625nm falls on a narrow slit and the resulting diffraction pattern is
observed on a screen 80cm away. It is observed that the first minimum is at a distance of 2.5mm from the
center of the screen. Find the (i) width of the slit (ii) angular width of central maximum.(iii) linear width of
central maximum.
(D)

Answer: Given  = 625  10−9 m , D = 80  10−2 m , ( y1 ) min = 2.5  10−3 m

D
(i) = 2.5  10−3  d = 200  10−6 m
d
d = 0.2mm
2 2  625  10−9
(ii) Angular width = = = 0.00625rad
d 2  10−4
2 D
(iii) Linear width = = 2  2.5  10−3 = 5mm (5 marks)
d

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