10.

Wave Optics
A Quick Recapitulation of the Chapter
1. The locus of all those particles which are vibrating in the same phase at any
instant is called wavefront. Thus, wavefront is a surface having same phase of
vibrating particles at any instant at every point on it. For point source, shape of
wavefront is spherical.
2. Phase speed is the speed with which wavefront moves and it is equal to wave
speed.
Each point on any waefront acts as independent source which emits spherical
wave.
3. Huygens' principle is essentially a geometrical construction which gives the shape
of the wavefront at any time, allows us to determine the shape of the wavefront at
a later time.
4. The laws of reflection and refraction can be verified using Huygens' principle.
5. Wavelength is inversely proportional to refractive index (μ) of the medium
' λ
i.e., λ =
μ
6. Coherent sources of light are the sources which emit light waves of same
frequency, same wavelength and have a constant initial phase difference.
7. Two such sources of light, which do not emit light waves with a constant phase
difference are called incoherent sources.
8. The phenomenon of redistribution of energy in the region of superposition of
waves is called interference. The points of maximum intensity in the regions of
superposition of waves are said to be in constructive interference whereas the
points of minimum intensity are said to be in destructive interference.
9. Conditions for Constructive Interference If initial phase difference is zero, then the
interference waves must have
(i) phase difference ¿ 2 nπ ,
(ii) path difference ¿ nλ ,
where, n=0 , 1, 2 , 3 , …
where, n=0 , 1, 2 , 3 , …
10. Conditions for Destructive Interference
Assuming initial phase difference ¿ 0
Necessary conditions for interference of waves
(i) phase difference ¿(2n−1)π , where, n=1 ,2 , 3 , …
λ
(ii) path difference ¿(2n−1) , where, n=1 ,2 , 3 , …
2
11. Two waves of amplitudes 1 and a 2 interfere at a point where phase difference is φ ,
a
then resultant amplitude is given by
2 2 2
A =a1 +a2 +2 a 1 a 2 cos ⁡φ
2
For constructive interference, Amax =( a1+ a2 )
2
For destructive interference, Amin =( a1−a 2 )
Also, resultant intensity, l=l 1 + I 2+2 √ l 1 l 2 cos ⁡φ
12. When I 1=I 2=I 0
Then, resultant intensity,
 I =I 0 + I 0 +2 I 0 cos ⁡φ=2 I 0 (1+cos ⁡φ)
2 φ
I =4 I 0 cos ⁡
2
13. In Young's double slit experiment,

For point P , Δ x=d sin ⁡θ=d tan ⁡θ=dy / D

Dλ
(i) Fringe width of bright and dark fringe, β=
d
where, λ=¿ wavelength of wave
D=¿ distance between slit and screen and d=¿ distance between two slits
β λ
Angular fringe width, θ= =
D d
(ii) Separation of n th order bright fringe from central fringe
Dnλ
y n= , n=1 ,2 , 3 , …
d
(iii) Separation of n th order dark fringe from central fringe
Dλ
y n=(2 n−1) , n=1 ,2 , 3 , …
2d
(iv) Angular position of n th order
y n nλ
(a) Bright fringe = =
D d
yn λ
(b) Dark fringe = =(2 n−1) , where n=1 , 2, 3 , …
D d
14. The phenomenon of bending of light around the sharp corners and the spreading of
light within the geometrical shadow of the opaque obstacles is called diffraction of
light.

15. Diffraction due to a Single Slit of Width (d ) A parallel beam of light with a plane
wavefront W W ' is made to fall on a single slit AB. Width of the slit is of the order of
wavelength of light, therefore, diffraction occurs on passing through the slit. As
shown in the diagram given below.
 (i) nth order secondary minima is obtained when d sin ⁡θ=nλ , where, n=1 ,2 , 3 , …
(ii) nth order secondary maxima is obtained when
λ
d sin ⁡θ=(2 n+1) , where, n=0 ,1 , 2 , 3 ,…
2
(iii) Angular separation for n th minima,
nλ
θn = , where, n=1 , 2, 3 , …
d
(iv) Linear separation of nth secondary minima,
Dnλ
y n=
d
(v) Angular position of n th order secondary maxima,
λ
θn =(2 n+1)
2d
(vi) Angular width of central maxima ¿ 2 λ/d
(vii) Linear width of central maxima ¿ 2 Dλ/d
λ
(viii) Angular width of secondary maxima or minima ¿
d
Dλ
(ix) Linear width of secondary maxima or minima ¿
d
(x ) Intensity of central maxima is maximum and intensity of secondary maxima
decreases with the increase of their order. The diffraction pattern is graphically as
shown below.

16. Resolving Power of Optical Instruments Resolving power of an optical instrument

is the ability of the instrument to produce distinctly separate images of two close
objects.
1 D
(i) Resolving power of a telescope ¿ =
dθ 1.22 λ
1 2 μ sin ⁡β
(ii) Resolving power of microscope ¿ =
Δd 1.22 λ
17. Fresnel's Distance The distance at which diffraction spread equal to the size of
2
d
aperture, Z F = .
λ
The ray optics is applicable, when Z< Z F .
18. The phenomenon of restricting the vibrations of light in a particular direction,
perpendicular to the direction of wave motion is called polarisation of light.
19. Malus' Law According to law of Malus'
i.e. ∝ cos2 ⁡θ ⇒ I =I 0 cos 2 ⁡θ
This rule is also called cosine squared rule. where, I 0=¿ intensity of plane polarised
light I =¿ intensity of transmitted light from the analyser and θ=¿ angle between
axis of the polariser and the analyser.
20. The angle of incidence at which the reflected light is completely plane polarised is
called polarising angle or Brewster's angle ( i B ).
21. Brewster's Law According to this law, when unpolarised light is incident at
polarising angle, i B on an interface separation air from a medium of refractive
index μ, then the reflected light is plane polarised (perpendicular to the plane of
incidence), provided,
μ=tan ⁡i B
where, i B =¿ Brewster's angle and μ=¿ refractive index of denser medium.
At polarising angle, i B +r =90∘,
i.e. reflected plane polarised light is at right angle from refracted light.
Objective Questions Based on NCERT Text
Topic 1 : Huygens' Principle
1. In geometrical optics, a ray of light is defined as
(a) path of propagation of light
(b) path of propagation of shadows
(c) direction of formation of image
(d) path of propagation of energy for λ → 0
2. For a ray of light, which of the following statement holds true?
(a) A ray is defined as the path of energy propagation
(b) The wavelength for a ray of light in geometrical optics is assumed to be
negligible, standing to zero
(c) A ray of light travels in a straight line
(d) All of the above
3. Which of the given phenomenon is based on the fact that light waves are
transverse electromagnetic waves?
(a) Diffraction
(b) Interference
(c) Polarisation
(d) All of these
4. Huygens' principle of secondary wavelets may be used to
(a) find the velocity of light in vacuum
(b) explain the particle's behaviour of light
(c) find the new position of a wavefront
(d) explain photoelectric effect
5. Which one of the following phenomena is not explained by Huygens' construction
of wavefront?
(a) Refraction
 (b) Reflection
(c) Diffraction
(d) Origin of spectra
6. The direction of wavefront of a wave with the wave motion is
(a) parallel
(b) perpendicular
(c) opposite
(d) at an angle of θ
7. Ray diverging from a point source on a wavefront are
(a) cylindrical (b) spherical (c) plane (d) cubical
8. If a source is at infinity, then wavefronts reaching to observer are
(a) cylindrical
(b) spherical (c)
(c) plane
(d) conical
9. In Huygens' wave theory, the locus of all points in the same state of vibration is
called
(a) a half period zone
(b) oscillator
(c) a wavefronts
(d) a ray
10. According to Huygens' principle, each point of the wavefront is the source of
(a) secondary disturbance
(b) primary disturbance
(c) third disturbance
(d) fourth disturbance
Topic 2 : Refraction and Reflection of Plane
11. If AB is incident wavefront. Then, refracted wavefront is

12. When light is refracted into a denser medium

(a) its wavelength and frequency both increases
(b) its wavelength increases but frequency remains unchanged
(c) its wavelength decreases but frequency remains the same
(d) its wavelength and frequency both decreases
13. In given figure, light passes from denser medium 1 to rare medium 2.
 When i>i c (critical angle of incidence). Then, wavefronts EC is
(a) formed further deep in medium 2
(b) formed closer to surface line AC
(c) formed perpendicular to AC
(d) formed in medium 1 (on same side of AB )
14. If a source of light is moving away from a stationary observer, then the frequency
of light wave appears to change because of
(a) Doppler's effect
(b) interference
(c) diffraction
(d) None of these
15. In the context of Doppler effect in light, the term red shift signifies
(a) decrease in frequency
(b) increase in frequency
(c) decrease in intensity
(d) increase in intensity
16. If source and observer are moving towards each other with a velocity, v radial and c
indicates velocity of light, then fractional change in frequency of light due to
Doppler's effect will be
Δ v v radial
(a) =
v c
Δ v −v radial
(b) =
v c
Δv c
(c) =
v v radial
Δ v −c
(d) =
v v radial
17. The refractive index of glass is 1.5 . The speed of light in glass is
(a) 3 ×10 8 ms−1
(b) 2 ×108 ms−1
(c) 1 ×108 ms−1
(d) 4 ×108 ms−1
18. Which of the colours of light travels fastest in prism made up of glass?
(a) Red
(b) Violet
(c) Blue
(d) Speed of light in glass is independent of the colour of light
19. A planet moves with respect to us so that light of 475 nm is observed at 475.6 nm.
The speed of the planet is
(a) 206 kms−1
(b) 378 kms−1
(c) 108 kms−1
(d) 100 kms−1
20. The earth is moving towards a fixed star with a velocity of 50 kms−1. An observer on
the earth observes a shift of 0.50 Å in the wavelength of light coming from the star.
The actual wavelength of light emitted by the star is
(a) 3000 Å
(b) 2400 Å
(c) 6000 Å
(d) 5800 Å
21. The wavelength of spectral line coming from a distant star shift from 400 nm to
400.1 nm. The velocity of the star relative to earth is
(a) 75 kms−1
(b) 100 kms−1
(c) 50 kms−1
(d) 200 kms−1
22. The source of light is moving towards observer with relative velocity of 3 kms−1. The
fractional change in frequency of light observed is
(a) 3 ×10−3
(b) 3 ×10−5
(c) 10−5
(d) None of these
Topic 3 : Coherent and Incoherent Addition of Waves
23. Suppose displacement produced at some point P by a wave is y 1=a cos ⁡ωt and by
another wave is y 2=a cos ⁡ωt . Let I 0 represents intensity produced by each one of
individual wave, then resultant intensity due to overlapping of both waves is
(a) I 0
(b) 2 I 0
I
(c) 0
2
(d) I 0
4
24. Interference can be observed in
(a) only longitudinal waves
(b) only transverse waves
(c) only electromagnetic waves
(d) All of the above
25. Two light waves interfere constructively at a point P. The total phase difference
between the two waves at P may be
(a) 0
(b) 2 π
(c) 4 π
(d) All of these
26. Two waves interfere at point P having path difference 1.5 λ between them. The
interference is
 (a) constructive
(b) destructive
(c) no interference pattern
(d) None of the above
27. Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength λ .
The separation between the sources is 2 λ. The smallest distance from S2 on a line
passing through S2 and perpendicular to S1 S 2, where minimum of intensity occurs,
is
7λ
(a)
12
15 λ
(b)
4
λ
(c)
2
3λ
(d) 28. Two identical coherent sources placed on a diameter of a circle of radius
2
R at separation x (¿ R) symmetrically about the centre of the circle. The sources of
points on the circle with maximum intensity is (x=5 λ).
(a) 20
(b) 22
(c) 24
(d) 26
28. Two sources S1 and S2 emitting light of wavelength 600 nm are placed at a distance
1.0 ×10 cm apart. A detector can be moved on the line S1 P which is perpendicular
−2

to S1 S 2. The position of the farthest minimum detected is approximately

(a) 1.5 m
(b) 1.0 m
(c) 1.07 m
(d) 1.03 m
29. The phase difference between the two light waves reaching at a point P is 100 π .
Their path difference is equal to
(a) 10 λ
(b) 25 λ
(c) 50 λ
(d) 100 λ
30. In the phenomenon of interference, energy is
(a) destroyed at destructive interference
(b) created at constructive interference
(c) conserved but it is redistributed
(d) same at all points
31. Two distinct light bulbs as sources
(a) can produce an interference pattern
(b) cannot produce a sustained interference pattern
(c) can produce an interference pattern, if they produce light of same frequency
(d) can produce an interference pattern only when the light produced by them is
monochromatic in nature
32. Two light waves superimposing at the mid-point of the screen are coming from
coherent sources of light with phase difference π rad. Their amplitudes are 2 cm
 each. The resultant amplitude at the given point will be
(a) 8 cm
(b) 2 cm
(c) 4 cm
(d) zero
33. If two waves of equal intensities I 1=I 2=I 0 , meets at two locations P and Q with path
difference Δ 1 and Δ 2 respectively, then the ratio of resultant intensity at points P
IP
and Q , will be
IQ
2 Δ1
cos ⁡
λ
(a)
Δ
cos2 ⁡ 2
λ
2
cos ⁡Δ 1
(b) 2
cos ⁡Δ 2
2 π Δ1
cos ⁡
λ
(c)
πΔ
cos2 ⁡ 2
λ
Δ1
(d)
Δ2
34. The ratio of maximum and minimum intensities of two sources is 4 :1. The ratio of
their amplitudes is
(a) 1 :81
(b) 3 :1
(c) 1 :9
(d) 1 :16
35. When two coherent monochromatic beams of intensity I and 9 I interference, the
possible maximum and minimum intensities of the resulting beam are
(a) 9 I and I
(b) 9 I and 4 I
(c) 16 I and 4 I
(d) 16 I and I
36. Two slits in Young's double slit experiments have width in ratio 1 :25. The ratio of
I max
intensity at the maxima and minima in the interference pattern
I min
is
9
(a)
4
121
(b)
49
49
(c)
121
4
(d)
9
[CBSE AIPMT 2015]
38. Two coherent light sources of intensity ratio n are employed in an interference
experiment. The ratio of the intensities of maxima and minima is
 n+1
(a)
n−1
n+1 2
(b)
n−1

(c)
√ n+1
√ n−1
(d)
√ n+1
√ n−1
39. Light from two coherent sources of the same amplitude A and wavelength λ
interference. The maximum intensity recorded is I 0. If the sources were incoherent,
the intensity at the same point will be
(a) 4 I 0
(b) 2 I 0
(c) I 0
(d) I 0 /2
40. If two incoherent sources each of intensity I 0 produce wave which overlaps at some
common point, then resultant intensity obtained is
(a) 4 I 0
(b) 2 I 0
I
(c) 0
2
(d) dependent on phase difference
41. Two identical and independent sodium lamps act as
(a) coherent sources
(c) Either (a) and (b)
(b) incoherent sources
(d) None of these
42. In an experiment with two coherent sources, the amplitude of the intensity
variation is found to be 5 % of the average intensity. The relative intensities of the
light waves of interferring sources will be
(a) 1600:1
(b) 900 :1
(c) 40 :1
(d) 400:1
Topic 4 : Interference of Light Waves, Young's Double Slit Experiment
43. In Young's double slit experiment, if source S is shifted by an angle φ as shown in
figure. Then, central-bright fringe will be shifted by angle φ towards

(a) end A of screen

(b) end B of screen
 (c) does not shift at all
(d) Either end A and B depending on extra phase difference caused by shifting of
source
44. White light may be considered to be mixture of wave with wavelength ranging
between 3000 Å and 7800 Å . An oil film of thickness 10000 Å is examined normally by
the reflected light. If μ=1.4 , then the film appears bright for
(a) 4308 Å , 5091 Å , 6222 Å
(b) 4000 Å , 5091 Å , 5600 Å
(c) 4667 Å , 6222 Å , 7000 Å
(d) 4000 Å , 4667 Å ,5600 Å , 7000 Å
45. The figure shows a Young's double-slit experiment where, P and Q are the slits.
The path lengths PX and QX are nλ and (n+ 2) λ respectively, where n is a whole
number and λ is the wavelength. Taking, path difference at the central fringe as
zero, what is formed at X ?

(a) First bright

(b) First dark
(c) Second bright
(d) Second dark
46. In Young's double slit experiment, a glass plate is placed before a slit which
absorbs half the intensity of light. Under this case
(a) the brightness of fringes decreases
(b) the fringe width decreases
(c) no fringes will be observed
(d) the bright fringes become fainter and the dark fringes have finite light intensity
1
47. In Young's double slit experiment, intensity at a point is th of the maximum
4
intensity. Angular position of this point is
(a) sin−1 ⁡(λ /d)
(b) sin−1 ⁡( λ /2 d )
(c) sin−1 ⁡(λ /3 d )
(d) sin−1 ⁡( λ / 4 d )
48. The Young's double slit experiment is done in a medium of refractive index 4/3. A
light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The
lower slit S2 is covered by a thin glass sheet of thickness 10.4 μ m and refractive
index 1.5. The interference pattern is observed on a screen placed 1.5 m from the
slits as shown in the figure.
The location of the central bright fringes on the Y -axis is
(a) 4.0 mm
(b) 4.33 mm
(c) 5 mm
(d) 4.5 mm
49. With reference of the above question, if 600 nm light is replaced by white light of
range 400 nm to 700 nm , the wavelength of the light that form maxima exactly at
point O are
(a) 650 nm and 400 nm
(b) 650 nm and 433.33 nm
(c) 400 nm and 667 nm
(d) 650 nm and 667 nm
50. A parallel beam of sodium light of wavelength 6000 Å is incident on a thin glass
plate of refractive index 1.5 such that the angle of refraction in the plate is 60∘ . The
smallest thickness of the plate which will make it dark by reflection.
(a) 4000 Å
(b) 4200 Å
(c) 1390 Å
(d) 2220 Å
51. In Young's double slit experiment two disturbance arriving at a point P have phase
difference of π /2. The intensity of this point expressed as a fraction of maximum
intensity I 0 is
3
(a) I 0
2
1
(b) I 0
2
4
(c) I 0
3
3
(d) I 0
4
52. Two monochromatic light wave of same amplitudes of 2 A interferring at a point
have a phase difference of 60∘ . The intensity at that point will be proportional to
(a) 5 A 2
(b) 12 A 2
(c) 7 A2
(d) 19 A 2
53. The shape of the fringe obtained on the screen in case of Young's double slit
experiment is
(a) a straight line
(b) a parabola
 (c) a hyperbola
(d) a circle
54. In Young's double slit experiment, distance between slits in kept 1 mm and a screen
is kept 1 m apart from slits. If wavelength of light used is 500 nm , then fringe
spacing is
(a) 0.5 mm
(b) 0.5 cm
(c) 0.25 mm
(d) 0.25 cm
55. If the 8 th bright band due to light of wavelength λ 1 coincides with 9 th bright band
from light of wavelength λ 2 in Young's double slit experiment, then the possible
wavelengths of visible light are
(a) 400 nm and 450 nm
(b) 425 nm and 400 nm
(c) 400 nm and 425 nm
(d) 450 nm and 400 nm
56. The Young's double slit experiment is performed with light of wavelength 6000 Å,
where in 16 fringes occupy a certain region on the screen. If 24 fringes occupy the
same region with another light, of wavelength λ , then λ is
(a) 6000 Å
(b) 4500 Å
(c) 5000 Å
(d) 4000 Å
57. The maximum intensity of fringes in Young's double slit experiment is I . If one of
the slit is closed, then the intensity at that place becomes I 0. Which of the following
relations is correct?
(a) I =I 0
(b) I =2 I 0
(c) I =4 I 0
(d) There is no relation between I and I 0
58. Young's double slit experiment is performed with sodium (Yellow) light of
wavelength 589.3 nm and the interference pattern is observed on a screen 100 cm
away. The 10th bright fringe has its centre at a distance of 12 nm from the central
maximum. The separation between the slits is
(a) 0.49 mm
(b) 0.6 mm
(c) 0.7 mm
(d) 0.53 mm
59. If Young's double slit experiment, is performed in water, the fringe width recorded
is ω 2. If it is performed in air, the fringe width recorded is ω 1. Then, ω 1 /ω2 is
( μ water =4 /3 )
(a) 3/2
(b) 3/ 4
(c) 4 /3
(d) Data insufficient
60. In a Young's double slit experiment, the slit separation is 1 mm and the screen is
1 m from the slit. For a monochromatic light of wavelength 500 nm , the distance of
3 rd minima from the central maxima is
 (a) 0.50 mm
(b) 1.25 mm
(c) 1.50 mm
(d) 1.75 mm
61. In Young's double slit experiment the two slits are d distance apart. Interference
pattern is observed on a screen at a distance D from the slits. A dark fringe is
observed on the screen directly opposite to one of the slit. The wavelength of light
is
2
D
(a)
2d
2
d
(b)
2D
2
D
(c)
d
2
d
(d)
D
62. The Young's double slit experiment is performed with blue and with green light of
wavelengths 4360 Å and 5460 Å , respectively. If x is the distance of 4 th maximum
from the central one, then
(a) x (blue ¿=x (green )
(b) x (blue) ¿ x (green )
(c) x (blue ¿< x (green)
x ( blue ) 5460
(d) ¿=
x (green ¿ 4360
63. In a Young's double slit experiment, two coherent sources are placed 0.90 nm apart
and the fringes are observed 1 m away. If it produces the second dark fringe at a
distance of 1 mm from the central fringe, the wavelength of monochromatic light
used would be
(a) 60 ×10−4 cm
(b) 10 ×10−4 cm
(c) 10 ×10−5 cm
(d) 6 ×10−5 cm
64. In Young's double slit experiment, the slits are 2 mm apart and are illuminated by
photons of two wavelengths λ 1=12000 Å and λ 2=10000 Å . At what minimum distance
from the common central bright fringe on the screen 2 m from the slit will a bright
fringe from one interference pattern coincide with a bright fringe from the other?
(a) 8 mm
(b) 6 mm
(c) 4 mm
(d) 3 mm
[NEET 2013]
65. In a two slit experiment with monochromatic light fringes are obtained on a screen
placed at some distance from the slits. If the screen is moved by 5 ×10−2 m towards
the slits, the change in fringe width is 3 ×10−5 m. If separation between the slits is
−3
10 m, the wavelength light used is
(a) 6000 Å
(b) 5000 Å
 (c) 3000 Å
(d) 4500 Å
66. In Young's double slit experiment, the distance between two slits is 0.1 mm and
these are illuminated with light of wavelength 5460 Å . The angular positions of first
dark fringe on the screen distant 20 cm from slits will be
(a) 0.8∘
(b) 0.6 ∘
(c) 0.4 ∘
(d) 0.16 ∘
67. In Young's double slit experiment the fringe width is 1 ×10−4 m. If the distance
between the slit and screen is double and distance between the two slit is reduced
to half and wavelength is changed from 6.4 × 10−7 m to 4.0 × 10−7 m , the value of new
fringe width will be
(a) 0.15 ×10−4 m
(b) 2.0 ×10−4 m
(c) 1.25 ×10−4 m
(d) 2.5 ×10−4 m
68. Two beams of light having intensities I and 4 I interfere to produce a fringe pattern
on a screen in Young's double slit experiment. The phase difference between the
beams is π /2 at point A and π at point B. Then, the difference between the resultant
intensities at A and B is
(a) 2 I
(b) 4 I
(c) 5 I
(d) 7 I
69. In Young's double slit experiment, let S1 and S2 be the two slits and C be the centre
of the screen. If the ∠ S1 C S 2=θ and λ is the wavelength, then fringe width will be
λ
(a)
θ
(b) λθ
2λ
(c)
θ
λ
(d)
2θ
70. In Young's double slit experiment, 12 fringes are obtained to be formed in a certain
segment of the screen when light of wavelength 600 nm is used. If the wavelength
of light is changed to 400 nm, number of fringes observed in the same segment of
the screen is given by
(a) 12
(b) 18
(c) 24
(d) 30
71. Young's double slit experiment is made in a liquid. The 10th bright fringe in liquid
lies, where 6th dark fringes lies in vacuum. The refractive index of the liquid is
approximately
(a) 1.8
(b) 1.54
 (c) 1.67
(d) 1.2
72. In the Young's double slit experiment using a monochromatic light of wavelength λ
the path difference (in terms of an integer n ) corresponding to any point having
half the peak intensity is
[JEE Advanced 2013]
λ
(a) (2 n+1)
2
λ
(b) (2 n+1)
4
λ
(c) (2 n+1)
8
λ
(d) (2 n+1)
16
73. In a Young's double slit experiment, the separation between the slits ¿ 2.0 mm , the
wavelength of the light ¿ 600 nm and the distance of the screen from the slits ¿ 2.0 m .
If the intensity at the centre of the central maximum is 0.20 Wm−2, then the intensity
at a point 0.5 cm away from this centre along the width of the fringes will be
(a) 0.05 Wm−2
(b) 0.15 Wm−2
(c) 0.20 Wm−2
(d) 0.10 Wm−2
74. In Young's double slit experiment, the slit width and the distance of slits from the
screen both are double. The fringe width
(a) increases
(b) decreases
(c) remain unchanged
(d) None of these
75. Two sources of light of wavelengths 1500 Å and 2500 Å are used in Young's double
slit experiment simultaneously. Which orders of fringes of two wavelength patterns
coincide?
(a) 3rd order of 1 st source and 5th of 2 nd
(b) 7 th order of 1 st and 5 th order of 2 nd
(c) 5 th order of 1 st and 3 rd order of 2 nd
(d) 5 th order of 1 st and 7 th order of 2 nd
76. In Young's double slit experiment using monochromatic light of wavelength λ , the
path difference (in terms of an integer n ) corresponding to any point having half
the peak intensity is
λ
(a) (2 n+1)
2
(2 n+1) λ
(b)
4
λ
(c) (2 n+1)
8
(2 n+1) λ
(d)
16
77. In a Young's double slit experiment the distance between slits is increased five
times where as their distance from screen in halved, then the fringe width is
 1
(a) becomes
90
1
(b) becomes
20
1
(c) becomes
10
(d) it remains same
78. The fringe width in a Young's double slit interference pattern is 3.2 ×10−4 m, when
red light of wavelength 5600 Å is used. How much will it change, if blue light of
wavelength 4200 Å is used?
(a) 8 ×10−4 m
(b) 0.8 ×10−4 m
(c) 4.2 ×10−4 m
(d) 0.45 ×10−4 m
79. In the Young's double-slit experiment, the intensity of light at a point on the screen
(where the path difference is λ ¿ is K , ( λ being the wavelength of light used). The
intensity at a point where the path difference is λ /4 , will be
[CBSE AIPMT 2014]
(a) K
(b) K / 4
(c) K /2
(d) zero
80. If the intensities of the two interferring beams in Young's double slit experiment be
I 1 and I 2, then the contrast between the maximum and minimum intensity is good
when
(a) I 1 is much greater than I 2
(b) I 2 is smaller than I 2
(c) I 1=I 2
(d) Either I 1=0 or I 2=0
81. In Young's double slit experiment, a third slit is made in between the double slits.
Then
(a) fringes of unequal width are formed
(b) contrast between bright and dark fringes is reduced
(c) intensity of fringes totally disappears
(d) only bright light is observed on the screen
82. In Young's double slit experiment, the central bright fringes can be identified
(a) as it is greater intensity than the other bright fringes
(b) as it has wider than the other bright fringes
(c) as it is narrower than the other bright fringes
(d) by using white light instead of monochromatic light
83. Yellow light from atomic sodium with a wavelength of 589 nm illuminates a single-
slit. The dark fringes in the diffraction pattern are found to be seperated on either
side of central bright by 2.2 mm , on a screen 1.0 m from the slit. The slit width is
(a) 0.7 mm
(b) 0.54 mm
(c) 1.0 mm
(d) 0.24 mm
Topic 5
Diffraction
84. What should be the slit width to obtain 10 maxima of the double slit pattern within
the central maxima of the single slit pattern of slit width 0.4 mm ?
(a) 0.4 mm
(b) 0.2 mm
(c) 0.6 mm
(d) 0.8 mm
85. In a single slit diffraction of light of wavelengths λ is used and slit of width e , the
size of the central maxima on a screen at a distance b is
(a) 2 bλ+ e
2bλ
(b)
e
2bλ
(c) +e
e
2bλ
(d) −e
e
86. A parallel beam of light of wavelength 6000 Å gets diffracted by a single-slit width
0.3 mm. The angular position of the first minima of diffracted light is
(a) 6 ×10−3 rad
(b) 1.8 ×10−3 rad
(c) 3.2 ×10−3 rad
(d) 2 ×10−3 rad
87. In a single-slit diffraction pattern observed on a screen placed at D m distance
from the slit of width d m , the ratio of the width of the central maximum to the
width of other secondary maximum is
(a) 2 :1
(b) 1 :2
(c) 1 :1
(d) 3 :1
88. Consider diffraction pattern obtained with a single slit at normal incidence. At the
angular position of the first order diffraction maximum, the phase difference
between the wavelets from the opposite edges of the slit is
π
(a)
4
π
(b)
2
(c) 3 π
(d) 2 π
89. Yellow light is used in a single-slit diffraction experiment with slit width of 0.6 mm.
If yellow light is replaced by X-rays, then the observed pattern will reveal
(a) that the central maximum is narrower
(b) more number of fringes
(c) less number of fringes
(d) no diffraction pattern
90. In a diffraction pattern due to a single slit of width a , the first minimum is observed
at an angle 30∘ when light of wavelength 5000 Å is incident of the slit. The first
secondary maximum is observed at an angle of
[NEET 2016]
−1 2
(a) sin ⁡
3
−1 1
(b) sin ⁡
2
−1 3
(c) sin ⁡
4
−1 1
(d) sin ⁡
4
91. A parallel beam of fast moving electrons is incident normally on a narrow slit. A
fluorescent screen is placed at a large distance from the slit. If the speed of the
electrons is increased, then which of the following statements is correct?
[NEET 2013]
(a) Diffraction pattern is not observed on the screen in the case of electrons
(b) The angular width of the central maximum of the diffraction pattern will
increases
(c) The angular width of the central maximum will decreases
(d) The angular width of the central maximum will be unaffected
92. A single-slit diffraction pattern is formed with white light. For what wavelength of
light the third secondary maximum in the diffraction pattern coincides with the
second secondary maximum in the pattern for red light of wavelength 6500 Å.
(a) 4400 Å
(b) 4100 Å
(c) 4642.8 Å
(d) 9100 Å
93. A single-slit of width d is illuminated by violet light of wavelength 400 nm and the
width of the central maxima is measured as y . When half of the slit width is
covered and illuminated by yellow light of wavelength 600 nm, the width of the
central diffraction pattern is
(a) the pattern vanishes and the width is zero
(b) y /3
(c) 3 y
(d) None of the above
94. A beam of light of wavelength 600 nm from a distant source falls on a single-slit
1 mm wider and the resulting diffraction pattern is observed on a screen 2 m away.
The distance between the first dark fringes on either side of the central bright
fringes is
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm
95. If we observe the single-slit diffraction with wavelength λ and slit width e , the
width of the central maximum is 2 θ. On decreasing the slit width for the same λ ,
then
(a) θ increases
(b) θ remains unchanged
(c) θ decreases
(d) θ increases or decreases depending on the intensity of light
96. In diffraction from a single slit the angular width of the central maxima does not
depend on
(a) λ of light used
(b) width of slit
(c) distance of slits from the screen
(d) ratio of λ and slit width
97. A parallel beam of light of wavelength 4000 Å gets diffraction by a single slit of
width 0.2 mm . The angular position of the first minima of diffracted light is
(a) 2 ×10−3 rad
(b) 3 ×10−3 rad
(c) 1.8 ×10−3 rad
(d) 6 ×10−3 rad
98. A beam of light of λ=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. The
distance between first dark fringes on either side of the central bright fringe is
[CBSE AIPMT 2014]
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm
99. In a Young's double slit experiment the angular width of a fringe is found to be 0.2∘
on a screen placed 1 m away. The wavelength of light used is 600 nm. If the entire
experimental apparatus is immersed in water (Refractive index of water is 4 /3 ),
then angular width of the fringe will be
(a) 0.25∘
(b) 0.15∘
(c) 0.75∘
(d) 1∘
100. In a Young's double slit experiment, the screen is placed at a distance of 1.25 m
from the slits. When the apparatus is immersed in water (μω=4 /3), the angular
width of a fringe is found to be 0.2∘. When the experiment is performed in air with
same set up, the angular width of the fringe is
(a) 0.4 ∘
(b) 0.28∘
(c) 0.35∘
(d) 0.15∘
101. The angular resolution of the telescope is determined by the
(a) image produced by the telescope
(b) objective of the telescope
(c) Both (a) and (b)
(d) Neither (a) nor (b)
102. In telescope, the radius of the central bright region ( r 0) is
0.61 λf
(a)
a
0.75 λf
(b)
a
 1.94 λf
(c)
a
2.43 λf
(d)
a
103. For better resolution, a telescope must have a
(a) large diameter objective
(b) small diameter objective
(c) may be large
(d) Neither large nor small
104. The diameter of objective lens of a telescope is 6 cm and wavelength of light used
is 540 nm . The resolving power of telescope is
(a) 9.1 ×10 4 rad−1
(b) 105 rad−1
(c) 3 ×10 4 rad−1
(d) None of these
105. A telescope is used to resolve two stars separated by 3.2 ×10−6 rad . If the wavelength
of light used in 5600 Å , what should be the aperture of the objective of the
telescope?
(a) 0.2135 m
(c) 0.567 m
(b) 0.1488 m
(d) 1 m
106. What will be the ratio (D/ f ) in microscope, where D is the diameter of the aperture
and f is the focal length of the objective lens?
(a) tan ⁡β
β
(b) tan ⁡
2
(c) 2 tan ⁡β
β
(d) tan ⁡
6
107. If the medium between the object and the objective lens of a microscope is not air
but refractive index n , then minimum separation gets modified to
1.44 λ
(a)
2n sin ⁡β
1.22 λ
(b)
2n sin ⁡β
3.2 λ
(c)
2n sin ⁡β
1.49 λ
(d)
n sin ⁡β
108. The resolving power of a compound microscope increases when
(a) refractive index of the medium is increased keeping wavelength of light (λ)
constant (or same)
(b) for the same medium, wavelength of light is decreased
(c) refractive index of the medium and wavelength of light used both are decreased
(d) Both (a) and (b)
109. In a Fresnel biprism experiment, the two positions of lens give separation between
the slits as 25 cm and 16 cm respectively. The actual distance of separation is
(a) 20 cm
 (b) 16 cm
(c) 18 cm
(d) 20.5 cm
110. For what distance is ray optics a good approximately when the aperture is 3 mm
wide and wavelength is 500 nm ?
(a) 18 m
(b) 25 m
(c) 30 m
(d) 35 m
Topic 6
Polarisation
111. When the displacement of the wave is at right angles to the direction of its
propagation, it is known as
(a) transverse wave
(b) longitudinal wave
(c) Either (a) or (b)
(d) Both (b) and (c)
112. Light waves are
(a) longitudinal waves
(b) electromagnetic waves
(c) transverse wave
(d) Both (b) and (c)
113. Which of the following can be used to control the intensity, in sunglasses, window
pares etc?
(a) Transverse wave
(b) Polaroids
(c) Plane polarised wave
(d) Polarised wave
114. The phenomenon of polarisation of light indicates that
(a) light is a longitudinal wave
(b) light is a transverse electromagnetic wave
(c) light is a transverse wave only
(d) Either (b) or (c)
115. Which of the following cannot be polarised?
(a) Ultraviolet
(b) Ultrasonic waves
(c) X-rays
(d) Radio waves
116. Which of the following is a dichronic crystal?
(a) Quartz
(b) Tourmaline
(c) Mica
(d) Selenite
117. In the propagation of light waves, the angle between the direction of vibration and
plane of polarisation is
(a) 0∘
(b) 90∘
 (c) 45 ∘
(d) 80∘
118. When light passes through two polaroids P1 and P2, then transmitted polarisation is
the component parallel to the polaroid axis. Which of the following is correct?

119. At which angle the intensity of transmitted light is maximum when a polaroid sheet
is rotated between two crossed polaroids?
(a) π /4
(b) π /2
(c) π /3
(d) π
120. Unpolarised light is incident on a plane glasses surface. What should be the angle
of incidence so that the reflected and refracted rays are perpendicular to each
other?
¿ for glass ¿ 1.5 ¿
(a) 60∘
(b) 90∘
(c) 0∘
(d) 57∘
121. An unpolarised beam of light of intensity I 0 falls on a polariod. The intensity of the
emergent beam is
I
(a) 0
2
I
(b) 0
I
(c) 0
4
(d) zero
122. For good polariser in case of unpolarised light, we will observe
(a) reflection and no transmission of light
(b) no reflection and total transmission of light
(c) diffraction
(d) total internal reflection of light
123. When both the components of electric field of light waves are present such that
one is stronger than the other and such light is viewed through a rotating analyser,
one sees a maximum and a minimum of intensity but not complete darkness. This
kind of light is called
(a) polarised
(c) partially polarised
 (b) linearly polarised
(d) None of these
124. An unpolarised beam intensity I 0 is incident on a pair of nicols making an angle of
∘
60 with each other. The intensity of light emerging the pair is
(a) I 0
(b) I 0 /2
(c) I 0 /4
(d) I 0 /8
125. A beam of unpolarised light of intensity I 0 is passed through a polaroid A and then
through another polaroid B which is oriented so that its principal plane makes an
angle of 45 ∘ relative to that of A . The intensity of the emergent light is
[JEE Main 2013]
(a) I 0
(b) I 0 /2
(c) I 0 /4
(d) I 0 /8
126. A beam of ordinary unpolarised light passes through a tourmaline crystal C 1 and
then its passes through another tourmaline crystal C 2, which is oriented such that
its principle plane is parallel to that of C 2. The intensity of emergent light is I 0.
Now, C 2 is rotated by 60∘ about the ray. The emergent ray have an intensity.
(a) 2 I 0
(b) I 0 /2
(c) I 0 /4
(d) I 0 / √ 4
127. When unpolarised light beam is incident from air to glass (μ=1.5) at the polarising
angle.
(a) reflected beam is polarised 100 %
(b) reflected and refracted beams are partially polarised
(c) the reason for (a) is that almost all the light is reflected
(d) All of the above
128. At what angle should an unpolarised beam to incident on a crystal of μ= √ 3 , so that
reflected beam is polarised?
(a) 45 ∘
(b) 60∘
(c) 90∘
(d) 0∘
−1 4
129. The critical angle of a certain medium is sin ⁡ . The polarising angle of medium is
5
−1 5
(a) tan ⁡
4
−1 4
(b) sin ⁡
5
−1 5
(c) sin ⁡
4
−1 4
(d) tan ⁡
3
130. When the angle of incidence is 45 ∘ on the surface of a glass slab, it is found that the
reflected ray is completely polarised. The velocity is light in glass in
(a) √ 3 ×108 ms−1
(b) 3 ×10 8 ms−1
(c) 2 ×108 ms−1
(d) √ 2× 108 ms−1
131. The refractive index of a medium is 1 . If the unpolarised light is incident on it at
the polarising angle of the medium, the angle of refraction is
(a) 60∘
(b) 45 ∘
(c) 30∘
(d) 0∘
132. Find the angle of incidence at which light reflected from glass (μ=1.5) be
completely polarised.
(a) 72.8∘
(b) 51.6∘
(c) 40.3 ∘
(d) 56.3∘
Special Format Questions
Assertion and Reason
Directions (Q. Nos. 133-145) In the following questions, a statement of assertion is
followed by a corresponding statement of reason. Of the following statements,
choose the correct one.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of
Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct
explanation of Assertion.
(c) Assertion is correct but Reason is incorrect.
(d) Assertion is incorrect but Reason is correct.
133. Assertion In the field of geometrical optics, light can in assumed to approximately
travel in straight line.
Reason The wavelength of visible light is very small in comparison to the
dimensions of typical mirrors and lenses, then light can be assumed to
approximately travel in straight line.
134. Assertion When monochromatic light is incident on a surface separating two
media, the reflected and refracted light both have the same frequency as the
incident frequency.
Reason Speed of light and wavelength of light both changes in refraction and
hence, the ratio v=c / λ is a constant.
135. Assertion The emergent plane wavefront is tilted on refraction of a plane wave by a
thin prism.
Reason The speed of light waves is more in glass and the base of the prism is
thicker than the top.
136. Assertion If we have a point source emitting waves uniformly in all directions, the
locus of point which have the same amplitude and vibrate in the same phase are
spheres.
 Reason Each point of the wavefront is the source of a secondary disturbance and
the wavelets emanating from these points spread out in all directions with the
speed of the wave.
137. Assertion Increase in the wavelength of light due to Doppler's effect is red shift.
Reason When the wavelength increases, then wavelength in the middle of the
visible region of the spectrum moves towards the red end to the spectrum.
138. Assertion No interference pattern is detected when two coherent sources are
infinitely close to each other.
Reason The fringe width is inversely proportional to the distance between the two
slits.
139. Assertion If the initial phase difference between the light waves emerging from the
slits of Young's double slit experiment is π -radian, the central fringe will be dark.
2π
Reason Phase difference is equal to times the path difference.
λ
140. Assertion In Young's double slit experiment, for two coherent sources, the
resultant intensity by
2 φ
I =4 I 0 cos ⁡
2
Reason Ratio of maximum and minimum intensity
I max ( √ I 1 + √ I 2 )
2

=
I min ( √ I −√ I ) 2
1 2

141. Assertion In Young's double slit experiment, the fringes become in distinct, if one
of the slits is covered with cellophane paper.
Reason The cellophane paper decreases the wavelength of light.
142. Assertion In interference, the film which appear bright in reflected system will
appear dark in the transmitted system and vice-versa.
Reason The condition for film to appear bright or dark in reflected light are just
reverse to those in the transmitted light.
143. Assertion In Young's double slit experiment, the fringe width for dark fringes is
same as that for white fringes. Reason In Young's double slit experiment, when the
fringes are performed with a source of white light, then only dark and bright
fringes are observed.
144. Assertion In Young's doule slit experiment, the fringe width is directly proportional
to wavelength of the source used.
Reason When a thin transparent sheet is placed infront of both the slits of Young's
double slit experiment, the fringe width will increase.
145. Assertion To observe diffraction of light the size of obstacle aperture should be of
the order of 10−7 m. Reason 10−7 m is the order of wavelength of visible light.
Statement Based Questions Type I
Directions (Q. Nos. 146-154) In the following questions, a statement I is followed by a
corresponding statement II. Of the following statements, choose the correct one.
(a) Both Statement I and Statement II are correct and Statement II is the correct
explanation of statement I.
(b) Both Statement I and Statement II are correct but Statement II is not the
correct explanation of Statement I.
(c) Statement I is correct but Statement II is incorrect.
(d) Statement I is incorrect but Statement II is correct.
146. Statement I Maxwell's electromagnetic theory of light proved that light is an
 electromagnetic wave. Statement II Light waves propagates even in vacuum
according to wave theory of light.
147. Statement I When monochromatic light is incident on a surface separating two
media, the reflected light both have the same frequency as the incident frequency.
Statement II Reflection and refraction arise through interaction of incident light
with the atomic constituents of matter. Atoms may be viewed as oscillators, which
take up the frequency of the external agency (light) causing forced oscillations.
148. Statement I Speed of light is independent of its colour only in vacuum.
Statement II Red colour travels slower than violet in glass.
149. Statement I When light travels from a rarer to a denser medium, the speed
decreases but energy of the wave remains same.
Statement II Intensity of light wave is directly proportional to the square of the
amplitude of the wave.
150. Statement I Sound waves cannot be polarised.
Statement II Sound waves are longitudinal in nature.
151. Statement I The intensity at the bright band on the screen is maximum and equal
to 4 I 0 , where I 0 is the intensity of light from each sources.
Statement II The intensity at the dark band is always zero irrespective of the
intensity of light waves coming from the two sources.
152. Statement I In Young's double slit experiment, at centre line of screen, a bright
fringe is obtained.
Statement II Path difference between two waves is given by S2 P−S1 P which is zero
at centre line.
153. Statement I In Young's double slit experiment, the width of one of the slits is
slowly increased to make it twice the width of the other slit. The intensity of both
the maxima and minima increases.
Statement II Intensity of light from the slits is directly proportional to the width of
the slit.
154. Statement I Diffraction determines the limitations of the concepts of light rays.
Statement II A beam of width a starts to spread out due to diffraction after it has
travelled a distance ( 2 a2 / λ ).

Statement Based Questions Type II

155. According to Maxwell's electromagnetic theory following phenomenon can be
explained.
I. propagation of light in vaccum
II. interference of light
III. polarisation of light
IV. photoelectric effect
(a) I, II and III
(b) I, II, IV
(c) I, III and IV
(d) II, III and IV
156. Which of the following statement(s) is/are correct?
I. A point source emitting waves uniformly in all directions.
II. In spherical wave, the locus of point which have the some amplitude and vibrate
in same phase are spheres.
 III. At a small distance from the source, a small portion of sphere can be
considered as plane wave.
(a) Only I
(b) Both I and II
(c) Only III
(d) All of the above
157. In case of reflection of a wavefront from a reflecting surface,

I. points A and E are in same phase

II. points A and C are in same phase
III. points B and A are in same phase
IV. points C and E are in same phase
(a) I and II
(b) II and III
(c) III and IV
(d) I and IV
158. Figure shows behaviour of a wavefront when it passes through a prism.

Which of the following statement(s) is/are correct?

I. Lower portion of wavefront ( B' ) is delayed resulting in a tilt.
II. Time taken by light to reach A' from A is equal the time taken to reach B' from B.
III. Speed of wavefront is same everywhere.
IV. A particle on wavefront A' B' is in phase with a particle on wavefront AB.
(a) I and II
(b) II and III
(c) III and IV
(d) I and III
159. Monochromatic light of wavelength 589 nm is incident from air on a water surface.
(Refractive index of water is 1.33).
Which of the following statement(s) is/are correct?
I. Frequency of reflected light and refracted light are same.
II. Wavelength of reflected light is more than that of refracted light.
III. Speed of reflected light is equal to that of refracted light.
IV. Intensity of reflected light is always more than that of refracted light.
(a) I and III
(b) II and IV
(c) I and II
(d) III and IV
160. Shape of wavefront in case of
I. light diverging from a point source.
II. light emerging out of convex lens when a point source is placed at its focus.
III. the portion of the wavefront of light from a distant star intercepted by the
earth are respectively.
(a) cylindrical, concave, plane
(b) spherical, plane, plane
(c) spherical, convex, plane
(d) spherical, cylindrical, plane
161. Let S1 and S2 are two sources and if wave from S1 reaches some common point P be
covering seven time of wavelength (λ) and from S2 by covering nine times of
wavelength (λ) .

Which of the following statement(s) is/are correct?

I. S2 P−S1 P=2 λ
II. Waves from S1 arrives exactly two cycles earlier than waves from S2.
III. At P waves from S1 and S2 are in phase.
IV. At P waves from S1 and S2 are out of phase.
(a) I, II and III
(c) II, III and IV
(b) I, III and IV
(d) I, II and IV
162. The conditions for producing sustained interference are
I. phase difference between interferring waves remains constant with time.
II. interferring waves have nearly same amplitude levels
III. interferring waves are of same frequency.
IV. interferring waves are moving in opposite directions.
(a) I, II and III
(b) II and III
(c) III and IV
(d) I and IV
163. Which of the following statement is/are correct for coherent sources?
I. Two coherent sources emit light waves of same wavelength.
II. Two coherent sources emit light waves of same frequency.
III. Two coherent sources have zero or constant initial phase difference with
respect to time.
Choose the correct option from the codes given below.
(a) Only I
(b) I and III
(c) II and III
(d) I, II and III
164. Choose the correct option for the statements given below.
 I. The interference pattern has a number of equally spaced bright and dark bands,
while the diffraction pattern has a central bright maximum, which is twice as wide
as the other maxima.
II. The interference pattern is superposing two waves originating from the two
narrow slits, while the diffraction pattern is a superposition of a continuous family
of waves originating from each point on a single slit.
III. For a single slit of width a , the first null of the interference pattern occurs at an
angle of λ /a while at the same angle of λ /a, a maximum (not a null) for two narrow
slits separated by a distance a .
(a) I and II are correct, III is incorrect
(b) I and II are correct, II is incorrect
(c) I, II and III are correct
(d) I, II and III are incorrect
165. Which of the given statements is/are correct for phenomenon of diffraction?
I. For diffraction through a single-slit, the wavelength of wave must be comparable
to the size of slit.
II. The diffraction is very common in sound waves but not so common in light
waves.
III. Diffraction is only observed in electromagnetic waves.
(a) Only I
(b) II and III
(c) I and II
(d) I, II and III
166. Which of the following statement(s) is/are correct with reference to the figure
given below?
I. Dots and arrows indicates that both polarisations are present in the incident and
refracted waves.
II. The reflected light is not linearly polarised.
III. Transmitted intensity will be zero when the axis of the analyser is in the plane
of the figure i.e., the plane of incidence.

(a) Only I
(b) Only II
(c) Both I and III
(d) Both I and II
Matching Type
167. Light waves travels in vacuum along X -axis, Y -axis and Z -axis. Column I lists the
equation of the plane wavefront and Column II lists the direction of propagation of
the wave. Match the items in Column I with terms in Column II and choose the
correct options from the codes given below.
 Column Column II
I

A 1 Along Y -
X =C
axis

B 2 Along X -
Y =C
axis

C 3 Along Z -
Z=C
axis

ABC
ABC
(a) 1 23
(b) 3 2 1
(c) 2 13
(d) 2 3 1
168. Match the following columns and choose the correct options from the codes given
below.
Column I Column II

Constructive 1
A. nλ
interference

Destructive 2
B. (n+1) λ/2
interference

Path difference for 3 Waves are in phase at

C.
constructive interference point of interference

D Path difference for 4 Waves are out of phase at

destructive interference point of interference

(a) A → 4 , B →1 , C → 3 , D → 2
(b) A → 4 , B →3 ,C → 1 , D → 2
(c) A → 3 , B → 4 ,C → 1 , D → 2
(d) A → 3 , B → 4 ,C → 2 , D → 1
169. Two slits are made one millimetre apart and the screen is placed 1 m away and
blue-green light of wavelength 500 nm is used.
Now, match the activity given in Column I with the change in fringe pattern
obtained in Column II.
Column I Column II
A. Screen is moved away from
1.increase
the plane of slits
2.
B. Source is replaced by another width of fringes ( β)
Angular separation of
source of shorter wavelength
fringe remains constant.
 C. Set up of experiment is Fringe separation
3.
dipped completely in water decreases.
D. Distance between slits is
4. Fringe width become 3/4th.
reduced.

(a) A →1 , B →3 , C → 4 , D → 2
(b) A → 3 , B → 1, C → 4 , D → 2
(c) A → 2, B →3 ,C → 4 , D → 1
(d) A → 2, B →3 ,C → 1 , D → 4 170. Consider the arrangement shows in the figure. The
distance D is large compared to the separation d between the slits. For this
arrangement, match the items in Column I with terms in Column II and choose the
correct option from codes given below.

Column
Column I
II

√
The minimum value of d so that there is 1 λD
A.
a dark fringe at O for x=D is 3
For x=D and d minimum such that there
2
B. is dark fringe at O , the distance y at 2d
which next bright fringe is located is

3
C. fringe width for x=D d

√
The minimum value of d so that there is 4 λD
D.
a dark fringe at O for x=D /2 is 2

(a) A → 4 , B →3 ,C → 2 , D → 1
(b) A →1 , B →2 , C → 3 , D → 4
(c) A →1 , B →3 , C → 2 , D → 4
(d) A → 4 , B → 2, C → 3 , D → 1
171. Column I shows the changes introduced in Young's double-slit experiment while
Column II tells the changes in the fringe pattern while performing the experiment.
Match each situation given in Column I with the result given in Column II.
Column I Column II

If sodium light is replaced by 1 All fringes are colourec

A.
red light of same intensity except central fringe

Monochromatic light is 2 Fringe width will

B.
replaced by white light became quadrupled
 Distance between slits and
3 The bright fringe will
C. screen is doubled and the
bocame less bright
distance betwen slits is halved

If one of the slits is covered 4 The fringe width will

D.
by cellophone paper increase

A B C D
(a) 4 1 2 3
(b) 1 2 3 4
(c) 2 3 1 4
(d) 3 2 4 1
Passage Based Questions
Directions (Q.Nos. 172-173) Answer the following questions based on the given
passage. Choose the correct options from those given below.
The figure represents a wavefront emanating from a point source.

172. The phase difference between the two points P and Q on the wavefront is
(a) π /2
(b) 0
(c) π /3
(d) Data insufficient
173. The amplitude of point P and Q on the wavefront is
(a) same
(b) different
(c) zero
(d) Data sufficient
Directions (Q.Nos. 174-175) Answer the following questions based on the given
passage. Choose the correct options from those given below.
A point source emits wave diverging in all directions.
174. At a finite distance r from the source the shape of wavefront is
(a) spherical
(c) Either (a) or (b)
(b) plane
(d) None of these
175. At a very large distance from the source, the shape of the wavefront will be
(a) spherical
(c) Either (a) or (b)
(b) plane
(d) None of these
Directions (Q.Nos 176-178) Answer the following questions based on the given
passage. Choose the correct options from those given below.
 In the given figure,

AB is an incident wavefront and EC is refracted wavefront. Speed of light in

medium 1 is v 1 and speed of light in medium 2 is v 2.
sin ⁡i
176. The ratio of is equal to
sin ⁡r
BC
(a)
AC
AE
(b)
EC
BC
(c)
AE
AE
(d)
BC
177. When light travelling through medium 1 , passes through medium 2, which of the
following statements is correct.
sin ⁡i v 2
(a) =
sin ⁡r v 1
C
(b) η=
v1
v2
(c) η21=
v1
v1 v2
(d) =
λ 1 λ2
178. In case of refraction of a light beam, which of these remains constant?
(a) Speed
(b) Wavelength
(c) Frequency
(d) Intensity
Directions (Q. Nos 179-180) Answer the following questions based on the given
passage. Choose the correct options from those given below.
Monochromatic light of wavelength 589 nm in incident from air on a water surface.
λ 1 and λ 2 are the wavelength of reflected and refracted light respectively, v 1 and v 2
are the velocities of reflected and refractive light, respectively.
[Refractive index of water is 1.33]
v1
179. The velocities of reflected and refracted light is v 1 and v 2, respectively. Then, is
v2
(ratio of frequency) is
(a) 1
(b) 2
 (c) 3
(d) 4
180. Match the items in Column I with terms in Column II and choose the correct option
from the codes given below.

Directions (Q.Nos 181-182) Answer the following questions based on the given
passage. Choose the correct options from those given below.
The expression for Doppler's shift is given by
Δ v v radial
=
v c
[Consider the directions from observer to source as positive]
181. Here, v radial refers to
(a) the component of the source velocity along the line joining the source to
observer
(b) the component of the source velocity along the line joining the observer to the
source relative to the observer
(c) the frequency of light as observed by the observer
(d) None of the above
182. Statement I The Doppler's shift expression is valid only when the speed of the
source is small compared to that of light.
Statement II Doppler's effect in light can be used to estimate the velocity of
aeroplanes, rockets submarines etc.
(a) Both the Statements are correct
(b) Both the Statements are incorrect
(c) Statement I is correct, Statement II is incorrect
(d) Statement I is incorrect, Statement II is correct

Directions (Q. Nos 183-184) Answer the following questions based on the given
passage. Choose the correct options from those given below.
The string shown above is given an up and down jerk at one end of it while the
outer end is fixed at origin.
183. If the string always remains confined to the XY -plane, then it represents
(a) a plane olarized wave
(b) an olarizedd wave
(c) linearly olarized wave
(d) Both (a) and (c)
184. If the plane of the vibrations of the string is changed randomly in a very short
intervals of time, it is known as
(a) polarized wave
(c) polarized wave
(b) plane polarized wave
(d) Both (a) and (b)
Direction (Q. Nos. 185-186) Answer the following questions based on the given
passage. Choose the correct options from those given below.
Light passes through two polaroids P1 and P2 with pass axis of P2 making an angle θ
with the pass axis of P1.
185. The value of θ for which the intensity of emergent light is zero, is
(a) 45 ∘
(b) 90∘
(c) 60∘
(d) 30∘ 186. A third polaroid is placed between P1 and P2 with its pass axis making
an angle β with the pass axis of P1. The value of β for which the intensity of light
I
from P2 is 0 , where I 0 is the intensity of light on the polaroid P1 is
8
∘
(a) 0
(b) 30∘
(c) 45 ∘
(d) 60∘
More than One Option Correct
187. In the Young's double slit experiment, the ratio of intensities bright and dark
fringes is 9 . This mean that
(a) the intensities of individual sources are 5 and 4 units respectively
(b) the intensities of individual sources are 4 and 1 units respectively
(c) the ratio of their amplitudes is 3
(d) the ratio of their amplitudes is 2
188. A thin film of thickness t and index of refraction 1.33 coats a glass with index of
refraction 1.50. Which of the following thickness t will not reflect normally incident
light with wavelength 640 nm in air?
(a) 120 nm
(b) 240 nm
(c) 360 nm
(d) 480 nm
189. Two sources S1 and S2 of intensity I 1 and I 2 are placed infront of a screen Fig. (a).
The pattern of intensity distribution seen in the central portion is given by Fig. (b).
 In this case, which of the following statements are true?
(a) S1 and S2 have the same intensities
(b) S1 and S2 have a constant phase difference
(c) S1 and S2 have the same phase
(d) S1 and S2 have the same wavelength
190. Consider the sunlight incident on a pinhole of width 103 Å . The image of the pinhole
seen on a screen shall be
(a) a sharp white ring
(b) different from a geometrical image
(c) a diffused central spot, white in colour
(d) diffused coloured region around a sharp central white spot
191. For light from a point source,
(a) the wavefront is spherical
(b) the intensity decreases in proportion to the distance squared
(c) the wavefront is parabolic
(d) the intensity at the wavefront does not depend on the distance
[NCERT & NCERT Exemplar Questions
NCERT
192. Monochromatic light of wavelength 589 nm is incident from air on a water surface.
What are the wavelength, frequency and speed of (i) reflected and (ii) refracted
light?(Refractive index of water is 1.33).
(a) Reflected light - 589 ×10−9 m, 5.09 ×1014 Hz , 3 ×10 8 ms−1 Refracted light
−7 −1
−4.42×10 m ,5.09 ×10 Hz, 2.25 ×10 ms
14 8

(b) Reflected light −475× 10−9 m , 509× 10−14 Hz , 2 ×105 ms−1. Refracted light -
−5 14 8 −1
5 ×10 m, 2.09 ×10 , 3 ×10 ms
(c) Reflected light −1 m ,1 ms−1 Refracted light - 1 m ,2 Hz ,3 ×10 6 ms−1
(d) None of the above
193. In a Young's double slit experiment, the slits are separated by 0.28 mm and 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 cm .
Determine the wavelength of light used in the experiment.
(a) 5 ×10−7 m
(b) 6 ×10−7 m
(c) 0.05 ×10−7 m
(d) 0.06 ×10−7 m
194. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain
interference fringes in a Young's double slit experiment with slit width ¿ 2 mm and
distance of screen ¿ 1.2 m .
(i) Find the distance of the third bright fringe on the screen from the central
maximum for wavelength 650 nm.
(ii) What is the least distance from the central maximum where the bright fringes
due to both the wavelength coincide.
(a) 1.17 ×10−3 m, 1.56 ×10−3 m
 (b) 2.25 ×10−2 m, 1.25 m
(c) 0.05 ×10−2 m, 1.3 ×10−4 m
(d) None of the above
195. Light of wavelength 5000 Å falls on a plane reflecting surface. What are the
wavelength and frequency of the reflected light?
For what angle of incidence is the reflected ray normal to the incident ray?
(a) λ=3000 Å , v=5 ×1015 Hz , ∠ i=45 ∘
(b) λ=5000 Å , ν=6 ×10 14 Hz ,∠ i=45∘
(c) λ=8800 Å , v=5 ×1015 Hz , ∠ i=60∘
(d) None of the above
196. The 6563 Å H α sign line emitted by hydrogen in a star is found to be red shifted by
15 Å . Estimate the speed with which the star is receding from the earth.
(a) −5.04 × 102 ms−1
(b) −6.86 ×10 5 ms−1
(c) 5.84 × 102 ms−1
(d) 8.8 ×10 3 ms−1
197. In Young's double-slit experiment using light of wavelength 600 nm, the angular
width of a fringe formed on a distant screen is 0.1∘. What is the spacing between
the two slits?
(a) 1 m
(b) 1.5 ×10−2 m
(c) 3.44 × 10−4 m
(d) 0.05 ×10−2 m
198. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m
above a hill halfway between the towers. What is the longest wavelength of radio
waves which can be sent between the towers without appreciable diffraction
effects?
(a) 0.125 m
(b) 2.5 m
(c) 0.05 m
(d) 0 m
199. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the
resulting diffraction pattern is observed on a screen 1 m away. It is observed that
the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the
width of the slit.
(a) 2 mm
(b) 1 mm
(c) 0.2 mm
(d) 0.1 mm
NCERT Exemplar
200. Consider 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
201. Consider a ray of light incident from air onto a slab of glass (refractive index n ) of
width d , at an angle θ . The phase difference between the ray reflected by the top
surface of the glass and the bottom surface is
4 πd 1 2 1/ 2
(a) 1− 2 sin ⁡θ + π
λ n
4 πd 1 2 1/ 2
(b) 1− 2 sin ⁡θ
λ n
4 πd 1 2 π
(c) 1− 2 sin ⁡θ +
λ n 2
4 πd 1 2 1/ 2
(d) 1− 2 sin ⁡θ + 2 π
λ n
202. 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 for red distinct from that for blue
(c) there shall be no distinct interference fringes
(d) there shall be an interference pattern for red mixing with one for blue
203. Figure shows a standard two slit arrangement with slits S1 , S 2 , P1 , P2 are the two
minima points on either side of P as shown in figure.

At P2 on the screen, there is a hole and behind P2 is a second 2-slit arrangement

with slits S3 , S 4 and a second screen behind them.
(a) There would be no interference pattern on the second screen but it would be
lighted
(b) The second screen would be totally dark
(c) There would be a single bright point on the second screen
(d) There would be a regular two slit pattern on the second screen
204. The human eye has an approximate angular resolution of φ=5.8 ×10−4 rad and a
typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch ¿ 2.54 cm ). At
what minimum distance z should a printed page be held so that one does not see
the individual dots?
(a) 14.5 cm
(b) 12.5 cm
(c) 19.8 cm
(d) 10.25 cm 205. Consider a two slits interference arrangements(figure) such that
the distance of the screen from the slits is half the distance between the slits.
Obtain the value of D in terms of λ such that the first minima on the screen falls at
a distance D from the centre O .
 (a) 0.358 λ
(b) 0.404 λ
(c) 0.725 λ
(d) 0.80 λ
206. Four identical monochromatic sources A , B ,C , D as shown in the figure, produce
waves of the same wavelength λ and are coherent. Two receiver R1 and R2 are at
great but equal distances from B

Which of the two receivers picks up the larger signal?

(a) R1
(b) R2
(c) R1 and R2
(d) None of these
207. To ensure almost 100 % transmittivity, photographic lenses are often coated with a
thin layer of dielectric material. The refractive index of this material is
intermediated between that of air and glass (which makes the optical element of
the lens). A typically used dielectric film is MgF 2 (n=1.38). What should the thickness
of the film so that at the centre of the visible spectrum (5500 Å) there is maximum
transmission?
(a) 5000 Å
(b) 2800 Å
(c) 1000 Å
(d) 725 Å
208. A small transparent slab containing material of μ=1.5 is placed along A S2 (figure).
What will be the distance from O of the principal maxima and of minima on either
side of the principal maxima obtained in the absence of the glass slab?
 AC=CO=D , S1 C=S 2 C=d ≪ D
2D 5
(a) above point O and below point O
√247 √231
(b) 3 √ 247 above point O and 5 √ 231 below point O
5 3D
(c) below point O and below point O
√ 465 √247
(d) None of the above

Answers
1. (d) 2. (d) 3. (c) 4. (c) 5. (d) 6. (b) 7. (b) 8. (c) 9. (c) 10. (a)
11. (b) 12. (c) 13. (d) 14. (a) 15. (a) 16. (b) 17. (b) 18. (a) 19. (b) 20. (a)
21. (a) 22. (c) 23. (d) 24 (d) 25. (d) 26. (b) 27. (a) 28. (a) 29. (c) 30. (c)
31. (c) 32. (b) 33. (d) 34. (c) 35. (b) 36. (c) 37. (a) 38. (d) 39. (d) 40. (b)
41. (b) 42. (a) 43. (b) 44. (a) 45. (c) 46. (d) 47. (c) 48. (b) 49. (b) 50. (a)
51. (b) 52. (b) 53. (c) 54. (a) 55. (d) 56. (d) 57. (c) 58. (a) 59. (c) 60. (b)
61. (d) 62. (c) 63. (d) 64. (b) 65. (a) 66. (d) 67. (d) 68. (b) 69. (a) 70. (b)
71. (a) 72. (b) 73. (a) 74. (c) 75. (c) 76. (b) 77. (c) 78. (b) 79. (c) 80. (c)
81. (b) 82. (d) 83. (d) 84. (b) 85. (c) 86. (d) 87. (b) 88. (a) 89. (c) 90. (c)
91. (b) 92. (c) 93. (c) 94. (d) 95. (a) 96. (c) 97. (a) 98. (d) 99. (b) 100. (b)
101. (b) 102. (a) 103. (a) 104. (a) 105. (a) 106. (c) 107. (b) 108. (d) 109. (a) 110.
(a)
111. (a) 112. (d) 113. (b) 114. (b) 115. (b) 116. (b) 117. (a) 118. (a) 119. (a) 120.
(d)
121. (a) 122. (b) 123. (c) 124. (d) 125. (c) 126. (c) 127. (a) 128. (b) 129. (a) 130. (b)
131. (b) 132. (d) 133. (a) 134. (b) 135. (c) 136. (a) 137. (a) 138. (a) 139. (b) 140.
(b)
141. (c) 142. (a) 143. (c) 144. (c) 145. (a) 146. (c) 147. (a) 148. (c) 149. (b) 150. (a)
151. (c) 152. (a) 153. (a) 154. (c) 155. (a) 156. (b) 157. (c) 158. (a) 159. (c) 160. (b)
161. (a) 162. (a) 163. (d) 164. (c) 165. (c) 166. (c) 167. (c) 168. (c) 169. (c) 170. (a)
171. (a) 172. (b) 173. (a) 174. (a) 175. (b) 176. (c) 177. (d) 178. (c) 179. (a) 180.
(a)
181. (b) 182. (a) 183. (d) 184. (c) 185. (b) 186. (c) 187. (b,d) 188. (b,d) 189. (a,b,d)
190. (b,d)
191. (a,b) 192. (a) 193. (b) 194. (a) 195. (b) 196. (b) 197. (c) 198. (a) 199. (c) 200.
(a)
201. (a) 202. (c) 203. (d) 204. (a) 205. (b) 206. (b) 207. (c) 208. (a)
 Hints and Explanations
1. (d) In geometrical optics a ray is defined as the path of the energy propagation in
the limit of wavelength tending to zero.
2. (d) A ray is defined as the path of energy propagation in the limit of wavelength
tending to zero. It travels in a straight line and defined as the path of energy
propagation.
3. (c) The phenomenon of polarisation is based on the fact that the light waves are
transverse electromagnetic waves. Diffraction and interference can be explained
by wave theory of light.
4. (c) Every point on a given wavefront act as a secondary source of light and emits
secondary wavelets which travels in all directions with the speed of light in the
medium. A surface touching all these secondary wavelets tangentially in the
forward direction, gives new wavefront at that instant of time.
5. (d) Huygens' construction does not explains quantisation of energy and it is not
able to explain emission and absorption spectrum.
6. (b) Wavefront is a surface perpendicular to a ray but a wavefront moves in the
direction of the light.
7. (b) Wavefronts emitting from a point source are spherical wavefronts.

8. (c) Rays reaching from a source at infinity are parallel and when we draw a surface
perpendicular to each ray, we get a plane wavefront.

9. (c) In Huygens' wave theory, the locus of all points in the same state of vibration is
called a wavefront.
10. (a) According to Huygens' principle, each point of the wavefront is the source of a
secondary disturbance and the wavelength emanating from these points spread
out in all directions with the speed of the wave.
11. (b) When ABC wavefront passes through glass, its velocity is reduced.
As, points A and C remain in glass for a short duration, they move for a larger distance
and B covers a small distance as it remains in glass for a longer duration (middle
portion of glass is thick) and finally A' B' C' is position of new wavefront. It is
concave in shape.
12. (c) Wavelength is dependent on refractive index medium by,
λ1 μ 2
=
λ2 μ 1
So, in denser medium, μ2 > μ1 so λ 1> λ2 (i.e. wavelength decreases as the light travels from
rarer to denser medium)
∵ c=vλ
13. (d) We define an angle i c by the following equation.
n2
sin ⁡i c =
n1
Thus, if i=i c , then sin ⁡r =1 and r =90 . Obviously for i>i c, there cannot be any refracted
∘

wave. The angle i c is known as the critical angle and for all angles of incidence
greater than the critical angle, we will not have any wavefront in medium 2.
14. (a) According to Doppler's effect, wherever there is relative motion between source
and observer, the frequency observed is different from that given out by source.
15. (a) When source moves away from the observer, frequency observed is smaller
than that emitted from the source and (as if light emitted is yellow but it will be
observed as red) this shift is called red shift.
16. (b) For small velocities compared to the speed of light. The fractional change in
frequency Δ ν /ν is given by −v radial /c , where, v radial is the component of the source
velocity along the line joining the observer to the source relative to the observer,
v radial is considered positive when the source moves away from the observer. Thus,
the Doppler's shift can be expressed as
Δ v −v radial
=
v c
17. (b) From Snell's law of refraction
sin ⁡i μ 2 v 1
= =
sin ⁡r μ 1 v 2
Given,
8 −1
v 1=3 × 10 ms
μ2 1 v 1 3× 108 −1
= μ2=1.5 ⇒ v 2= 1 = ms
μ1 μ2 1.5
8
3 ×10 8 −1
∴ Speed of light, v 2= =2× 10 ms
3 /2
18. (a) We know from Cauchy's expression,
 b c
μ (λ) ¿ a+ + +..
λ λ4
2

or λred ¿ λ blue > λ violet

μred ¿ μ blue > μ violet
(for glass prism)
So, refractive index of prism for violet colour is more hence from Eq. (i), the velocity of
violet colour in medium 2 ( v 2 ) will be less than the red colour. Red colour light will
travel fastest in glass prism.
19. (b) The relation between v , c and λ is vλ=c
For small changes in v and λ
Δv −Δ λ −v radial
¿ =
v λ c
as or Δ λ ¿ 475.6−475.0=+0.6 nm
0.6 0.6
v radial ¿c = × 3 ×108
475 475
¿ ¿
v
20. (a) Using, Δ λ= λ
c
Here,
−1 3 −1
v=50 kms =50× 10 ms
c=3 ×108 ms−1
Wavelength, Δ λ=0.50 Å
8
c 3 ×10
∴ Wavelength, λ= Δ λ= 3
×0.50=3000 Å
v 50 ×10
21. (a) Here, λ=400 nm
Δ λ ¿ 400.1 nm−400 nm=0.1 nm
vs Δλ
¿
as c λ
Δλ 0.1 nm
vs ¿ c= ×3 × 108 ms−1
λ 400 nm
¿ ¿
22. (c) Doppler's shift is given by
Δ v v radial
3
Δ v 3 ×10 −5
= ⇒ = 8
=10
v c v 3 ×10
23. (d) The displacement produced by the source S1 at the point P is given by
y 1=a cos ⁡ωt
The displacement produced by the sources S2 (at the point P ¿ is also given by
y 2=a cos ⁡ωt
Thus, the resultant of displacement at P would be given by
y= y 1+ y 2 =2 a cos ⁡ωt
Since, the intensity is the proportional to the square of the amplitude, the resultant
intensity will be given by
I =4 I 0
where, I 0 represents the intensity produced by each one of the individual waves.
24. (d) Interference is a wave phenomenon. Longitudinal waves like sound, transverse
waves like wave on a string or electromagnetic waves like light show interference.
25. (d) For constructive interference,
Phase difference (Δ φ)=2 nπ (even multiple of π ¿
 For n=0 , Δ φ=0
For n=1 , Δ φ=2 π
For n=2 , Δ φ=4 π and so on
3
26. (b) Given, path difference (Δ x )=1.5 λ= λ
2
Phase difference (Δ φ) and path difference (Δ x ) are related by the relation,
2π
⇒ Δφ ¿ ×Δx
λ
2π 3
⇒ Δφ ¿ × λ=3 π
λ 2
Δφ ¿ ¿
So, destructive interference occurs.
27. (a) Given, separation between sources S1 and S2=2 λ . For minimum intensity at P,
destructive interference must take place at P.

So, S1 P−S2 P=Δ x

(path difference)
λ
¿(2n+ 1) ( for destructive interference )
2
For minimum distance,
3λ λ
S1 P−S2 P= ≠
2 2 √x +¿ ¿
2

¿
λ 15 λ 7λ
Note If we proceed with Eq. (i) taking S1 P−S2 P= , x= which is more than . 28. (
2 4 12
a) From the figure, path difference ¿ S1 M =P

(∵ x≪ R)
For maximum intensity, P=nλ
⇒ x cos ⁡θ=nλ
nλ
⇒ cos ⁡θ=
x
nλ $$
⇒ cos ⁡θ=
5λ
n
⇒ cos ⁡θ=
5
(\because x=5 \lambda)
$$
We know, −1 ≤cos ⁡θ ≤1
n
⇒ −1 ≤ ≤1
5
⇒ −5≤ n ≤ 5
Possible values of n={0 , ± 1 ,± 2 ,± 3 , ± 4 , ± 5}
Let us analysis each value of n for θ in range.
θ ∈(0 , 2 π )
1
For n=1 ,cos ⁡θ=
5

Here, negative value of n means the path difference ( S1 P−S2 P ) is negative, i.e., for those
points S1 P< S 2 P .
For n=0 , ±1 , ±2 , ± 3 ,± 4,
From the given graph of cosine function, it can be observed that in interval θ ∈ 0 ,2 π ¿ for
above values of n there are in total 18 points, i.e., 2 points for n=0 , 4 points each for
n=±1 , ± 2 ,± 3 , ± 4.
For n=+5 , cos ⁡θ=+1,
One value of θ i.e., θ=0∘ is possible as for θ=2 π , the points will coincide.
For n=−5 , cos ⁡θ=−1, i.e., θ=π .
Thus, in total 20 points of maxima's are possible in all 4 quadrants.
29. (c) The position of farthest minimum detection occurs when the path difference
λ
is least and odd multiple of , i.e.,
2
condition for destructive interference and approaches zero as P moves to infinity.

So, if S2 P=D
 λ
S1 P−S2 P ¿(2n+ 1) for destructive interference.
2
¿ ¿
λ
so S1 P−S2 P=
2
λ
⇒ √ D +d −D=
2 2
2
2
2 2 λ
⇒ D +d =D+
2
2
2 λ
⇒ d =Dλ+
4
2
d λ
⇒ D= −
λ 4
2
( 1.0 ×10−4 m )
¿ −150× 10−9 m
( 600 × 10 m )
−9

⇒ D=1.07 m
30. (c) Given, Δ φ=100 π
We know, change in phase difference,
2π
i.e., Δ φ= ×Δx
λ
where, Δ x =¿ path difference
λ λ
⇒ Δ x=Δ φ× =100 π × =50 λ
2π 2π
31. (c) In the phenomenon of interference, energy in conserved but it is redistributed.
32. (b) As two distinct sources are incoherent, so phase changes are random, so no
fixed pattern of maxima or minima.
33. (d) Resultant amplitude
A=√ A 21+ A 22+ 2 A 1 A2 cos ⁡φ
Here
A1 ¿ A 2=2 cm ⇒φ=π rad
A ¿
34. (c) Resultant intensity is given by I =4 I 0 cos 2 ⁡φ/2.
2π
Now, phase difference (φ)= × Δ (path difference ¿
λ
As a path difference of one wavelength corresponds to a phase difference of 2 π radius.
2 2π Δ 2 π Δ
⇒ I ¿ 4 I 0 cos ⁡ =4 I 0 cos ⁡
2λ λ
IP 2 π Δ1 2 π Δ2
∴ ¿ cos ⁡ /cos ⁡
IQ λ λ
I max 4
35. (b) Given, =
I min 1
I max r +1 4
We know, = =
I min r−1 1
r +1 2
⇒ = ⇒ r +1=2 r−2 or r =3
r−1 1
A1
∴ The ratio of amplitudes =r=3
A2
36. (c) Given, I 1=I and I 2=9 I
 I max
⇒
I1
I2
I 1
9I 9
A √I
A2 √ I 2
1 1
¿ = ⇒ r= 1 = 1 = =
9 3 √
r +1
¿
I min r−1
¿ ¿
37. (a) Given, Young's double slit experiment, having two slits of width are in the ratio
of 1 :25.
So, ratio of intensity,
I1 W 1 1 I 2 25
= = ⇒ =
I 2 W 2 25 I 1 1

∴
2
I max ( √ I 2 + √ I 1)
=
√
=
I2 2
I1
+1

√
I min ( √ I − √ I )2 ) I2
2 1
−1
I1
2 2
5+1 6 36 9
⇒ = = =
5−1 4 16 4
I 9
Thus, max =
I min 4

38. (d) Given,

I1
I2
I
√
=n ⇒ r = 1 =√ n
I2

'
¿I0 Here, ¿ 4 I0
39. (d) We know, I max=( √ I 1 + √ I 2 ) =4 I 0 ( if I 1 =I 2=I '0 )
2 '
' I0
⇒ ¿ I0 ¿
4
For incoherent source, the interference has uniform intensity throughout given by
I =I 1 + I 2
I I
I =2 I '0 =2× 0 = 0
4 2
40. (b) When the sources are incoherent there is no interference and resultant
intensity is I 1+ I 2. For sources of same intensity I 0, resultant intensity will be 2 I 0 .
41. (b) Two identical and independent sodium lamps (i.e., two independent sources of
light) can never be coherent. Hence, no coherence between the light emitted by
different atoms.
42. (a) Let the average intensity be I av.
The amplitude of intensity variation means.
 I =I av ±0.05 I av
⇒ I max =I av (1+ 0.05)=1.05 I av
⇒ I min =I av (1−0.05)=0.95 I av
I max 1.05 105
So, = =
I min 0.95 95
2
r + 1 105
⇒ =
r −1 95
⇒ ( r +1+2 r ) 95=105 ( r +1−2 r )
2 2

2
⇒ 10 r +10−200⋅ 2r =0
⇒ 10 r 2 −400 r +10=0
⇒ r 2−40 r +1=0
¿
43. (b) Consider that the source is moved to some new point S' and suppose that Q is
the mid-point of S1 and S2. If the angle S' QS is φ , then the central bright fringe
occurs at an angle −φ , on the other side. Thus, if the source S is on the
perpendicular bisector, then the central fringe occurs at O , also on the
perpendicular bisector.
If S is shifted by an angle φ to point S' , then the central fringe appears at a point O' at an
angle −φ , which means that it is shifted by the same angle on the other side of the
bisector.
This also means that the source S' , the mid-point Q and the point O' of the central fringe
are in a straight line. 44. (a) The film appears bright if the path difference is
λ
2 μt ¿(2 n−1) , where, n=1 ,2 , 3 , …
2
4 μt
∴λ ¿
(2 n−1) $$
−10
4 × 1.4 ×10000 ×10 56000
λ ¿ = Å
(2 n−1) (2 n−1)
∴ λ ¿56000 Å , 18666 Å , 11200 Å , 8000 Å ,
6222 \AA, 5091 \AA, 4308 \AA, 3733 \AA
$$
The wavelengths which are not within specified ranges produce minima.
45. (c) Path difference ¿ QX −PX=(n+2)λ−nλ=2 λ
For constructive interference or bright band,
Path difference ¿ Δ x=nλ ¿ where, n=1 ,2 , … ¿ From Eq. (i), it is obvious that second bright
band is formed as n=2.
46. (d) We know,
Intensity of bright band, I max=( √ I 1 + √ I 2 )
2

Intensity of dark band, I min =( √ I 1− √ I 2 )

2

Case I When there is no glass slab

⇒ I 1=I 2=I 0 or
I max =4 I 0 (Complete brightness)
and I min =0
(Complete darkness)
Case II When glass slab is inserted,
I
⇒ I 1 < I 2 or I 1= 0 and I 2=I 0
2
or
 I max < 4 I 0
I min >0
[from Eq. (i)]
and
[from Eq. (ii)]
Hence, the bright band becomes less bright and dark band becomes less dark.
2 Δφ
47. (c) Using relation, I =I max cos ⁡ ,
2
where, Δ φ=¿ total phase difference
I
Given, I = max at certain point
4
I max 2 Δφ 1 2 Δφ
⇒ =I max cos ⁡ or =cos ⁡
4 2 4 2
Δφ π 2π
⇒ cos ⁡ =cos ⁡ or Δ φ=
2 3 3
λ λ 2π λ
Path difference, Δ x = × Δ φ= × =
2π 2π 3 3
For Young's double slit experiment we know, path difference ¿ d sin ⁡θ
where, θ=¿ angular separation of the point
Using Eqs. (i) and (ii), we get
λ λ
¿ d sin ⁡θ ⇒ sin ⁡θ=
3 3d
−1 λ
⇒θ ¿ sin ⁡
3d
48. (b) Shift produced due to insertion of slab
μg D
Δ x ¿ t −1
μm d
¿ ¿
Thus, the central maximum is obtained at a distance 4.33 mm below point O on the
screen as the slab is placed in the path of lower slit.
μg
49. (b) At O , path difference, P= −1 t
μm
For maximum intensity at O
P
∴ λ=
n
μg 10.4 ×10 nm λ=1300 nm¿ For n=2 , ¿ λ=650 nm ¿ For n=3 , ¿ λ=433.33 nm ¿
3
t 1.5
or λ= −1 = −1 ×
μm n 4 /3 n
¿ ¿
$$
P=n \lambda(n=1,2,3, \cdots)
$$
Thus, the wavelength in the range 400 to 700 nm are 650 nm and 433.33 nm.
50. (a) The condition for minimum thickness corresponding to a dark band in reflection
2 μt cos ⁡r =λ
−10
λ 6000 × 10
∴ t= = =4000 Å
2 μ cos ⁡r 2 ×1.5 × cos ⁡60∘
51. (b) The resultant internsity
2 φ
I =I 0 cos ⁡
2
 π
Here, I 0 is the maximum intensity and φ=
2
2 π 2 π
I =I 0 cos ⁡ =I 0 cos ⁡
2 ×2 4
I0
I=
2
52. (b) Here, A1=2 A , A2=2 A , φ=60∘
R ¿ √ A 1+ A 2+ 2 A 1 A2 cos ⁡φ
2 2

¿ ¿
as intensity ∝ ¿
2
Therefore, I ∝ 12 A
54. (a) Fringe spacing
−7
Dλ 1 ×5 ×10
m ( 1 nm=10 m)
−9
β ¿ = −3
d 1 ×10
¿ ¿
nλD
55. (d) Position of n th bright fringe from central maxima is .
d
8 λ1 D 9 λ2 D
∴ ¿
d d
λ1 9
⇒ ¿=
λ2 8
Hence, the possible wavelengths of visible light is of the ratio of 9 :8 .
56. (d) λ 1=6000 Å , n1=16 fringes and n2 =24 fringes
nDλ
Position of nth fringe, ⇒ n 1 λ1 =n2 λ2
d
λ1 n2 6000 24
⇒ = ⇒ =
λ 2 n1 λ2 16
6000 × 16 96000
⇒ λ 2= = =4000 Å
24 24
57. (c) Suppose slit width are equal, so they produces wave of equal intensity say I '.
Resultant intensity at any point I R =4 I ' cos 2 ⁡φ , where φ is the phase difference
between the moves at the point of observation. For maximum intensity.
'
φ=0 ⇒ I max =4 I =I
Also, when one slit is closed
'
I =I 0
From Eqs. (i) and (ii), we get 4 I 0 =I
10 λD
58. ( a) Position of 10 th bright fringe ¿
d
10 λD
Also, ¿ 12 10 λD
d = ¿
12
⇒ ¿
The separation between the slits
−9
10 ×589.3 ×10 × 1
¿ −3
12 ×10
−4
¿ 4.9 × 10 m=0.49 mm
λ air
59. (c) μ=
λ water
Using Eq. (iii), we get
 ω1 μ water 4
= =
ω2 μ air 3
60. (b) Here, λ=500 nm , d=1 mm , D=1 m
1 λD
Distance of 3 rd minima i.e., x n=n+
2 d
1 λD
x3 ¿2+
2 d
5 λD
x ¿
2d
−4
¿ ¿ 12.5 ×10 m=1.25 mm
61. (d) Since, dark fringe is directly opposite to one of the slits,

d
∴ Distance of the dark fringe from central maxima ¿
2
λD
Position of nth dark fringe ¿ (2n−1)
2d
d λD
or ¿(2n−1)
2 2d
d2
⇒ λ ¿
D(2 n−1)
d2
⇒ For n=1 , λ ¿
D

62. (c) Position of n th maximum from central maxima

nλD
¿ ⇒ xn ∝ λ
d
So, x (blue) ¿ x (green) as
λ blue < λgreen
63. (d) Given, d=0.90 mm=0.90 ×10−3 , D=1 m Position of 2 nd dark fringe from central
fringe
3 λD −3 −6
¿ =1× 10 m= 3 × λ ×1 ¿ ∴ λ ¿=
1.8 ×10
=0.6 ×10−6 m=6 ×10−5 cm¿
2d −3
3
¿ 2× 0.90 ×10
64. (b) Given, λ 1=12000 Å and λ 2=10000 Å ,
 −3
D=2 cm and d=2 mm=2 ×10 cm.
λ1 12000 6 n 2
We have = = =
λ2 10000 5 n 1
as
n1 λ1 D 5 ×12000 ×10−10 × 2
x ¿ = −3
d 2× 10
¿ ¿
λ D1 λ D2
65. (a) The fringe width i.e., β 1= and β 2=
d d
λ −8
β 1−β 2 ¿ ( D1 −D2 ) λ 3 10
= −3 ( 5 × 10 ) ( ∵ D1−D2=5 ×10 m ) ¿ ⇒¿ λ ¿= × −2 =0.6 × 10 m=6000 Å ¿
−2 −2 −6
So, d
10 5 10
⇒ ¿
66. (d )

x
Angular position of first dark fringe ¿ tan ⁡θ ≈θ=
D
−10
λ 5460 ×10 λD
⇒θ ¿ = −3
∵ x=
2d 0.1 ×10 2d
¿ ¿
180 −5
As, we know, θ ¿ in degree ¿= ×546 ×10
π
180 −5 0.32 ∘
¿ ×7 × 546 ×10 ≈ =0.16
22 2
λD
67. (d) β=
d
β2 λ2 D2 d1
∴ =
β1 λ 1 D 1 d 2
β1 × λ 2 × 2 D1 ×d 1
⇒ β 2=
λ 1 × D1 × d 1 /2
λ
⇒ β 2=β × 2 ×4
λ1
−4
⇒ β 2=2.5 ×10 m
68. (b) Resultant intensity, I R =I 1 + I 2+ 2 √ I 1 I 2 cos ⁡Δ φ
Case I (at A ¿ Δ φ=π /2
I R =I 1 + I 2=I + 4 I =5 I
1

Case II (at B ) Δ φ=π

I R =I 1 + I 2−2 √ I 1 I 2=5 I −2 ×2 I =I $$
2

\therefore \quad I_{R_{1}}-I_{R_{2}}=5 I-I=4 I

$$
69. (a)
So, distance between two slits i.e., S1 and S2
d=(2 tan ⁡θ /2) D
For small angles θ , tan ⁡θ ≈ θ
θ D 1
⇒ d=2× × D=Dθ or =
2 d θ
λD λ
Fringe width, β= =
d θ
70. (b) Length of segment ¿ constant
⇒ n 1 ω 1 ¿ n2 ω2 ⇒ n1 λ1=n2 λ2
λ 600
or n 2 ¿ n1 1 =12 × =18
λ2 400
nλ D
71. (a) In liquid position of 10th bright fringe, x n= l
d
10 λl D
⇒ x=
d
where, λ l=¿ wavelength in liquid.
λD
∵ Position of dark fringe ¿(2n−1)
2d
11 λair D
In vacuum position of 6 th dark fringe ¿
2d
[put n=6 in Eq (i)]
Since, 10th bright fringe in liquid is located at 6th dark fringe in air,
10 λl D 11⋅ λair D λ l 5.5
⇒ = ⇒ =
d 2d λair 10
Also,
λl μair 1 5.5
= ⇒ =
λair μl μ l 10
or
10 20
μl = = =1.8
5.5 11
72. (b)
2 φ
I =I max cos ⁡
2
I
Given, I = max
2
 π 3π 5π
∴ From Eqs. (i) and (ii), we have, φ= , ,
2 2 2
λ
Or path difference, Δ x = ⋅φ
2π
λ 3 λ 5 λ 2 n+1
∴ Δ x= , , … λ
4 4 4 4
n=0 , 1, 2 , … ..
73. (a) Fringe width
−9
λD 600 ×10 × 2 −4 −2
i.e., ¿ β= = =6 ×10 m 2 πx 2 π 0.5 ×10 100
d 2 ×10
−3
=I max cos ⁡ ¿ ⇒ ¿ I ¿=0.20 cos ⁡ ¿ ⇒ ¿ I ¿=0.20 cos2 ⁡
β 6 ×10
−4
12
Using, ¿
λD
74. (c) As we know, fringe width β 1=
d
2 D λD
β2 ¿ λ =
2d d
β1 ¿ β2
75. (c) Let n th fringe of 1500 Å coincide with (n−2) th fringe of 2500 Å
∴ 1500 ×n ¿ 2500×(n−2)
3n ¿ 5(n−2)
3n ¿ 5 n−10
2n ¿ 10
⇒n ¿5
(n−2) ¿3
∴ 5 th order of 1 st and 3 rd order of 2 nd .
76. (b) In YDSE, as we know
2 Δφ
Intensity, I ¿ I max cos ⁡
2
I max 2 Δφ
⇒ ¿ I max cos ⁡
2 2
2 Δ φ 1 Δ φ 2n+ 1
cos ⁡ ¿ ⇒ = π ,n=0 , 1 ,2 , …
2 2 2 4
2n+1
Δφ ¿ π
2
Δφ
⇒ Δx ¿ ×λ
2π
¿ ¿
λD
77. (c) Fringe width, β=
d
According to the question,
' D '
D ¿ and d =5 d
2
D
' λ
' D λ 2 1 Dλ
∴β ¿ ' = =
d 5 d 10 d
' β
β ¿
10
−4
78. (b) Here, β 1=3.2 ×10 m
 λ1 ¿5600 Å , λ2=4200 Å
β2 λ 4200
¿ 2=
β1 λ 1 5600
6
or β2 ¿ × β1
8
¿ ¿
Decrease in fringe width,
Δ β ¿ β1 − β 2
¿ ¿
2 φ 2π
79. (c) For net intensity, I =4 I 0 cos ⁡ φ= ×λ
2 λ
For the first case,
2
K=4 I 0 cos ⁡(π )
K=4 I 0
For the second case,
' 2 π /2 2π λ
K ¿ 4 I 0 cos ⁡ φ= × =2 I ¿
2 λ 4 0
¿ ¿
Comparing Eqs. (i) and (ii),
'
K =K /2
80. (c) The contrast interference will occur when there is absolute darkness at the
dark band due to destructive interference i.e., I R =I min =0 and there is complete
(max.) brightness at the bright band due to constructive interference i.e.,
I R =I max =4 I 0, which is possible only when individual intensities are same,
So, I 1=I 2=I 0
81. (b) Contrast between the bright and dark fringes will be reduced.
82. (d) By using white light, the central maxima will be white while the fringes closest
on either side of central fringe is red and farthest will appear blue.
83. (d) Diffraction is observed when slit width is of the order of wavelength of light (or
any electromagnetic wave) used. ∴ λ X -rays (1−100 Å)≪ slit width (0.6 mm)
⇒ So, no pattern of diffraction will be observed.
84. (b) As, the path difference aθ is λ ,
λ
then θ=
a
10 λ 2 λ d 10
⇒ = ⇒ a= = =0.2 mm
d a 5 5
So, the width of each slit is 0.2 mm .
85. (c) The direction in which the first minima occurs is θ (say).
λ
Then, e sin ⁡θ= λ or eθ=λ or θ=
e
¿
2 λb
Width of the central maxima ¿ 2 bθ+e= ±e
e
86. (d) Given, λ=6000 Å=6000 ×10−10 m, d=0.3 mm
For minima, d sin ⁡θ=mλ
First minima means (m=1),
λ
⇒ sin ⁡θ=
d
Angular position of 1 st minima,
 −10
λ 6000 ×10 −3
sin ⁡θ=θ= = −3
=2 ×10 rad
d 0.3 ×10
So, angular position of first minima is 2 ×10−3 rad .
87. (b) Given, λ=589 nm=589 ×10−9 m , D=1.0 m
Slit width ¿ d=¿ ?
Here, given the distance between two dark fringes (i.e., dark fringes for m=± 1 )
¿ width of central maximum ¿ 2 y=2.2 mm
dy
or y=1.1 mm=1.1 ×10−3 m Using, for zero intensities path difference ¿ mλ , we have =mλ
D
mλD (1) ( 589 ×10 m ) ×(1.0 m)
−9
Slit width i.e., d= = −3
y 1.1× 10 m
¿ 0.54 mm
88. (a) Width of central maximum
2 λD
( Δ y 0 )=2 y= d
Width of 1 st order secondary maxima =Distance between D1 and D2 (consecutive dark
bands) ¿ y 2− y 1

For secondary minima (or dark band) path difference ¿ d sin ⁡θ=mλ (where, m=1 , 2 ,3 , … )
Position of 1 st dark band
y d λD
Path difference ¿ 1 =λ or y 1=
D d
Position of 2nd dark band
y d 2 λD
Path difference ¿ 2 =2 λ ⇒ y 2=
D d
∴ Width of secondary maxima ( Δ y 1 )
λD
⇒ Δ y 1= y 2− y 1 or Δ y1 = = y
d
Thus, width of other secondary maxima is half that of central maxima.
or
Δ y0 2
=
Δ y1 1
2π
89. (c) Phase difference ¿ Δ φ= ×(d sin ⁡θ)
λ
(for two end of slit)
For first order diffraction maximum,
 λ
d sin ⁡θ=(2 m+1) ,
2
3λ 2π 3 λ
where m=1= ⇒ Δ φ= × =3 π
2 λ 2
90. (c) For minima,
a sin ⁡θ ¿ nλ ¿(n=1)
∘
⇒ a sin ⁡30 ¿=(1)λ
a ¿ ¿
For 1st secondary maxima
3λ 3λ
⇒ a sin ⁡θ1= ⇒ sin ⁡θ1=
2 2a
Substitute value of a from Eq. (i) to Eq. (ii), we get
3λ 3
sin ⁡θ 1 ¿ ⇒ sin ⁡θ1=
4λ 4
−1 3
θ1 ¿ sin ⁡
4
92. (c) For 2nd secondary maxima using red light
( 2m+1) λ1
d sin ⁡θ1= , where m=2
2
5λ 5λ
d sin ⁡θ 1= 1 ⇒ sin ⁡θ 1= 1
2 2d
When white light is used, for position of 3rd secondary maxima (m=3)
7 λ2 7 λ2
d sin ⁡θ2= ⇒ sin ⁡θ2=
2 2d
Since, the position coincide with each other for white and red light
5 λ1 7 λ2
sin ⁡θ 1=sin ⁡θ2 ⇒ =
2d 2d
5 5
⇒ λ2= λ1= × 6500 Å
7 7
Wavelength of white light, λ 2=4642.85 Å
93. (c) Using violet light
Slit width ¿ d , λ1=400 ×10−9 m
Width of diffraction pattern (central maxima)
2 λ1 D
¿ 2 y= d −9
d = 2 ¿ λ 2 ¿=600 ×10 m¿
¿
Width of diffraction pattern(Central maxima)
2λ D
¿ 2 y'= 2
(d /2)
y λ1
From Eqs. (i) and (ii), ' =
y 2 λ2
y 400 2 1 '
⇒ '= = = ⇒ y =3 y
y 600 ×2 3 ×2 3
94. (d) Distance between the first dark fringes on either side of the central bright
fringe ¿ Width of central maxima
−9
2 λD 2 ×600 ×10 ×2
2 y= ¿ −3
d 1 ×10
¿ ¿
2λ 2 λ
95. (a) Angular width of central maxima (2 θ)= =
d e
 λ 1
⇒ 2 θ ∝ ⇒ θ ∝ (for λ= constant)
e e
Thus, on decreasing slit width (e ), then θ will increases. 96. (c) Angular width of central
2λ
maxima ¿ 2 θ= .
d
Thus, θ does not depend on D i.e., distance between the slits and the screen.
97. (a) Here, λ=4000 Å=4000 ×10−10 m=4 × 10−7 m
−3 −4
a ¿ 0.2 mm=0.2 ×10 m=2× 10 m
λ 4 × 10−7 m −3
sin ⁡θ ¿ = =2 ×10
a 2 ×10−4 m
As sin ⁡θ is very small
−3
∴ θ ≅ sin ⁡θ=2× 10 rad
x /2
98. (d) From the figure, tan ⁡θ=
2
For small θ and when θ is counted in rad , tan ⁡θ ≃ θ

2 λD
Width of central maximum ¿ =2.4 mm
d
x /2 λ x
So, θ ≃ ⇒ ≃
2 9 4
−9
4 λ 4 × 600 ×10
x ≃ ≃ −3
a 10
¿ ¿
⇒ ≃ 2.4 mm
λ
99. (b) Angular width, θ=
d
θ1 λ1 θ1 λ1
∴ = ⇒ =
θ 2 λ 2 θ 2 λ 1 /μ
θ1
⇒ =μ
θ2
θ 3
∴ θ2= 1 = × 0.2=0.15∘
μ 4
∘
100. (b) In water angular width θ w =0.2
λ
We know, θ w = water
d
Let μ water =¿ refractive index of water In air,
λair
θair =
d
On dividing Eq. (i) from Eq. (ii), we get
θ w λ water
=
θair λair
or
θw μ air 1 λair μ water
= = ∵ = $$
θair μwater μ water λ water μ air
\Rightarrow \quad \theta_{\text {air }}=\mu_{\text {water }} \theta_{w}=\frac{4}
{3} \times 0.2^{\circ} \approx 0.28^{\circ}
$$
101. (b) The angular resolution of the telescope is determined by the objective of the
telescope.
102. (a) Radius of the central bright region is approximately given by

103. (a) Thus, Δ θ will be small if the diameter of the objective is large. This implies that
the telescope will have better resolving power, if a is large. It is for this reason that
for better resolution, a telescope must have a large diameter objective.
1 D
104. (a) Resolving power of telescope (RP)= = where, D=¿ diameter of objective,
Δ θ 1.22 λ
λ=¿ wavelength of light Given, D=6 cm=6 × 10−2 m , λ=540 nm=540 ×10−9 m
−2
6 ×10 −1
⇒ RP ¿ −9
rad
1.22 ×540 ×10
¿ ¿
105. (a) Aperture of the telescope
1.22 λ
D=
dθ
−10 −6
Here λ=5600 Å=5600 ×10 m, dθ=3.2× 10 rad
1.22 ×5600 × 10−10
∴ D= ⇒ D=0.2135 m
3.2× 10−6
106. (c) The objective lens of a microscope, the object is placed slightly beyond f , so
that a real image is formed at a distance v [figure]. The magnification ratio of
image size to object size is given by m ≈ v / f . It can be seen from figure that
D/ f ≈ 2 tan ⁡β
where, 2 β is the angle subtended by the diameter of the objective lens at the focus of
the microscope.
107. (b) If the medium between the object and the objective lens is not air but a medium
of refractive index n,
1.22 λ
d min =
2 nsin ⁡β
108. (d) For compound microscope,
Resolving power
2 μ sin ⁡β
i.e., RP=
1.22 λ
(i) ∵ RP∝ μ
If the refractive index (μ) of the medium between the object and the objective lens
increases, the resolving power increases.
1
(ii) ∵ RP∝
λ
On increasing the wavelength of light used, the resolving power of microscope
decreases and vice-versa.
109. (a) In Fresnel biprism experiment, the actual distance of separation between the
two slits,
d= √ d1 d 2=√ 25 ×16=20 cm
2
a
110. (a) According to Fresnel distance, Z F i.e., Z F =
λ
2
( 3 ×10−3 )
9 ×10−6
¿ −7 −7
=
=18 m
5 ×10 5 ×10
112. (d) Light waves are transverse in nature; i.e., the electric field associated with a
propagating light wave is always at right angles to the direction of propagation of
the wave. We can say light waves are transverse electromagnetic waves.
113. (b) Polaroids can be used to control the intensity in sunglasses windowpanes, etc.
The intensity can be further controlled from 50 % to zero of the incident intensity by
adjusting the angle between the pass-axes of two polaroids.
114. (b) The phenomenon of polarisation is based on the fact that light waves are
transverse electromagnetic waves.
Light waves are transverse in nature i.e., the electric field associated with a
propagating light wave is always at right angles to the direction of propagation of
the wave.
115. (b) Ultrasonic waves being sound waves are longitudinal and hence cannot be
polarised.
116. (b) Some crystals such as tourmaline and sheets of iodosulphate of quinine have
the property of strongly absorbing the light with vibrations perpendicular to a
 specific direction (called pass axis), transmitting the light with vibration parallel to
it. This selective absorption of light called dichroism.
117. (a) Plane of vibration is perpendicular to the direction of propagation and also
perpendicular to plane of polarisation. Thus, the angle between plane of
polarisation and direction of vibration is 0∘ i.e, they are parallel.
118. (a) If an identical piece of polaroid P2 be placed before P1. As expected, the light
from the lamp is reduced in intensity on passing through P2 alone. But now
rotating P1 has a dramatic effect on the light coming from P2. In one position, the
intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90∘ from
this position, P1 transmits nearly the full intensity emerging from P2 as shown
figure.

(a)

(b)

119. (a) Suppose I 0 be the intensity of polarised light after passing through the first
polariser P1. Then, the intensity of light after passing through second polariser will
be
2
I =I 0 cos ⁡θ
where, θ is the angle between pass axes of P1 and P2. Since, P1 and P3 are crossed the
π
angle between the pass axes of P2 and P3 will be −θ .
2
Hence, the intensity of light emerging from P3 will be
2 2 π
I ¿ I 0 cos ⁡θ cos ⁡ −θ
2
¿ ¿
Therefore, the transmitted intensity will be maximum when θ=π /4 .
120. (d) Given, i+r=π /2
According to Brewster's law, we get
tan ⁡i B ¿ μ=1.5
So, i B ¿ tan ⁡(1.5)⇒ i B=57 ∘
−1

i.e., this is the Brewster's angle for air to glass interface.

121. (a) In unpolarised beam, vibrations are probable in all directions in a plane
perpendicular to the direction of propagation. Therefore, θ can have any value
from 0 to 2 π .
2π 2π
1
[ cos 2 ⁡θ ]av =
2π 0
∫ ❑ cos 2 ⁡θdθ= 21π ∫ ❑ 1+ cos
2
⁡2 θ
dθ=
1
2
0

1 I
So, using law of Malus, I =I 0 cos 2 ⁡θ ⇒ I 0=I 0 × = 0
2 2
122. (b) When an unpolarised beam of light is incident at the Brewster's angle on an
interface of two media, only part of light with electric field vector perpendicular to
the plane of incidence will be reflected. Now, by using a good polariser, if we
completely remove all the light with its electric vector perpendicular to the plane
of incidence and let this light be incident on the surface of the prism at Brewster's
angle, we will observe no reflection and there will be total transmission of light.
123. (c) In the special situation, one of the two perpendicular components of the electric
field is zero. At other angles, both components are present but one is stronger than
the other.
There is no stable phase relationship between the two perpendicular components, since
these are derived from two perpendicular components of an unpolarised beam.
When such light is viewed through a rotating analyser, one sees a maximum and a
minimum of intensity but not complete darkness. This kind of light is called
partially polarised.
124. (d )

I0 2
Intensity of light emerging from P2 is I = cos ⁡θ where, θ=∠ angle between P1 and P2
2
so,
I0 2 ∘ I0
I= cos ⁡60 =
2 8
125. (c)
According to Malus law,
I0 I 0 1 I0
cos ⁡( 45 ) = × =
2 ∘
I R=
2 2 2 4
C
126. (c) Case I Since, ray emergent from 2 has intensity.

I1 2 I1 2 ∘ I1
I ¿ cos ⁡θ= cos ⁡0 =
2 2 2
I1
or ¿ I 0 ⇒ I 1=2 I 0
2
Case II Angle between C 1 and C 2=60∘

I 1=⃗ ⃗⃗
2 I 0 C1 ⟶ C 2
2 I =I 0 cos ⁡θ
2 I0
= I0
2

1 I
2
Intensity of emergent ray ¿ I 0 cos 2 ⁡60∘=I 0 =
0
2 4
127. (a) If unpolarised light is incident at polarising angle, then reflected light is
completely i.e., 100% polarised perpendicular to the plane of incidence.
128. (b) tan ⁡i B=μ, where i B =¿ polarising or Brewster's angle
⇒ i B =tan ⁡( μ)=tan ⁡( √ 3)=60
−1 −1 ∘

−1 4
129. (a) Here, Critical angle, i c =sin ⁡
5
4
∴ sin ⁡i c ¿
5
1 5
As μ ¿= =
sin ⁡i c 4
According to Brewster's law,
tan ⁡i p =μ
where, i p is the polarising angle
5 −1 5
∴ tan ⁡i p= ⇒ i p =tan ⁡
4 4
130. (b) As reflected light is completely polarised, therefore
∘
i p=45
μ=tan ⁡i p =tan ⁡45∘=1
c
As μ=
v
c 3 ×10 8
⇒ v= = ⇒ v=3 ×108 ms−1
μ 1
131. (b) Using tan i p=μ
tan ⁡i p ¿1
ip ¿ tan ⁡( 1)=45∘
−1

As r ¿ 90∘−i p=90 ∘−45∘ ⇒ r=45∘

132. (d) Using, tan ⁡i p =μ=1.5 ⇒ i p=tan −1 ⁡(1.5)=56.3∘
133. (a) The branch of optics in which one completely neglects the finiteness of the
wavelength is called geometrical optics. The wavelength of light is very small as
compared to the dimensions of objects (such as mirror, lenses etc.) and hence, it
can be neglected and assumed to travel in a straight line.
134. (b) Reflection and refraction arise through interaction of incident light with
constituents of matter. Atoms may be viewed as oscillators, which take up the
frequency of the external agency (light) causing forced oscillations. The frequency
of light emitted by a charged oscillator equals its frequency of oscillation. Thus,
the frequency of scattered light equals the frequency of incident light.
135. (c) 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 wavefront.

136. (a) According to Huygens' principle each point of the wavefront is the source of a
secondary disturbance and the wavelets emanating from these point spread out in
all directions with the space of wave.
These wavelets emanating from the wavefront are usually referred to as secondary
wavelets and if we draw a common tangent to all these spheres, we obtain the new
position of the wavefront at a later time.
137. (a) Increase in wavelength of light when the source move away from the observer
due to Doppler's effect is called red shift. The visible regions shifts towards red
end of electromagnetic spectrum and hence called red shift.
λD
138. (a) As, we know, fringe width β i.e., ¿
d
So, smaller the distance between the slits (d ), then larger will be fringe width (β ).
Hence, single fringe will cover whole screen and pattern will not be visible.
139. (b) Given, initial phase difference ¿ φ S −φS =π
1 2

At central maximum, Δ x =0 ¿ path difference ¿ Δ x ¿ ⇒ Total phase difference

2π
¿ φ S −φS + Δx
1 2
λ
At central maximum,
2π
Δφ ¿ π+ × Δ x=π +0
λ
Δ φ ¿ π = odd multiple of π
Hence, at central maximum dark band is obtained.
140. (b)
( √ I 1 +√ I 2 )
2
I
∴ max =
I min ( √ I −√ I ) 2
2 2

141. (c) When one of slits is covered with cellophane paper, the intensity of light
emerging from the slit is decreased (because this medium is translucent).
Now, the two interferring beam have different intensities or amplitudes.
Hence, intensity at minima will not be zero and fringes will become indistinct.
142. (a) For reflecting system of the film, the condition for maxima or constructive
(2 n−1)
interference is 2 μt cos ⁡r = λ , while the maxima for transmitted system of film is
2
given by equation 2 μt cos ⁡r =nλ , where t is thickness of the film and r is angle of
refraction.
From these two equations, we can see that condition for maxima in reflected system
and transmitted system are just opposite.
143. (c) In Young's double slit experiment fringe width for dark and white fringes are
same while in the same experiment, when a white light as source is used, the
central fringe is white around while few coloured fringes are observed on either
side.
λD
144. (c) Fringe width, β= shall remain the same as the waves travel in air only, after
d
passing through the thin transparent sheet. Due to introduction of this sheet, only
path difference is changed, due to which there is shift of position of fringes only,
D(μ−1)t
which is given as Δ x = , where, μ is refractive index of thin sheet and t is
d
thickness.
145. (a) For diffraction to occur, the size of an obstacle/aperture is comparable to the
wavelength of light wave. The order of wavelength of light wave is 10−7 , so
diffraction occurs.
146. (c) Maxwell proposed that light must be an electromagnetic wave. Thus, according
to Maxwell, light waves are associated with changing electric and magnetic fields.
The changing electric and magnetic field result in the propagation of
electromagnetic waves (or light waves) even in vacuum.
147. (a) The frequency of light emitted by a charged oscillator equal to its frequency of
oscillation. So, the frequency of scattered light equals to the frequency of incident
light.
148. (c) Red colour travels faster than violet in glass. Speed of light is independent of its
colour only in vacuum. For light travelling from medium 1 to medium 2,
v 1 λ1 μ 2
= =
v 2 λ2 μ 1
c
149. (b) μ=
v
Hence, the speed of light decreases in denser medium.
Energy carried by a wave depends on the amplitude of the wave, not on the speed of
wave propagation. Energy remains same.
Also, intensity of wave ∝ ¿ or I ∝ A 2
150. (a) Only transverse waves can be polarised. Sound waves are longitudinal waves,
so these waves cannot be polarised.
151. (c) I max=( √ I 1 + √ I 2 ) =( 2 √ I 0 ) =4 I 0 The minimum intensity observed at dark band is given
2 2

by
I min ¿ [ √ I 1− √ I 2 ]
2
=I 2=I 0 , I min=0 ¿ If ¿¿ I 1 ¿≠ I 2 ; I min ≠ 0 ¿
If ¿
152. (a) For centre of screen,
S1 P−S2 P=0 ⇒ Δ L=0
2π
Δ φ= × Δ L=0
λ
So, waves meet in phase and results in intensity maxima or bright fringe due to
constructive interference. 153. (a) Intensity is the amount of light energy falling
per unit area per unit time. So, when a slit width is increased, area over which
light falls increases and hence, more light energy falls and hence, intensity
increases.
(Intensity from slit) ∝ slit width of each slit
⇒ I max =( √ I 1 + √ I 2) ⇒ I min =( √ I 1 −√ I 2 )
2 2

So, maximum and minimum intensities both increase.

154. (c) Diffraction determines the limitations of the concept of light rays. A beam of
2
a
width a travels a distance , called the fresnal distance, before it starts to spread
λ
out due to diffraction.
155. (a) Except photoelectric effect, all others phenomenon such as propagation of light
in vacuum, interference and polarisation of light can be explained by wave theory
of light. In photoelectric effect light behaves as it is made up of particles.
156. (b) For a point emitting waves uniformly in all to direction, the locus of points
which have the same amplitude and vibrate in the same phase are spheres. But at
a large distance from the source, the small portion of the sphere can be considered
as plane wave as shown in Figure

157. (c) Figure shows AB as incident wavefront, so A and B are in same phase.
By the time B reaches C , secondary wavelet from A reaches E . So, points C and E are
same time intervals apart as they are in same phase.
158. (a) When incident wave fronts passes through a prism, then lower portion of
wavefront (B) is delayed resulting in a tilt. So, time taken by light to reach A' from
A is equal to the time taken to reach B' from B.
159. (c) Frequency does not changes in reflection, According to Snell's law of refraction,
we get
v air λair
ηw = =
v water λ water
λair λ air 3
⇒ As wavelength i.e., λwater = = = λ
ηw 4 /3 4 air
So, wavelength of reflected light is more than that of refracted light.
160. (b) Case I A light rays diverging from a point source.

Case II A light ray emerging out of convex lens when a point source is placed at its
focus.
Case III A portion of the wavefront of light from a distant star intercepted by the earth.

161. (a) It is given that S1 P=7 λ and S2 P=9 λ

We have, S2 P−S1 P=9 λ−7 λ
⇒ S2 P−S1 P=2 λ
The waves emanating from S1 will arrive exactly two cycles earlier than the waves from
S2 and will again be in phase.

162. (a) When interferring sources have same frequency and their phase difference
remains constant with time, interference is sustained (stayed for a finite time
 interval). If amplitudes are of nearby values, then contrast will be more
pronounced.
163. (d) Light sources which emit light waves of same wavelength (or frequency) having
either zero or a constant originating phase difference are called coherent sources
of light.
164. (c) For a single slit of width a , the first null of the interference pattern occurs at an
angle of λ /a. At the same angle of λ /a, we get a maximum (not a null) for two
narrow slits separated by a distance a . 165. (c )
I. For diffraction pattern, the size of slit should be comparable to the wavelength of
wave used.
II. Diffraction phenomenon is commonly observed in our daily routine in case of sound
waves (which is a longitudinal wave) because wavelength of sound waves is large
(0.1-1 m). However, as wavelength of light waves is extremely small ( 10−6−10−7 m) ,
we do not observe diffraction of light in daily routine.
III. Diffraction is a wave phenomenon. It is observed in electromagnetic and
longitudinal waves as well.
166. (c) Figure shows light reflected from a transparent medium, say, water. As before,
the dots and arrows indicate that both polarisations (E) are present in the incident
and refracted waves.
As the figure shows, the reflected light is therefore, linearly polarised perpendicular to
the plane of the figure (represented by dots). This can be checked by looking at the
reflected light through an analyser.
The transmitted intensity will be zero when the axis of the analyser is in the plane of
the figure, i.e., the plane of incidence

167. (c) Since, light wave travels along the direction perpendicular to its wavefront, for
rays travelling along X -axis,
i.e., plane, X =C is the perpendicular plane.
Similarly, for rays along Y and Z -axes plane wavefronts Y =C and Z=C represent the
wavefront, respectively.
168. (c) A constructive interference is produced when waves overlaps such that a crest
meets a crest and waves are in phase.
For maxima,
For minima,
S1 P ∼ S2 P=nλ (n=0 , 1, 2 , 3 , …)
1
S1 P ∼ S2 P=n+ λ(n=0 , 1 ,2 , 3 ,⋯)
2
we will have destructive interference and the resultant intensity will be zero.
169. (c )
A. Angular separation of the fringes remains constant (¿ λ /d) . The actual separation of
the fringes increases in proportion to the distance of the screen from the plane of
the two slits.
B. The separation of the fringes (and also angular separation) decreases.
λ 3
C. When medium is water, λ ' = air = λ air
4/3 4
'
' λ D 3 λD 3
∴β = = = β
d 4 d 4
Dλ
As we know, fringe width β=
d
1
D. When d is reduced, β ∝
d
So, β is increased.
170. (a) Path difference, P=( S S 2+ S 2 O )− ( S S 1+ S 1 O )
2 2
d d
S S 2=√ x + d =x 1+ 2 =x 1+ 2
2 2

x 2x

(∵ d ≪< x )
 2
d
Similarly, S2 O=√ ( D +d )=D 1+
2 2
2
2D
Also, S S 1 O=x+ D
2 2
d d
∴ P ¿ x 1+ 2
+ D 1+ 2
−(x + D)
2x 2D
¿ ¿
λ
For dark fringe, P=
2

[for minimum d , P=
(2 n−1) λ
2
; n=1 ]
λ d2 1 1
⇒ =
2 2 x D
+ or d=
√ λxD
(x+ D)

Put x=D , d=
√ λD
2
⇒ Put x=D/2 , d=
λD λD
λD
3 √
Fringe width =β= = =2 d
d √ λD/2
Distance of next bright fringe from O .
Distance of consecutive bright and dark band
Fringe width
¿ =d
2
Dλ
171. (a) Fringe width, W =
d
where, D=¿ distance between slits and screen
d=¿ distance between slits and
λ =wavelength of light
A. λ increase so W also increase (∴ A → 4)
B. White light produces coloured fringes (∵ B → 1)
C. If D is doubled and d is halved, then W becomes four times (∵ C → 2)
D. If intensity of either slit is reduced, the bright fringes became less bright.
(∵ D →3)
172. (b) A wavefront is locus of points, which oscillate in phase i.e., it is a surface of
constant phase.
173. (a) If we have a point source emitting waves uniformly in all directions, then the
locus of points which have the same amplitude and vibrate in same phase are
spheres.
174. (a) At a finite distance r the shape of the wavefront is spherical.
175. (b) At a large distance from the source, a small portion of the spherical wave can
be approximated by a plane wave.

176. (c )

AE ν 2 τ
and sin ⁡r = =
AC AC
BC v 1 τ
sin ⁡i= =
AC AC
where, i and r are the angles of incidence and refraction, respectively.
sin ⁡i BC / AC BC
= =
sin ⁡r AE / AC AE
177. (d) According to Snell's law of refraction, we have
sin ⁡i v 1 τ / AC v 1
= =
sin ⁡r v 2 τ / AC v 2
v 1 f λ1 λ1 v1 v2
= = or =
v 2 f λ2 λ2 λ1 λ 2
178. (c) In refraction, speed and wavelength changes but frequency remains constant.
As part of light is always reflected (and also absorbed,) there is change in intensity
of light also.
179. (a) On reflection and refraction, frequency of light remains the same.
v2
⇒ ' =1
v2
180. (a) According to Snell's law of refraction,
sin ⁡i μ 2 v 1 λ 1
= = =
sin ⁡r μ 1 v '2 λ'2
Given, wavelength of incident light, λ 1=589 nm
For reflection, λ 2=λ1=589 nm
Also, v 2=v 1=3 × 108 ms−1=300 × 106 ms−1
For refraction, using Eq. (i)
μ2 λ1 v1 ' λ1 589 nm
= ' ¿ ' ⇒ λ2 = 1 =
μ1 λ2 v2 μ2 4 /3
¿ ¿
' v 1
8
3 ×10 ms
−1
8 −1
Also, v 2= 1 = =2.25 ×10 ms
μ2 4 /3
6 −1
¿ 2.25 ×10 ms
Δ v v radial
181. (b) In Doppler's shift given by, =
v c
Δv
=¿ fractional change in frequency
v
v radial =¿ the component of the source velocity along the line joining the observer to the
source relative to the observer c=¿ speed of light in vacuum ¿ 3 ×108 ms−1.
182. (a)
I. The Doppler's shift is valid only when the speed of source is small compared to that of
light. When speeds are close to that of light, the concept of Einstein's special
theory of relativity is used.
II. Doppler's effect finds application in estimation of the velocity of aeroplanes, rockets,
submarines etc. 183. (d ) In transverse wave the displacement is in the y -direction,
it is often referred to as a y -polarised wave. Since, each point on the string moves
on a straight line, the wave is also referred to as a linearly polarised wave.
Further, the string always remains confined to the XY -plane and therefore, it is
also referred to as a plane polarised wave.
184. (c) If the plane of vibration of the string is changed randomly in very short
intervals of time, then we have what is known as an unpolarised wave. Thus for an
unpolarised wave the displacement will be randomly changing with time though it
will always be perpendicular to the direction of propagation.
185. (b) By law of Malus, intensity of emergent light from P2 is I =I 0 cos 2 ⁡θ, where θ=¿
angle between P1 and P2 pass axis. ⇒ I =0 when θ=90∘
186. (c) Let P3 be the new polaroid inserted.
β=¿ angle between the pass axis of P1 and P3 (given)
I 0=¿ Intensity of light on polaroid P1 (given)
Let α be the angle between P3 and P2 pass axis.
I
Intensity of light from P1= 0
2
 I0 2
Intensity of light from P3= cos ⁡β
2
I0
Intensity of light from P2= cos 2 ⁡β cos2 ⁡α
2
π
∵ α = −β (as P1 and P2 are perpendicular)
2
I0 2 2
I = cos ⁡β cos ⁡( π /2−β)$$
2
=\frac{I_{0}}{2} \cos ^{2} \beta \sin ^{2} \beta
$$
I0 2
I = sin ⁡2 β
8
I
Also, I = 0
8
I0 I0 2
⇒ = sin ⁡2 β or sin2 ⁡2 β=1⇒ β=π /4=45∘
8 8
I max a 1+ a2
187. (b , d ) =9 ⇒
I min a1 −a2
a1 +a 2 a1 3+1 4 a 1
=√ 9=3 ⇒ = = ⇒ =2
a1−a2 a2 3−1 2 a 2
Therefore I 1 : I 2=4 :1
188. (b , d ) We have, for minima is reflection
nλ 640 × 3
2 μ1 t ¿ nλ ⇒ t= =n =240 nm
2 μ1 2×4
t ¿ 240 nm , 480 nm ,…
189. (a ,b ,d ) Consider the pattern of the intensity shown in the figure of question.
(i) As intensities of all successive minima is zero, hence we can say that two sources S1
and S2 are having same intensities. (ii) Regular pattern shows constant phase
difference.
(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very
clear pattern on the screen.
190. (b , d ) Given, width of pinhole ¿ 103 Å=1000 Å
We know that wavelength of sunlight ranges from 4000 Å to 8000 Å .
Clearly, wavelength λ< ¿ width of the slit.
Hence, light is diffracted from the hole. Due to diffraction from the sunlight the image
formed on the screen will be different from the geometrical image and overlaping
of colour v
191. (a ,b) Consider the diagram in which light diverges from a point source (O).
Due to the point source light propagates in all directions symmetrically and hence,
wavefront will be spherical as shown in the diagram.
If power of the source is P, then intensity of the source will be
P
I= 2
4πr
where, r is radius of the wavefront at any time.
192. (a) Given, wavelength of light, λ=589 nm=589 ×10−9 m Refractive index of water
μw =1.33
(i) For reflected light
(a) Wavelength of reflected light, λ=589 ×10−9 m
8
c 3 ×10
(b) Frequency of reflected of light, v= = where c is velocity of light
λ 589× 10−9
( ∵ Speed of light, c=3 × 108 ms−1 )
v=5.09 ×1014 Hz
(c) As the reflection takes place in the same medium so Speed of reflected light
8 −1
c=3 × 10 ms
(ii) For refracted light (In this process wavelength and speed changes but frequency
remains the same)
Wavelength of refracted light
−9
' λ 589 ×10 −7
λ= = =4.42 ×10 m
μ 1.33
8
c 3 ×10 8 −1
∴ Velocity of refracted li, v= = =2.25× 10 ms
μ 1.33
193. (b) Given, separation between slits
−3
d=0.28 mm=0.28 ×10 m
Distance between screen and slit D=1.4 m Distance between central bright and fourth
fringe
−2
x=1.2 cm=1.2× 10 m
Number of fringes n=4
Dλ
For constructive interference x=n
d
−2 4 × 1.4 × λ
1.2 ×10 = −3
0.28 × 10
−2 −3
1.2×10 × 0.28 ×10 −7
Wavelength, λ= ⇒ λ=6 ×10 m
4 ×1.4
194. (a) Given, wavelength λ 1=650 nm=650 × 10−9 m
and λ 2=520 nm=520 ×10−9 m , d=2× 10−3 m
(i) For third bright fringe, n=3 , D=1.2 m
The distance of third bright fringe from central maximum.
nλD −9 D
x ¿ =3 ×650 ×10 × m
d d
¿ ¿
(ii) Let n th bright fringe due to wavelength λ 2=520 nm, coincide with (n+1) th bright
fringe due to wavelength λ 1=650 nm.
D D
i.e., ¿ n λ2 ¿(n−1)λ 1
d d ¿ 4 n ¿=5 n−5 or n=5 ¿
−9
n ×520 ×10 ¿ ¿
 D −9 D
Thus, the least distance, x=n λ 2 =5 ×520 ×10
d d
−9
D −9 1.2× 10 −3
x=2600 ×10 m=2600 × −3
m=1.56 ×10 m
d 2 ×10
195. (b) Given, wavelength of light λ=5000 Å=5000 ×10−10 m On the reflection there is no
change in wavelength and frequency. So, wavelength of reflected light will be
5000 Å .

Frequency of the incident light

8
c 3 ×10 14
v= = −7
=6 ×10 Hz
λ 5× 10
When reflected ray is normal to the incident ray.
AO and BO are the incident and reflected rays.
BO ¿ ⊥ AO ∘ ∘ ∘
=90 ¿ For reflection, ¿ ∠ i ¿=∠ r ¿ ∴ ¿ 2∠ i ¿ ¿=45 ¿ ∠ i ¿ ¿ ¿=45 ¿
∴ ¿
Thus, the angle of incidence is 45 ∘.
196. (b) Given, wavelength of H α , λ=6563 Å=6563× 10−10 m Red shift Δ λ=15 Å
Since, the star is found to be red-shifed, hence star is receding away from earth and
Doppler's shift is negative.
8
vλ −Δ λ ⋅ c −15 ×3 ×10
Δ λ ¿− ⇒ v= =
c λ 6563
5 −1
v ¿−6.86 × 10 ms
Negative sign shows that the star is receding away from earth.
197. (c) Given, wavelength of light, λ=600 nm=600× 10−9 m
∘ 0.1 π
Angular width of fringe, θ=0.1 = rad
180
λ
Using the formula, θ=
d
−9
λ 600 ×10 ×180
Spacing between the slits, d= =
θ 0.1× π
−4
d=3.44 ×10 m
Thus, the spacing between the two slits is 3.44 × 10−4 m.
198. (a) There is no obstruction by the hill to spreading the radio beams, the radial
spread of the beam over the hill 20 km away must not exceed 50 m .
i.e., Z F (Fresnel's distance) ¿ 20 km=20 ×10−3 m⇒ a=50 m
2 2
a a 50 ×50 −4
Z F = ⇒ λ= = =1250 ×10 m
λ Z F 20× 103
Thus, the longest wavelength of radio waves is 0.125 m.
199. (c) Given, wavelength of light λ=500 nm=500 × 10−9 m
−3
D=1 m, n=1, x=2.5 mm=2.5 ×10 m
 nDλ
Distance of n th minimum from the centre, x=
d
−9
nDλ 1× 1× 500× 10 −4
d= = −3
=2 ×10 m ⇒ d=0.2 mm
x 2.5 ×10
Thus, the width of slit of 0.2 mm .
200. (a) Given, width of the slit ¿ 104 Å
4 −10 −6
¿ 10 × 10 m=10 m=1 μ m
Wavelength of (visible) sunlight varies from 4000 Å to 8000 Å . As the width of slit is
comparable to that of wavelength, hence diffraction occurs with maxima at centre.
So, at the centre all colours appear i.e., mixing of colours form white patch at the
centre.
201. (a) Consider the diagram, the ray (P) is incident at an angle θ and gets reflected in
the direction P' and refracted in the direction P' '. Due to reflection from the glass
medium, there is a phase change of π .

Time taken to travel along O P ' '

''
OP d /cos ⁡r nd
Δ t= = =
v c /n c cos ⁡r
sin ⁡θ
From Snell's law, n=
sin ⁡r
sin ⁡θ
⇒ sin ⁡r=
n

√
2
sin ⁡θ
cos ⁡r= √ 1−sin ⁡r= 1−
2
2
n
nd n2 d sin2 ⁡θ
∴ Δ t= = 1−
2
sin ⁡θ c n
2
c 1− 2
n
2 −1/ 2
2π 2 πnd sin ⁡θ
Phase difference ¿ Δ φ= × Δ t= 1− 2
λ λ n
So, net phase difference ¿ Δ φ+ π
4 πd 1 2 −1/ 2
¿ 1− 2 sin ⁡θ +π
λ n
202. (c) 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 filteration only red and blue
lights are present. In Young's double slit monochromatic light is used for the
formation of fringes on the screen. Hence, in this case there shall be no
interference fringes.
203. (d) There is a hole at point P2 (minima). The hole will act as a source of fresh light
for the slits S3 and S4 . Therefore, there will be a regular two slit pattern on the
second screen.
204. (a) Given, angular resolution of human eye, φ=5.8 ×10−4 rad. and printer prints 300
dots per inch. The linear distance between two dots is
2.54 −2
l= =0.84 ×10 cm.
300
l
At a distance of z cm, this subtends an angle, φ=
z
−2
l 0.84 ×10 cm
∴ z= = −4
=14.5 cm.
φ 5.8 ×10
205. (b) For nth minima to be formed on the screen path difference between the rays
λ
coming from S1 and S2 must be (2 n−1) .
2
From the given figure of two slit interference arrangements, we can write
T 2 P=T 2 O+ OP=D+ x
and
T 1 P=T 1 O−OP=D−x
√
S1 P= ( S1 T 1 ) + ( P T 1 ) =√ D +¿ ¿
2 2 2

and
√
S2 P= ( S2 T 2 ) + ( T 2 P ) =√ D + ¿¿
2 2 2

λ
The minima will occur when S2 P−S1 P=(2n−1)
2
i.e., ¿ ¿
(for first minima n=1 )
If x=D
2 1/ 2 1 /2 λ
we can write [ D + 4 D ] −[ D +0 ] =
2 2
2
2 1 /2 2 1 /2 λ
⇒ [5 D ] −[D ] =
2
λ
⇒ √ 5 D−D=
2
λ
⇒ D ( √ 5−1)=λ /2 or D=
2( √ 5−1)
Putting √ 5=2.236
⇒ √ 5−1 ¿ 2.236−1=1.236
λ
D ¿ =0.404 λ
2(1.236)
206. (b) The resultant disturbance at a point will be calculated by sun of disturbances
due to individual sources.
Consider the disturbances at the receiver R1 which is at a distance d from B.
Let the wave at R1 because of A be Y A =a cos ⁡ωt . The path difference of the signal from A
with that from B is λ /2 and hence, the phase difference is π .
Thus, the wave at R1 because of B is
y B =a cos ⁡(ωt −π )=−a cos ⁡ωt
The path difference of the signal from C with that from A is λ and hence the phase
difference is 2 π .
Thus, the wave at R1 because of C is Y C =a cos ⁡(ωt−2 π)
The path difference between the signal from D with that of A is

√ d2+
λ2
2
λ 1/ 2
−( d−λ /2)=d 1+ 2 −d +

λ2 1/ 2
4d
λ λ
λ
2

¿ d 1+ 2
−d + ≈ (∵ d > λ)
8d 2 2
Therefore, phase difference is π .
∴ Y D =a cos ⁡(ωt−π)=−a cos ⁡ωt
Thus, the signal picked up at R1 from all the four sources is Y R = y A + y B + y C + y D
1

¿ a cos ⁡ωt−a cos ⁡ωt +a cos ⁡ωt−a cos ⁡ωt=0

Let the signal picked up at R2 from B be y B =a 1 cos ⁡ωt . The path difference between signal
at D and that at B is λ /2. ∴ y D =−a 1 cos ⁡ωt
The path difference between signal at A and that at B is
√¿¿
As d ≫ λ, therefore this path difference → 0
2
2π 1 λ
and phase difference ¿ →0
λ 8 d2
Hence, y A =a1 cos ⁡(ωt−φ)
Similarly, y C =a1 cos ⁡( ωt−φ)
∴ Signal picked up by R2 is
y A + y B + y C + y D= y =2 a1 cos ⁡(ωt−φ) 2 2 2 2
∨ y ¿ =4 a1 cos ⁡(ωt−φ)∴< I ≥2 a1 ¿
¿
Thus, R1 picks up the larger signal.
207. (c) In this figure, we have shown a dielectric film of thickness d deposited on a
glass lens.
Refractive index of film ¿ 1.38
and refractive index of glass ¿ 1.5.
Given, λ=5500 Å .
Consider a ray incident at an angle i . A part of this ray is reflected from the air-film
interface and a part refracted inside. This is partly reflected at the film-glass
interface and a part transmitted. A part of the reflected ray is reflected at the film-
air interface and a part transmitted as r 2 parallel to r 1. Of course successive
reflections and transmissions will keep on decreasing the amplitude of the wave.
The optical path difference between r 2 and r 1 is
n( AD +CD )−AB
If d is the thickness of the film, then
d
AD ¿ CD= ⇒ AB= AC sin ⁡i
cos ⁡r
AC
¿ d tan ⁡r
2
∴ AC ¿ 2 d tan ⁡r
Hence, AB=2 d tan ⁡r sin ⁡i .
2 nd
Thus, the optical path difference ¿ −2 d tan ⁡r sin ⁡i
cos ⁡r
2
sin ⁡id sin ⁡r 1−sin ⁡r
¿2⋅ −2 d sin ⁡i=2d sin ⁡
sin ⁡r cos ⁡r cos ⁡r sin ⁡r cos ⁡r
¿ 2 nd cos ⁡r
λ
For these waves to interfere destructively path difference ¿ .
2
λ
⇒ ¿ 2 nd cos ⁡r ¿
2
λ
⇒ nd cos ⁡r ¿=
4
For photographic lenses, the sources are normally in vertical plane
∘
∴ i=r=0
∘ λ
From Eq. (i), nd cos ⁡0 =
4
λ 5500 Å
⇒ d= = ≈ 1000 Å
4 n 4 ×1.38
208. (a) In case of transparent glass slab of refractive index μ, the path difference
¿ 2 d sin ⁡θ+(μ−1) L, slit width ¿ 2 d For the principle maxima, (path difference is zero)
i.e., 2 d sin ⁡θ0 +(μ−1)L=0
or
or
L(μ−1) −L(0.5)
sin ⁡θ 0 ¿− = [∵ L=d /4 ]
2d 2d
−1
sin ⁡θ 0 ¿
16
−D
OP ¿ D tan ⁡θ0=D sin ⁡θ 0=
16
λ
For the first minima, the path difference is ± .
2
λ
2 d sin ⁡θ 1+0.5 L ¿±
2
or ± λ /2−0.5 L
sin ⁡θ 1 ¿
2d
¿ ¿
¿ The diffraction occurs if the wavelength of waves in nearly equal to the slit width (d ) ¿.
' +1 1 3
On the positive side, sin ⁡θ 1= − =
4 16 16
'' −1 1 −5
On the negative side, sin ⁡θ 1 = − =
4 16 16
The first principal maxima on the positive side is at distance.
'
' sin ⁡θ1 3 3D
D tan ⁡θ 1=D =D = above
√ 1−sin ⁡θ1 √16 −3 √247
2 ' 2 2

point O .
The first principal minima on the negative side is at distance.
'' 5 5
D tan ⁡θ 1 = = below point O .
√ 16 −5 √ 231
2 2