Chapter-wise DPP of Selected Questions for NEET

Wave Optics

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1. The phenomenon of polarization of light indicates that

a) Light is a longitudinal wave
b) Light is a transverse wave
c) Light is not a wave
d) Light travels with the velocity of 3 × 108 ms−1
2. The fringe width in Young’s double slit experiment increases when
a) Wavelength increases
b) Distance between the slits increases
c) Distance between the source and screen decreases
d) The width of the slits increases
3. When the angle of incidence on a material is 60°, the reflected light is completely polarized. The velocity of
the refracted ray inside the material is (in 𝑚𝑠 −1 )
3
a) 3 × 108 b) ( ) × 108 c) √3 × 108 d) 0.5 × 108
√2
4. The principle of superposition is basic to the phenomenon of
a) Total internal reflection b) Interference
c) Reflection d) Refraction
5. Wavefront means
a) All particles in it have same phase
b) All particles have opposite phase of vibrations
c) Few particles are in same phase, rest are in opposite phase
d) None of these
6. In Young’s double slit experiment we get 60 fringes in the field of view of monochromatic light of
wavelength 4000Å. If we use monochromatic light of wavelength 6000 Å, then the number of fringes
obtained in the same field of view are
a) 60 b) 90 c) 40 d) 1.5
7. An unpolarised beam of intensity 𝐼0 is incident on a pair of nicols making an angle of 60° with each other.
The intensity of light emerging from the pair is
a) 𝐼0 b) 𝐼0 /2 c) 𝐼0 /4 d) 𝐼0 /8
8. In Young’s double slit experiment, the 8𝑡ℎ maximum with wavelength 𝜆1 is at a distance 𝑑1 from the
central maximum and the 6𝑡ℎ maximum with a wavelength 𝜆2 is at a distance 𝑑2 . Then (𝑑1 /𝑑2 ) is equal to
4 𝜆2 4 𝜆1 3 𝜆2 3 𝜆1
a) ( ) b) ( ) c) ( ) d) ( )
3 𝜆1 3 𝜆2 4 𝜆1 4 𝜆2
9. The phenomenon which does not take place in sound waves is
a) Scattering b) Diffraction c) Interference d) Polarisation

10. Intensities of the two waves of light are 𝐼 and 4𝐼. The maximum intensity of the resultant wave after
superposition is
a) 5 𝐼 b) 9 𝐼 c) 16 𝐼 d) 25 𝐼
11. Two identical radiators have a separation of 𝑑 = 𝜆/4 where 𝜆 is the wavelength of the waves emitted by
either source. The initial phase difference between the sources is 𝜋/4. Then the intensity on the screen at a
distant point situated at an angle 𝜃 = 30° from the radiators is (here 𝐼𝑜 is intensity at that point due to one
radiator alone)
a) 𝐼𝑜 b) 2𝐼𝑜 c) 3𝐼𝑜 d) 4𝐼𝑜
12. Huygen’s principle of secondary wavelets may be used to
a) Find the velocity of light in vacuum b) Explain the particle behavior of light
c) Find the new position of the wavefront d) Explain photoelectric effect
13. The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the
distance between two fringe from the centre is
a) 2 b) 1/2 c) 4 d) 16
14. 𝑛th Bright fringe if red light (𝜆1 = 7500 Å) coincides with (𝑛 + 1)𝑡ℎ bright fringe of green
light(𝜆2 = 6000 Å). The value of 𝑛 =?
a) 4 b) 5 c) 3 d) 2
15. The Young’ experiment is performed with the lights of blue (𝜆 = 4360 Å) and green colour (𝜆 = 5460Å), if
the distance of the 4th fringe from the centre is 𝑥, then
𝑥(𝐵𝑙𝑢𝑒) 5460
a) 𝑥 (Blue) = 𝑥 (Green) b) 𝑥(Blue) > 𝑥(Green) c) 𝑥(Blue) < 𝑥(Green) d) =
𝑥(𝐺𝑟𝑒𝑒𝑛) 4360
16. Refractive index of material is equal to tangent of polarizing angle. It is called
a) Brewster’s law b) Lambert’s law c) Malus’s law d) Bragg’s law
17. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with
the wavelength 𝜆1 is found to be coincident with the third maximum at 𝜆2 . So
a) 3𝜆1 = 0.3𝜆2 b) 3𝜆1 = 𝜆2 c) 𝜆1 = 3.5𝜆2 d) 0.3𝜆1 = 3𝜆2
18. If two waves represented by 𝑦1 = 4 sin 𝜔𝑡 and 𝑦2 = 3 sin (𝜔𝑡 + ) interfere at a point, the amplitude of
𝜋
3
the resulting wave will be about
a) 7 b) 6 c) 5 d) 3.5
19. Two light rays having the same wavelength 𝜆 in vacuum are in phase initially. Then the first ray travels a
path 𝐿1 through a medium of refractive index 𝑛1 while the second ray travels a path of length 𝐿2 through a
medium of refractive index 𝑛2 . The two waves are then combined to produce interference. The two waves
are then combined to produce interference. The phase difference between the two waves is
2𝜋 2𝜋 2𝜋 2𝜋 𝐿1 − 𝐿2
a) (𝐿2 − 𝐿1 ) b) (𝑛1 𝐿1 − 𝑛2 𝐿2 ) c) (𝑛2 𝐿1 − 𝑛1 𝐿2 ) d) ( )
𝜆 𝜆 𝜆 𝜆 𝑛1 − 𝑛2
20. Two polaroids are placed in the path of unpolarised beam of intensity 𝐼0 such that no light is emitted from
the second polaroid. If a third polaroid whose polarization axis makes an angle 𝜃 with the polarization axis
of first polaroid, is placed between these polaroids then the intensity of light emerging from the last
polaroid will be
𝐼 𝐼 𝐼
a) ( 0 ) sin2 2𝜃 b) ( 0 ) sin2 2𝜃 c) ( 0 ) cos4 2𝜃 d) 𝐼0 cos4 𝜃
8 4 2
21. What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength
6000Å is used and slit width is 12 × 10−5 𝑐𝑚
a) 2 𝑟𝑎𝑑 b) 3 𝑟𝑎𝑑 c) 1 𝑟𝑎𝑑 d) 8 𝑟𝑎𝑑
22. Two sources of same intensity interfere at a point and produced resultant 𝐼. When one source is removed,
the intensity at that point will be
a) 𝐼 b) 𝐼/2 c) 𝐼/4 d) 𝐼/3
23. If the two waves represented by 𝑦1 = 4 sin 𝜔𝑡 and 𝑦2 = 3 sin(𝜔𝑡 + 𝜋/3) interfere at a point, the
amplitude of the resulting wave will be about
a) 7 b) 5 c) 6 d) 3.5
24. In Young’s double slit experiment, if the widths of the slits are in the ratio 4 : 9, the ratio of the intensity at
maxima to the intensity at minima will be
a) 169 : 25 b) 81 : 16 c) 25 : 1 d) 9 : 4
25. In the phenomenon of diffraction of light, when blue light is used in the experiment instead of red light,
then
a) Fringes will become narrower b) Fringes will become broader
c) No change in fringe width d) None of the above
26. The angle of polarization for any medium is 60°, what will be critical angle for this
1
a) sin−1 √3 b) tan−1 √3 c) cos −1 √3 d) sin−1
√3
27. An unpolarised beam of intensity 2 𝑎 2 passes through a thin Polaroid. Assuming zero absorption in the
Polaroid, the intensity of emergent plane polarized light is
𝑎2
a) 2 𝑎 2 b) 𝑎 2 c) √2𝑎 2 d)
2
28. Ray diverging from a point source form a wave front that is
a) Cylindrical b) Spherical c) Plane d) Cubical
29. Two light sources are said to be of coherent nature
a) When they have same frequency and a varying phase difference
b) When they have same frequency and a constant phase difference
c) When they have constant phase difference and different frequencies
d) When they have varying phase difference and different frequencies
30. Two waves 𝑦1 = 𝐴1 sin(𝜔𝑡 − 𝛽1 ) and 𝑦2 = 𝐴2 sin(𝜔𝑡 − 𝛽2 ) superimpose to form a resultant wave whose
amplitude is
a) √𝐴21 + 𝐴22 + 2𝐴1 𝐴2 cos(𝛽1 − 𝛽2 ) b) √𝐴21 + 𝐴22 + 2𝐴1 𝐴2 sin(𝛽1 − 𝛽2 )
c) 𝐴1 + 𝐴2 d) |𝐴1 + 𝐴2 |
31. Wave nature of light is verified by
a) Interference b) Photoelectric effect c) Reflection d) Refraction
32. In young’s double slit experiment, the intensity of the maxima is𝐼. If the width of each slit is doubled, the
intensity if the maxima will be
a) 𝐼/2 b) 2𝐼 c) 4𝐼 d) 𝐼
33. An unpolarised beam of intensity 𝐼0 falls on a polariod. The intensity of the emergent light is
𝐼 𝐼 d) Zero
a) 0 b) 𝐼0 c) 0
2 4
34. 100 𝜋 phase difference = ……. Path difference.
a) 10𝜆 b) 25𝜆 c) 50𝜆 d) 100𝜆
35. A narrow slit of width 2 mm is illuminated by monochromatic light of wavelength 500 nm. The distance
between the first minima on either side on a screen at a distance of 1 m is
a) 5 mm b) 0.5 mm c) 1 mm d) 10 mm
36. In a Young’s double slit experiment using red and blue lights of wavelengths 600 nm and 480 nm
respectively, the value of 𝑛 from which the 𝑛𝑡ℎ red fringe coincides with (𝑛 + 1) the blue fringe is
a) 5 b) 4 c) 3 d) 2
37. By corpuscular theory of light, the phenomenon which can be explained is
a) Refraction b) Interference c) Diffraction d) Polarization
38. When two coherent monochromatic beams of intensity I and 9I interface, the possible maximum and
minimum intensities of the resulting beam are
a) 9I and I b) 9I and 4I c) 16I and 4I d) 16I and I
39. In Young’s double slit experiment if monochromatic light used is replaced by white light, then
a) No fringes are observed
b) Only central fringe is white, all other fringes are coloured
c) All bright fringes become white
d) All bright fringes have colours between violet and red
40. In an interference pattern the position of zeroth order maxima is 4.8 mm from a certain point 𝑃 on the
screen. The fringe width is 0.2 mm. The position of second maxima from point 𝑃 is
 a) 5.1 mm b) 5 mm c) 40 mm d) 5.2 mm
41. Two coherent monochromatic light beams of intensities 𝐼 and 4𝐼 are superposed. The maximum and
minimum possible intensities in the resulting beam are
a) 5𝐼 and 𝐼 b) 5𝐼 and 3𝐼 c) 9𝐼 and 𝐼 d) 9𝐼 and 3𝐼
42. In a Young’s double slit experiment, the fringe width is found to be 0.4 𝑚𝑚. If the whole apparatus is
immersed in water of refractive index 4/3 without disturbing the geometrical arrangement, the new fringe
width will be
a) 0.30 𝑚𝑚 b) 0.40 𝑚𝑚 c) 0.53 𝑚𝑚 d) 450 𝑚𝑖𝑐𝑟𝑜𝑛
43. The diffraction effect can be observed in
a) Only sound waves b) Only light waves
c) Only ultrasonic waves d) Sound as well as light waves
44. Plane polarized light is passed through a polaroid. On viewing through the polaroid we find that when the
polariod is given one complete rotation about the direction of the light, one of the following is observed
a) The intensity of light gradually decreases to zero and remains at zero
b) The intensity of light gradually increases to a maximum and remains at maximum
c) There is no change in intensity
d) The intensity of light is twice maximum and twice zero
45. What causes change in the colours of the soap or oil films for the given beam of light
a) Angle of incidence b) Angle of reflection c) Thickness of film d) None of these
 : HINTS AND SOLUTIONS :
1 (b)
The polarization is the property of 10 (b)
electromagnetic waves such as light which 𝐼max = 𝐼1 + 𝐼2 + 2√𝐼1 𝐼2
describes the direction of their transverse electric So, 𝐼max = 𝐼 + 4𝐼 + 2√𝐼. 4𝐼 = 9𝐼
field. More generally, the polarization of 11 (b)
transverse wave describes the direction of The intensity at a point on screen is given by
oscillation, in the plane perpendicular to the 𝐼 = 4𝐼0 cos2 (𝜙/2)
direction of travel. Longitudinal waves such as Where 𝜙 is the phase difference. In this problem
sound waves do not exhibit polarization, becomes 𝜙 arises (i) due to initial phase difference of 𝜋/4
for these waves the direction of oscillation is and (ii) due to path difference for the observation
along the direction of travel. point situated at 𝜃 = 30°. Thus
𝜋 2𝜋 𝜋 2𝜋 𝜆
2 (a) 𝜙= + (𝑑 sin 𝜃) = + . (sin 30°)
𝜆𝐷 4 𝜆 4 𝜆 4
𝛽= ⇒𝛽∝𝜆 𝜋 𝜋 𝜋
𝑑 = + =
4 4 2
3 (c) 𝜙 𝜋
𝑐 Thus 2 = 4 and 𝐼 = 4𝐼0 cos2 (𝜋/4) = 2𝐼0
From Brewster’s law 𝜇 = tan 𝑖𝑝 ⇒ 𝑣 = tan 60° =
12 (c)
√3
𝑐 3 × 108 13 (a)
⇒𝑣= = = √3 × 108 𝑚/𝑠
√3 √3 2
𝜙
4 (b) 𝐼 = 4𝐼 0 cos
2
At central position 𝐼1 = 4𝐼0 …(i)
5 (a) Since the phase difference between two
Wavefront is the locus of all the particles which successive fringes is 2𝜋, the phase difference
vibrates in the same phase between two points separated by a distance equal
6 (c) to one quarter of the distance between the two,
As 𝑥 = 𝑛1 𝛽1 = 𝑛2 𝛽2 = 𝑛2 𝜆1 = 𝑛2 𝜆2 successive fringes is equal to
1 𝜋
𝛿 = (2𝜋) (4) = 2 radian
𝑛1 𝜆1 60 × 4000
∴ 𝑛2 = = = 40 𝜋
𝜆2 6000 ⇒ 𝐼2 = 4𝐼0 cos2 ( 22 ) = 2𝐼0 …(ii)
7 (c) 𝐼 4𝐼
Using (i) and (ii), 𝐼1 = 2𝐼0 = 2
2 0
According to Malus’ law
14 (a)
2
1 𝐼0
𝐼 = 𝐼0 cos2 θ = 𝐼0 (cos2 60°) = 𝐼0 × ( ) = 15 (c)
2 4
𝑛𝜆𝐷
Distance of 𝑛𝑡ℎ bright fringe 𝑦𝑛 = , 𝑖. 𝑒. , 𝑦𝑛 ∝𝜆
8 (b) 𝑑
𝑥𝑛 𝜆1 𝑥(Blue) 4360
Position of 𝑛th maxima from central maxima is ∴ 1= ⇒ =
𝑛𝜆𝐷 𝑥𝑛2 𝜆2 𝑥(Green) 5460
given by 𝑥𝑛 = 𝑑 ∴ 𝑥(Green) > 𝑥(Blue)
𝑑1 𝑛1 𝜆1 8𝜆1 4 𝜆1 16 (a)
⇒ 𝑥𝑛 ∝ 𝑛𝜆 ⇒ = = = ( )
𝑑2 𝑛2 𝜆2 6𝜆2 3 𝜆2
9 (d) 17 (c)
As sound waves are longitudinal, therefore, Position of first minima = position of third
polarization of sound waves is not possible. maxima 𝑖. 𝑒.,
 1 × 𝜆1 𝐷 (2 × 3 + 1) 𝜆2 𝐷 Let the initial intensity of light is 𝐼0 . So Intensity of
= ⇒ 𝜆1 = 3.5𝜆2 𝐼0
𝑑 2 𝑑 light after transmission from first polaroid = 2
𝐼0
Intensity of light emitted from 𝑃3 𝐼1 = cos2 𝜃
2
18 (b) Intensity of light transmitted from last polaroid
𝜙 = 𝜋/3, 𝑎1 = 4, 𝑎2 = 3 𝑖. 𝑒. from
So, 𝐴 = √𝑎12 + 𝑎22 + 2𝑎1 . 𝑎2 cos 𝜙 ⇒ 𝐴 = 6 𝐼0
𝑃2 = 𝐼1 cos2 (90° − 𝜃) = cos2 𝜃. sin2 𝜃
19 (b) 2
𝐼0 𝐼0
The optical path between any two points is = (2 sin 𝜃 cos 𝜃)2 = sin2 2𝜃
8 8
proportional to the time of travel. 21 (c)
2𝜆 2×6000×10−10
The distance traversed by light in a medium of Angular width = = 12×10−5×10−2 = 1𝑟𝑎𝑑
𝑑
refractive index 𝜇 in time t is given by 22 (c)

𝑑 = 𝑣𝑡 … … … . (𝑖)
23 (c)
Where v is velocity of light in the medium. The 𝑦1 = 4 sin 𝜔𝑡
distance traversed by light in a vacuum in this
𝑦2 = 3 sin(𝜔𝑡 + 𝜋/3)
time.
Here, 𝑎 = 4, 𝑏 = 3, ϕ = 𝜋/3
Δ = 𝑐𝑡

𝑑 𝑅 = √𝑎 2 + 𝑏 2 + 2𝑎𝑏 cos ϕ
=𝑐. [𝐹𝑟𝑜𝑚 𝐸𝑞. (𝑖)]
𝑣
= √42 + 32 + 2 × 4 × 3 cos 𝜋/3
𝑐
=𝑑 = 𝜇𝑑 (𝑖𝑖)
𝑣 = √37 = 6
𝑐
(Since, 𝜇 = 𝑣) 24 (c)
Slit width ratio = 4: 9; hence 𝐼1 : 𝐼2 = 4: 9
This distance is the equivalent distance in vacuum 𝑎12 4 𝑎1 2
and is called optical path. ∴ 2= ⇒ =
𝑎2 9 𝑎2 3
𝐼max (𝑎1 + 𝑎2 )2 25
Here, optical path for first ray = 𝑛1 𝐿1 ∴ = =
𝐼min (𝑎1 − 𝑎2 )2 1
Optical path for second ray = 𝑛2 𝐿2 25 (a)
Width of central maximum is given by
Path difference = 𝑛1 𝐿1 − 𝑛2 𝐿2
2𝑓𝜆
Now, phase difference 𝑤= … (𝑖)
𝑎
2𝜋
= × path difference Where 𝑓 is focal length of lens, a is width of slit
𝜆
and 𝜆 is wavelength of light used.
2𝜋
= × (𝑛1 𝐿1 − 𝑛2 𝐿2 ) From Eq. (i), it is clear that fringe width
𝜆

20 (a) 𝑤∝𝜆
No light is emitted from the second polaroid, so 𝑃1
So, when blue light is used in the experiment
and 𝑃2 are perpendicular to each other
instead of red light, the fringes will become
narrower.

26 (d)
By using 𝜇 = tan 𝜃𝑝 ⇒ 𝜇 = tan 60 = √3
1 1
Also 𝐶 = sin−1 (𝜇 ) ⇒ 𝐶 = sin−1 ( )
√3

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27 (b) 35 (b)
The intensity of plane polarised light is=2𝑎 2 . Distance =
2𝜆
×𝑑
𝑏

∴ Intensity of polarised light from first nicol prism 2 × 0.5 × 10−4

= × 100 = 0.5 mm
2
𝐼0 1
= = × 2𝑎 2 = 𝑎 2
2 2 36 (b)
𝑛𝜆𝑟 = (𝑛 + 1)𝜆𝑏
28 (b)
𝑛 + 1 𝜆𝑟 600 4
= = =
𝑛 𝜆𝑏 480 5
S
1 4 1
Point Spherical = −1= 𝑛 =4
source 𝑛 5 4
wave front

37 (a)
29 (b) Corpuscular theory explains refraction of light
38 (c)
30 (a) Given, 𝐼1 = 𝐼 and 𝐼2 = 9𝐼
Amplitude 𝐴1 and 𝐴2 are added as vector. Angle 2
Maximum intensity = (√𝐼1 + √𝐼2 )
between these vectors is the phase difference 2
(𝛽1 − 𝛽2 ) between them = (√𝐼 + √9𝐼) = 16𝐼
Minimum intensity
2 2
∴ 𝑅 = √𝐴21 + 𝐴22 + 2𝐴1 𝐴2 cos(𝛽1 − 𝛽2 ) = (√𝐼1 − √𝐼2 ) = (√𝐼 − √9𝐼) = 4𝐼
39 (b)
31 (a) In Young’s double slit experiment if white light is
Photoelectric effect verifies particle nature of used instead of monochromatic light, then we
light. Reflection and refraction verify both particle shall get a white fringe at the centre surrounded
nature and wave nature of light on either side with some coloured fringes, with
32 (b) violet fringe in the beginning and red fringe in the
𝐼𝑚𝑎𝑥 = 𝐼 = 𝐼1 + 𝐼2 + 2√𝐼1 𝐼2 last.

When width of each slit is doubled, intensity from 40 (a)

each slit becomes twice ie, The distance between zeroth order maxima and
second order minima is
𝐼1′ = 2𝐼1 and 𝐼2′ = 2𝐼2
𝛽 3
𝑦1 = + 𝛽 = 𝛽
′ ′ ′ ′ ′
∴ 𝐼𝑚𝑎𝑥 = 𝐼 = 𝐼1 + 𝐼2 + 2√𝐼1 × 𝐼2 ′ 2 2
3
= 2𝐼1 + 2𝐼2 + 2√2𝐼1 × 2𝐼2 = × 0.2 mm = 0.3 mm
2
= 2(𝐼1 + 𝐼2 + 2√𝐼1 × 𝐼2 ) = 2𝐼 ∴ The distance of second maxima from point P is

33 (a) 𝑦 = (4.8 + 0.3)mm = 5.1 mm

If an unpolarised light is converted into plane
polarized light by passing through a polaroid its 41 (c)
2 2
intensity becomes half. 𝐼max = (√𝐼1 + √𝐼2 ) = (√𝐼 + √4𝐼) = 9𝐼
2 2
34 (c) 𝐼min = (√𝐼1 − √𝐼2 ) = (√𝐼 − √4𝐼) = 𝐼
𝜆
Path difference = 2𝜋 × phase difference
𝜆
= × 100𝜋 = 50𝜆
2𝜋

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42 (a)
𝐵𝑎𝑖𝑟 0.4 43 (d)
𝛽water = = = 0.3𝑚𝑚
𝜇 4/3
44 (d)

45 (c)
For viewing interference in oil films or soap
bubble, thickness of film is of the order of
wavelength of light

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