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NEET • PHYSICS • WAVE OPTICS
Wave optics practice · Mixed difficulty
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61
Medium
61 of 150
Remembering
+4 | -1
Brewster’s angle in terms of refractive index (n) of the medium
View answer and explanation
Correct option: D
Brewster's angle is given by the formula , where n is the refractive index of the medium. Therefore, option D is correct.
62
Easy
62 of 150
Understanding
+4 | -1
Assertion: Distance between position of bright and dark fringe remain same in YDSE
Reason: Fringe width
Reason: Fringe width
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
Both the Assertion and Reason are correct, and the Reason is the correct explanation of the Assertion because the fringe width determines the distance between bright and dark fringes in YDSE.
63
Medium
63 of 150
Understanding
+4 | -1
represent the velocities, the refractive indices of ordinary and extraordinary rays for a doubly refracting crystal. Then
View answer and explanation
Correct option: C
if the crystal is calcite
For a calcite crystal, the refractive index for the ordinary ray is greater than that for the extraordinary ray (), which implies that the velocity of the ordinary ray is less than that of the extraordinary ray ().
64
Medium
64 of 150
Analyzing
+4 | -1
Assertion: If Fraunhoffer diffraction through single slit is observed under monochromatic light, then central maxima cannot be identified from other maxima.
Reason: In Fraunhoffer single slit diffraction under monochromatic light, all maxima has equal width.
Reason: In Fraunhoffer single slit diffraction under monochromatic light, all maxima has equal width.
View answer and explanation
Correct option: D
If both Assertion and Reason are false statements.
Option D (If both Assertion and Reason are false statements) is correct. The central maxima *can* be identified, and the maxima do *not* have equal width.
65
Easy
65 of 150
Remembering
+4 | -1
According to Huygen’s theory of light
View answer and explanation
Correct option: B
Light is a wave
Option B is correct. Huygens' theory of light describes light as a wave, not as a stream of photons or particles.
66
Medium
66 of 150
Analyzing
+4 | -1
In YDSE, let and be two slits. Films of thicknesses and and refractive indices and are placed in front of and , respectively. If , then the central maxima will:
View answer and explanation
Correct option: D
If and if
Option A (Not shift) is correct because if , the optical path lengths introduced by the films are equal, resulting in no shift of the central maxima.
67
Medium
67 of 150
Application
+4 | -1
In young’s two slit experiment the distance between the two coherent sources is and the screen is at a distance of . If the fringe width is found to be , then the wavelength of the light used is
View answer and explanation
Correct option: D
Option D () is correct because it is calculated using the formula for fringe width in Young's double-slit experiment.
68
Medium
68 of 150
Application
+4 | -1
Two coherent sources are 0.3 mm apart. They are 1 m away from the screen. The second dark fringe is at a distance of 0.3 cm from the centre. Distance of fourth bright fringe from the centre is
View answer and explanation
Correct option: B
0.8 cm
Option B (0.8 cm) is correct because it is calculated using the formula for the position of bright fringes in a double-slit experiment.
69
Medium
69 of 150
Understanding
+4 | -1
If a source of light is moving away from a stationary observer, then the frequency of light wave appears to change because of
View answer and explanation
Correct option: A
Doppler’s effect
Option A (Doppler’s effect) is correct because the Doppler effect explains the change in frequency of a wave due to the relative motion between the source and the observer.
70
Medium
70 of 150
Application
+4 | -1
A wavefront presents one, two and three HPZ at points respectively. If the ratio of consecutive amplitudes of HPZ is , then the ratio of resultant intensities at these point will be
View answer and explanation
Correct option: B
Option B () is correct because the intensity is proportional to the square of the amplitude, and the resultant amplitude is calculated based on the number of half-period zones.
71
Medium
71 of 150
Remembering
+4 | -1
Colours in thin films are due to
View answer and explanation
Correct option: C
Interference phenomenon
Option C is correct. Colors in thin films are a result of the interference of light waves reflected from the different surfaces of the film.
72
Medium
72 of 150
Analyzing
+4 | -1
In Young’s double slit experiment, a third slit is made in between the double slits. Then
View answer and explanation
Correct option: B
Contrast between bright and dark fringes is reduced
Option B (Contrast between bright and dark fringes is reduced) is correct because the third slit introduces additional light, increasing the minimum intensity and thus reducing the fringe visibility.
73
Medium
73 of 150
Understanding
+4 | -1
The fringe width in Young’s double slit experiment increases when
View answer and explanation
Correct option: D
Distance between slits plane and screen increases
Option D is correct. The fringe width in Young's double-slit experiment is directly proportional to the distance between the slits and the screen.
74
Medium
74 of 150
Understanding
+4 | -1
In Young’s double slit experiment, if one of the slits is closed fully, then in the interference pattern
View answer and explanation
Correct option: A
A bright slit will be observed, no interference pattern will exist
Option A (A bright slit will be observed, no interference pattern will exist) is correct because interference requires two coherent sources, and closing one slit eliminates the interference.
75
Easy
75 of 150
Analyzing
+4 | -1
Statement-1: In Young's double slit experiment the two slits are at distance apart. Interference pattern is observed on a screen at distance from the slits. At a point on the screen when it is directly opposite to one of the slits, a dark fringe is observed. Then the wavelength of wave is proportional to square of distance between the two slits.
Statement-2: For a dark fringe intensity is zero.
Statement-2: For a dark fringe intensity is zero.
View answer and explanation
Correct option: C
Statement-1 is true, Statement-2 is false
Option D (Statement-1 is false, Statement-2 is true) is correct. Statement 1 is false because the wavelength is proportional to the square of the fringe width, not the distance between the slits. Statement 2 is true because the intensity is indeed zero for a dark fringe.
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76
Hard
76 of 150
Application
+4 | -1
A slit of width ‘a’ is illuminated by white light. For red light , the first minima is obtained at , then the value of ‘a’ will be
View answer and explanation
Correct option: C
1.24 microns
Option C (1.24 microns) is correct because it is the calculated width of the slit using the single-slit diffraction formula for the first minima.
77
Medium
77 of 150
Application
+4 | -1
A narrow stream of electrons of energy 100eV is fired at two parallel slits very close to each other. The distance between the slits is 10A. The electron waves after passing through the slits interfere on a screen, 3m away from slits. The fringe width is
View answer and explanation
Correct option: A
0.36m
Option A (0.36m) is correct because it is calculated using the de Broglie wavelength and the fringe width formula for electron interference.
78
Medium
78 of 150
Analyzing
+4 | -1
Assertion: The clouds in sky generally appear to be whitish.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.
Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.
View answer and explanation
Correct option: C
Assertion is true and Reason is false
Option C (Assertion is true and Reason is false) is correct. Clouds appear whitish due to scattering, not diffraction, and scattering is not equally efficient at all wavelengths.
79
Medium
79 of 150
Application
+4 | -1
The two slits are apart from each other and illuminated with a light of wavelength . If the distance of the screen is from the slits, then the distance between third dark fringe and fifth bright fringe is
View answer and explanation
Correct option: C
Option C () is correct because it is calculated by finding the positions of the third dark fringe and fifth bright fringe and then subtracting them.
80
Easy
80 of 150
Application
+4 | -1
An objective lens of a telescope has diameter 200 cm. The limit of resolution of telescope for wavelength 600 nm is nearly
View answer and explanation
Correct option: B
Option B () is correct because it is calculated using the formula for the limit of resolution of a telescope.
81
Medium
81 of 150
Understanding
+4 | -1
Through which character we can distinguish the light waves from sound waves
View answer and explanation
Correct option: C
Polarization
The correct answer is C (Polarization) because polarization is a property unique to transverse waves, like light, and cannot occur in longitudinal waves, like sound.
82
Medium
82 of 150
Application
+4 | -1
If a star emitting light of wavelength 500 nm is moving towards earth with a velocity of then the shift in the wavelength due to doppler’s effect will be:
View answer and explanation
Correct option: A
The correct answer is A () because it is the calculated wavelength shift due to the Doppler effect.
83
Medium
83 of 150
Application
+4 | -1
A star is moving away from the earth with a velocity of . If the velocity of light is then the shift of its spectral line of wavelength due to Doppler’s effect will be
View answer and explanation
Correct option: B
Option B () is correct because the intensity is proportional to the square of the amplitude, and the resultant amplitude depends on the number of half-period zones.
84
Medium
84 of 150
Understanding
+4 | -1
Resolving power of a microscope depends upon
View answer and explanation
Correct option: D
The wavelength of light illuminating the object
Option D is correct. The resolving power of a microscope depends on the wavelength of light illuminating the object, as shorter wavelengths provide better resolution.
85
Easy
85 of 150
Application
+4 | -1
The diffraction pattern of a light of wavelength diffracting from a salt of width is focused on the focal plane of a convex lens of focal length . The width of the secondary
View answer and explanation
Correct option: B
The correct answer is B (2 mm) because the width of the first secondary maximum in a single-slit diffraction pattern is twice the distance from the central maximum to the first minimum.
86
Medium
86 of 150
Remembering
+4 | -1
Soap bubble appears coloured due to the phenomenon of
View answer and explanation
Correct option: A
Interference
Option A is correct. The colors seen in soap bubbles are due to the interference of light waves reflected from the inner and outer surfaces of the thin film.
87
Medium
87 of 150
Remembering
+4 | -1
If microwaves, X rays, infrared, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum are denoted by M, X, I, G, U, R and V then which of the following is the arrangement in ascending order of wavelength?
View answer and explanation
Correct option: C
G, X, U, V, I, M and R
Option C is correct. The electromagnetic spectrum in ascending order of wavelength is: Gamma rays, X-rays, Ultraviolet, Visible, Infrared, Microwaves, and Radio waves.
88
Easy
88 of 150
Application
+4 | -1
In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:
View answer and explanation
Correct option: A
12
Option A (12) is correct because the number of fringes is inversely proportional to the wavelength.
89
Medium
89 of 150
Application
+4 | -1
The radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction is atomic hydrogen. the energy of the emitted wave is nearly
View answer and explanation
Correct option: D
Option D () is correct because the energy of the emitted wave is very small, corresponding to the hyperfine interaction in atomic hydrogen, and can be calculated using .
90
Medium
90 of 150
Remembering
+4 | -1
For the sustained interference of light, the necessary condition is that the two sources should
View answer and explanation
Correct option: A
Have constant phase difference
Option A is correct. For sustained interference, the two sources must have a constant phase difference to create a stable interference pattern.
Revision notes and key takeaways
Review WAVE OPTICS in your textbook or class notes, then use this set to check your understanding.
Showing 61-90 of 150 questions
