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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 396 questions

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61 Medium
61 of 396
Remembering +4 | -1
For the constructive interference the path difference between the two interfering waves must be equal to
View answer and explanation
Correct option: C
nλn\lambda
Option C (nλn\lambda) is correct because for constructive interference, the path difference between two interfering waves must be an integer multiple of the wavelength.
62 Medium
62 of 396
Understanding +4 | -1
Two coherent sources of intensities, I1  and  I2  {I_1}\;and\;{I_2}\; produce an interference pattern. The maximum intensity in the interference pattern will be
View answer and explanation
Correct option: D
(I1+I2)2{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}
Option D ((I1+I2)2{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}) is correct because the maximum intensity in an interference pattern occurs when the phase difference is zero, leading to constructive interference.
63 Medium
63 of 396
Understanding +4 | -1
The phenomenon of polarization of light indicates that
View answer and explanation
Correct option: B
Light is a transverse wave
Option B (Light is a transverse wave) is correct because polarization is a phenomenon exhibited only by transverse waves, not longitudinal waves.
64 Medium
64 of 396
Understanding +4 | -1
In the diffraction pattern of a single slit
View answer and explanation
Correct option: D
Central band is wider
Option D (Central band is wider) is correct because the central maximum in a single-slit diffraction pattern is twice the width of the other fringes.
65 Medium
65 of 396
Understanding +4 | -1
An unpolarised beam of light is incident on a group of four polarising sheets which are arranged in such a way that the characteristic direction of each polarising sheet makes an angle of 30∘30^{\circ} with that of the preceding sheet.The percentage of incident light transmitted by the first polariser will be
View answer and explanation
Correct option: B
50%50 \%
Option B (50 %) is correct because the first polariser reduces the intensity of unpolarised light by half.
66 Medium
66 of 396
Application +4 | -1
Two light waves of same intensity on superposition gives a maximum of 400 units. If amplitude of both waves are reduced by 50%50 \% then the maximum intensity will be-
View answer and explanation
Correct option: C
100
Option C (100) is correct because the maximum intensity is proportional to the square of the amplitude, and reducing the amplitude by 50% reduces the intensity to 1/4 of the original.
67 Medium
67 of 396
Application +4 | -1
A ray of light travelling in water is incident on a glass plate immersed in it. When the angle of incidence is 51∘51^{\circ}, the reflected ray is totally polarized. Find out the refractive index of glass.

[The refractive index of water is 1.3 and tan⁡51∘=1.235\tan 51^{\circ}=1.235 ]
View answer and explanation
Correct option: A
1.605
Option A (1.605) is correct because it is calculated using Brewster's law, where the tangent of the polarizing angle equals the ratio of refractive indices.
68 Medium
68 of 396
Understanding +4 | -1
The ratio of intensities of successive maxima in the diffraction pattern due to the single slit is
View answer and explanation
Correct option: C
1:49π2:425π21:\frac{4}{{9{\pi ^2}}}:\frac{4}{{25{\pi ^2}}}
Option C (1:49π2:425π21:\frac{4}{{9{\pi ^2}}}:\frac{4}{{25{\pi ^2}}}) is correct because it represents the relative intensities of the maxima in a single-slit diffraction pattern.
69 Medium
69 of 396
Application +4 | -1
In a Young’s experiment, two coherent sources are placed 0.90  mm0.90\;mm apart and the fringes are observed one meter away. If it produces the second dark fringe at a distance of 1  mm1\;mm from the central fringe, the wavelength of monochromatic light used will be
View answer and explanation
Correct option: D
6×10−5  cm6 \times {10^{ - 5}}\;\mathrm{cm}
Option D (6×10−5  cm6 \times {10^{ - 5}}\;\mathrm{cm}) is correct because it is the wavelength calculated using the formula for the position of dark fringes in Young's experiment.
70 Medium
70 of 396
Understanding +4 | -1
When a ray of light is incident normally on a surface, then
View answer and explanation
Correct option: B
It passes undeviated
Option B (It passes undeviated) is correct because when a ray of light is incident normally on a surface, the angle of incidence is zero, so there is no deviation.
71 Medium
71 of 396
Understanding +4 | -1
Wave which cannot travel in vacuum is
View answer and explanation
Correct option: B
Infrasonic
Option B (Infrasonic) is correct because infrasonic waves are mechanical waves and require a medium to propagate, unlike electromagnetic waves.
72 Medium
72 of 396
Application +4 | -1
In Young's double slit experiment, 50 fringes are observed in the central view zone with light of wavelength 4000 A∘.4000 \, \mathop {\text{A}}\limits^ \circ. The number of fringes that will be observed in the same view zone with the light of wavelength 5000 A∘5000 \, \mathop {\text{A}}\limits^ \circ is
View answer and explanation
Correct option: D
40
Option D (40) is correct because the number of fringes is inversely proportional to the wavelength of light used in Young's double slit experiment.
73 Medium
73 of 396
Understanding +4 | -1
If the shift in a star light is towards red end, then
View answer and explanation
Correct option: D
Both (b) and (c )
Option D (Both (b) and (c )) is correct because a redshift indicates that the star is moving away from the Earth, and the observed frequency is lower than the actual frequency.
74 Medium
74 of 396
Understanding +4 | -1
The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is
View answer and explanation
Correct option: C
1:49π2:425π21:\frac{4}{{9{\pi ^2}}}:\frac{4}{{25{\pi ^2}}}
Option C (1:1\begin{array}{llllllllllllllllllll}{\mathrm{1 : 1}}\end{array}) is correct because specific resistance is a material property and only depends on the material and temperature, not on dimensions.
75 Medium
75 of 396
Understanding +4 | -1
By Huygen’s wave theory of light, we cannot explain the phenomenon of
View answer and explanation
Correct option: C
Photoelectric effect
Option C (Photoelectric effect) is correct because Huygen's wave theory cannot explain the particle nature of light, which is fundamental to the photoelectric effect.
76 Medium
76 of 396
Application +4 | -1
In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be
View answer and explanation
Correct option: C
0.375694444
Option C (09:01) is correct because the ratio of intensities is calculated based on the ratio of slit widths in Young's double-slit experiment.
77 Medium
77 of 396
Application +4 | -1
Two Nicols are oriented with their principal planes making an angle of 60∘60^\circ. The percentage of incident unpolarised light which passes through the system is
View answer and explanation
Correct option: C
12.5%12.5\%
Option D (37.5%37.5\%) is correct because the intensity of light passing through two polarizers at 60 degrees is calculated by Malus's law.
78 Medium
78 of 396
Application +4 | -1
In a Young’s double slit experiment, the slit separation is 0.2  cm0.2\;cm, the distance between the screen and slit is 1  m1\;m. Wavelength of the light used is 5000A∘5000\mathop \mathrm{A}\limits^ \circ. The distance between two consecutive dark fringes (in mmmm) is
View answer and explanation
Correct option: A
0.25
Option A, 0.25, is correct because it is the calculated fringe width using the formula β=λDd\beta = \frac{{\lambda D}}{d} with the given values.
79 Medium
79 of 396
Application +4 | -1
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm coming from a distant object, the limit of resolution of the telescope is close to
View answer and explanation
Correct option: A
3.0×10−7rad3.0 \times {10^{ - 7}}\mathrm{rad}
Option A (3.0×10−7rad3.0 \times {10^{ - 7}}\mathrm{rad}) is correct because it is calculated using the formula for the limit of resolution of a telescope: θR=1.22λD\theta_R = \frac{{1.22\lambda}}{D}.
80 Medium
80 of 396
Remembering +4 | -1
The condition for the first minimum in case of diffraction due to a single slit is
View answer and explanation
Correct option: A
asin⁡θ=λ\operatorname{asin} \theta=\lambda
Option A (asin⁡θ=λ\operatorname{asin} \theta=\lambda) is correct because this equation represents the condition for the first minimum in single-slit diffraction, where 'a' is the slit width, 'θ\theta' is the angle, and 'λ\lambda' is the wavelength.
81 Medium
81 of 396
Analyzing +4 | -1
Assertion (A) : For two sources to be coherent phase difference between two waves at all points should be same.

Reason (R) : Two different light sources are never coherent.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D is correct because the assertion is false, as phase difference needs to be constant, not necessarily the same at all points, and the reason is true, as two different light sources are never perfectly coherent.
82 Medium
82 of 396
Application +4 | -1
In a double slit experiment λ=405 nm,\lambda = 405\,nm, the slit separation is 20 μm20\,\mu m and slit width is 4 μm .4\,\mu m\,. Number of bright fringes within central peak of diffraction envelop is:-
View answer and explanation
Correct option: C
10
Option C (10) is correct because it is calculated by finding the ratio of the width of the central diffraction peak to the fringe width in the double-slit experiment.
83 Medium
83 of 396
Understanding +4 | -1
Huygen’s principle does not use
View answer and explanation
Correct option: D
Point of spectra origin
Option D (Point of spectra origin) is correct because Huygens' principle is used to explain reflection, refraction, and diffraction, but it doesn't deal with the origin of spectra.
84 Medium
84 of 396
Application +4 | -1
In a Young’s experiment, one of the slits is covered with a transparent sheet of thickness 3.6×10−3  cm3.6 \times {10^{ - 3}}\;cm due to which position of central fringe shifts to a position originally occupied by 30th{30^{th}} fringe. The refractive index of the sheet, if λ=6000A∘\lambda = 6000\mathop \mathrm{A}\limits^ \circ, is
View answer and explanation
Correct option: A
1.5
Option A (1.5) is correct because the refractive index is calculated using the shift in fringe position due to the transparent sheet in Young's experiment.
85 Medium
85 of 396
Application +4 | -1
Path difference between two wavefronts emitted from coherent sources is 2.1μm2.1 \mu \mathrm{m}. Phase difference between the wavefronts at that point is 7.692π7.692 \pi. Wavelength of light emitted by source will be
View answer and explanation
Correct option: C
5460A05460 A^0
Option C (5460A05460 A^0) is correct because it is the wavelength calculated using the path difference and phase difference.
86 Medium
86 of 396
Understanding +4 | -1
In Fresnel’s biprism experiment, on increasing the prism angle, fringe width will
View answer and explanation
Correct option: B
Decrease
Option B (Decrease) is correct because the fringe width in Fresnel's biprism experiment is inversely proportional to the prism angle.
87 Medium
87 of 396
Remembering +4 | -1
The wave theory of light, in its original form, was first postulated by
View answer and explanation
Correct option: B
Christian Huygens
Option B (Christian Huygens) is correct because Christian Huygens first postulated the wave theory of light in its original form.
88 Medium
88 of 396
Analyzing +4 | -1
Assertion: If the whole apparatuis of Young's experiement is immersed in liquid, the fringe width will decrease.

Reason: The wavelength of light in water is more than that in air.
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 because the fringe width decreases when the apparatus is immersed in a liquid, but the wavelength of light in water is less than that in air.
89 Medium
89 of 396
Application +4 | -1
The wavelength of a distant star is 5700  A∘5700\,\,\mathop {\text{A}}\limits^ \circ and the spectral light has a shift 1.9  A∘1.9\,\,\mathop {\text{A}}\limits^ \circ towards red end then the velocity of star relative to earth will be:
View answer and explanation
Correct option: D
1×105 m/sec1 \times 10^5 \mathrm{~m} / \mathrm{sec}
Option D (1×105 m/sec1 \times 10^5 \mathrm{~m} / \mathrm{sec}) is correct because it is calculated using the Doppler shift formula.
90 Medium
90 of 396
Understanding +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labeled as Reason R.
Assertion A: An electron microscope can achieve better resolving power than an optical microscope.
Reason R: The de-Brogilie’s wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: B
Both A and R are true and R is the correct explanation of A.
Option B (Both A and R are true and R is the correct explanation of A) is correct because electron microscopes have better resolving power due to the shorter de Broglie wavelength of electrons compared to visible light.
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