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NEET • PHYSICS • WAVE OPTICS

Wave optics practice · Mixed difficulty

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Showing 31-60 of 150 questions

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31 Easy
31 of 150
Application +4 | -1
In Young's double slit experiment, the width of fringe obtained from a source of light of wavelength 5000A∘5000\mathop {\text{A}}\limits^ \circ is 3.9 mm\mathrm{mm}. if the apparatus is immersed in a liquid of μ=1.3\mu=1.3, the fringe width will be
View answer and explanation
Correct option: A
3
The correct answer is A (3) because the fringe width is inversely proportional to the refractive index of the medium.
32 Hard
32 of 150
Understanding +4 | -1
Ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is
View answer and explanation
Correct option: D
1:49π2:425π21: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}
Option D (1:49π2:425π21: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}) is correct because the intensities of the secondary maxima in single-slit diffraction decrease rapidly and are proportional to the inverse square of the sine of the angle.
33 Hard
33 of 150
Application +4 | -1
In a Lloyd's mirror experiment if the mirror reflects 75%75 \% of light incident on it the ratio of intensity at interference maxima and minima will be:
View answer and explanation
Correct option: C
194
Option C (194) is correct because it is the closest integer to the calculated ratio of maximum to minimum intensity, considering the reflection coefficient.
34 Medium
34 of 150
Analyzing +4 | -1
In YDSE, if one slit is covered by a transparent film so that it transmitts only one half the light intensity of other then choose the incorrect statement :
View answer and explanation
Correct option: B
Fringe pattern shifts away from the covered slit
Option B is the incorrect statement. When one slit is covered with a transparent film, the fringe pattern shifts towards the covered slit because it introduces an additional path difference.
35 Medium
35 of 150
Application +4 | -1
In young’s double slit experiment, the fringe pattern is observed on screen placed at a distance D. The slits are illuminated by light of wavelength λ\lambda. The distance from the central point where intensity becomes 34th\frac{3}{4}th of the maxima will be
View answer and explanation
Correct option: A
λD6d\frac{{\lambda D}}{{6d}}
Option A (λD6d\frac{{\lambda D}}{{6d}}) is correct because it is derived from the intensity formula in Young's double-slit experiment.
36 Medium
36 of 150
Analyzing +4 | -1
Assertion : Light ray passing through water droplet is partially polarized.
Reason : Water is a polar molecule.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option A is correct. Light scattering from water droplets is partially polarized due to the polar nature of water molecules, making both the assertion and reason true, and the reason the correct explanation.
37 Medium
37 of 150
Understanding +4 | -1
Which one of the following property of light does not support wave theory of light?
View answer and explanation
Correct option: C
Light shows photoelectric effect
Option C (Light shows photoelectric effect) is correct because the photoelectric effect demonstrates the particle nature of light, not the wave nature.
38 Easy
38 of 150
Understanding +4 | -1
If the monochromatic source in Young's double slit experiment is replaced by white light, then
View answer and explanation
Correct option: B
There will be a central bright white fringe surrounded by a few coloured fringes
Option B is correct. With white light, the central fringe is white because all colors interfere constructively. Colored fringes appear on either side due to different wavelengths interfering at different locations.
39 Medium
39 of 150
Application +4 | -1
In Young’s double slit experiment, when violet light of wavelength 4358A∘4358\mathop \mathrm{A}\limits^ \circ is used, the 84 fringe are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen in the field of view, the wavelength of sodium light is
View answer and explanation
Correct option: B
5904A∘5904\mathop \mathrm{A}\limits^ \circ
Option B (5904A∘5904\mathop \mathrm{A}\limits^ \circ) is correct because the ratio of the wavelengths is inversely proportional to the number of fringes observed.
40 Hard
40 of 150
Application +4 | -1
Two coherent sources emit light waves, which superimpose at a point where these can be expressed as

E1=E0sin⁡(ωt+π/4)E2=2E0sin⁡(ωt−π/4)\begin{aligned} & \mathrm{E}_1=\mathrm{E}_0 \sin (\omega \mathrm{t}+\pi / 4) \\ & \mathrm{E}_2=2 \mathrm{E}_0 \sin (\omega \mathrm{t}-\pi / 4) \end{aligned}

here E1\mathrm{E}_1 and E2\mathrm{E}_2 are the electric field strengths of the two waves at the given point. If I\mathrm{I} is the intensity of wave expressed by field strength E1\mathrm{E}_1, find the resultant intensity.
View answer and explanation
Correct option: D
5I5 \mathrm{I}
Option D (5I5 \mathrm{I}) is correct because the resultant intensity is calculated by considering the phase difference and amplitudes of the two interfering waves.
41 Easy
41 of 150
Understanding +4 | -1
Statement-1: In interference experiment, the fringe system will shift, if a glass plate is interposed in the path of the one of the interfering beams and the shift will be on the side of the glass plate.
Statement-2: The velocity of light in glass is smaller than in air.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A is correct. Statement-1 is true because the introduction of a glass plate shifts the fringe system. Statement-2 is true because light travels slower in glass than in air, and it is the correct explanation for the shift.
42 Medium
42 of 150
Remembering +4 | -1
The wave theory of light was given by
View answer and explanation
Correct option: C
Huygen
Option C (Huygen) is correct because Christian Huygens is credited with proposing the wave theory of light.
43 Hard
43 of 150
Application +4 | -1
Two light rays having the same wavelength λ\lambda in vacuum are in phase initially. Then the first ray travels a path L1{{L_1}} through a medium of refractive index n1{{n_1}} while the second ray travels a path of length L2{{L_2}} through a medium of refractive index n2{{n_2}}. 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
View answer and explanation
Correct option: B
2πλ(n1L1−n2L2)\frac{{2\pi }}{\lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)
Option B (2πλ(n1L1−n2L2)\frac{{2\pi }}{\lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)) is correct because the phase difference is proportional to the optical path difference.
44 Medium
44 of 150
Analyzing +4 | -1
Two monochromatic coherent point sources S1S_1 and S2S_2 are separated by a distance LL. Each source emits light of wavelength λ\lambda, where L>>λL >> \lambda. The line S1S2S_1 S_2 when extended meets a screen perpendicular to it at a point A.A.
View answer and explanation
Correct option: A
The interference fringes on the screen are circular in shape
Option A (The interference fringes on the screen are circular in shape) is correct because the path difference remains constant along circles centered at the midpoint of S1S_1 and S2S_2.
45 Medium
45 of 150
Remembering +4 | -1
The resolving power of a telescope depends on
View answer and explanation
Correct option: D
Diameter of the objective lens
Option D is correct. The resolving power of a telescope is directly proportional to the diameter of the objective lens.
46 Medium
46 of 150
Remembering +4 | -1
The condition for diffraction of mth{m^{th}} order minima is
View answer and explanation
Correct option: A
dsin⁡θm=mλ,  m=1,2,3,…d\sin {\mathrm{\theta }_m} = m\lambda ,\;m = 1,2,3, \ldots
Option A is correct. The condition for diffraction minima is given by dsinθ = mλ, where m = 1, 2, 3,...
47 Easy
47 of 150
Application +4 | -1
A parallel beam of monochromatic unpolarised
light is incident on a transparent plate of
refractive index 3.\sqrt{3}. The reflected beam is
completely polarized. The angle of incidence
is
View answer and explanation
Correct option: B
60∘60^\circ
Option B (60∘60^\circ) is correct because at the Brewster angle, the reflected light is completely polarized, and the tangent of this angle equals the refractive index.
48 Medium
48 of 150
Application +4 | -1
A parallel beam of light of intensity I0{I_0} is incident on a glass plate, 25%25\% of light is reflected by upper surface and 50%50\% of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is
View answer and explanation
Correct option: A
(12+3812−38)2{\left( {\frac{{\frac{1}{2} + \sqrt {\frac{3}{8}} }}{{\frac{1}{2} - \sqrt {\frac{3}{8}} }}} \right)^2}
The correct answer is A ((12+3812−38)2{\left( {\frac{{\frac{1}{2} + \sqrt {\frac{3}{8}} }}{{\frac{1}{2} - \sqrt {\frac{3}{8}} }}} \right)^2}) because it is the ratio of maximum to minimum intensity calculated from the reflection coefficients.
49 Medium
49 of 150
Understanding +4 | -1
To increase the resolving power of a telescope
View answer and explanation
Correct option: C
Both (1) & (2)
Option C is correct. The resolving power of a telescope increases when the wavelength of light is decreased and the aperture of the objective lens is increased.
50 Medium
50 of 150
Application +4 | -1
Find the thickness of a plate which will produce a change in optical path equal to half the wavelength λ\lambda of the light passing through it normally. The refractive index of the plate μ\mu is
View answer and explanation
Correct option: D
λ2(μ−1)\frac{\lambda }{{2\left( {\mu - 1} \right)}}
Option D (λ2(μ−1)\frac{\lambda }{{2\left( {\mu - 1} \right)}}) is correct because it represents the thickness required to introduce an optical path difference of half the wavelength.
51 Easy
51 of 150
Application +4 | -1
Two light waves of the same intensity are interfering. The intensity of light at a bright fringe compared to the intensity of one of the waves will
View answer and explanation
Correct option: C
Be four times
Option C (Be four times) is correct because the intensity of light is proportional to the square of the amplitude, and in constructive interference, the amplitudes add up.
52 Medium
52 of 150
Application +4 | -1
A beam of ordinary unpolarised light passes through a tourmaline crystal C1{C_1} and then it passes through another tourmaline crystal C2{C_2}, which is oriented such that its principal plane is parallel to that of C2{C_2}. The intensity if emergent light is I0{I_0}. Now C2{C_2} is rotated by 60∘60^\circ about the ray. The emergent ray will have an intensity
View answer and explanation
Correct option: C
I0/4{I_0}/4
The correct answer is C (I0/4{I_0}/4) because the intensity is reduced by a factor of cos²(θ) after passing through the second polarizer, and the initial intensity is already halved by the first polarizer.
53 Medium
53 of 150
Understanding +4 | -1
In Young’s double slit experiment, the distance between the two slits is made half, then the fringe width will become
View answer and explanation
Correct option: B
Double
Option B (Double) is correct because fringe width is inversely proportional to the distance between the slits. Halving the distance doubles the fringe width.
54 Medium
54 of 150
Application +4 | -1
The wavelength of light observed on the earth from a moving star is found to decrease by 0.05  %0.05\,\,\%. the star is
View answer and explanation
Correct option: C
Coming closer with a velocity of 1.5×105  ms−11.5 \times {10^5}\,\,\mathrm{m}{\mathrm{s}^{ - 1}}
Option C (Coming closer with a velocity of 1.5×105 ms−11.5 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}) is correct because a decrease in wavelength indicates a blueshift, meaning the star is approaching, and the velocity can be calculated using the Doppler shift formula.
55 Medium
55 of 150
Analyzing +4 | -1
Statement A: In the interference pattern the intensity is same at all points in a bright band

Statement B: In Young's double slit experiment, as we move away from the central maximum, the third maximum always comes before the third minimum.
View answer and explanation
Correct option: A
A and B are correct
Option A (A and B are correct) is correct because both statements about interference patterns and Young's double-slit experiment are true.
56 Medium
56 of 150
Understanding +4 | -1
The phenomenon of polarization of light indicates that
View answer and explanation
Correct option: B
Light is a transverse wave
The phenomenon of polarization of light indicates that light is a transverse wave. Only transverse waves can be polarized, as their oscillations are perpendicular to the direction of propagation.
57 Medium
57 of 150
Application +4 | -1
The kk line of singly ionized calcium has a wavelength of 393.3 nm393.3\,nm as measured on earth. In the spectrum of one of the observed galaxies, this spectral line is located at 401.8 nm401.8\,nm. The speed with which the galaxy is moving away from us, will be
View answer and explanation
Correct option: A
6480 km/s6480\,km/s
Option D (0.052777778) is correct because the intensity of scattered light is proportional to the fourth power of the frequency.
58 Medium
58 of 150
Analyzing +4 | -1
Statement A: In Fraunhofer diffraction central observation point is always bright where as in Fresnel diffraction observation point may bright or dark.

Statement B: In Fresnel light waves approaches at observation point are always in same phase, whereas in Fraunhofer light waves approaches at central observation point may be in same phase or in opposite phase.
View answer and explanation
Correct option: C
A is correct and B is wrong
Option C (A is correct and B is wrong) is correct. In Fraunhofer diffraction, the central observation point is always bright. In Fresnel diffraction, it can be bright or dark. In Fresnel diffraction, the light waves arriving at the observation point are not always in the same phase.
59 Medium
59 of 150
Understanding +4 | -1
Interference may be seen using two independent
View answer and explanation
Correct option: C
Lasers
Option C (Lasers) is correct because lasers are highly coherent light sources, making them suitable for observing interference.
60 Easy
60 of 150
Understanding +4 | -1
In young's double slit experiment with a source of light of wavelength 6320 A∘6320\,\mathop {\text{A}}\limits^ \circ, the first maxima will occur when
View answer and explanation
Correct option: C
Path difference is 6320 A∘6320\,\mathop {\text{A}}\limits^ \circ
Option C is correct. The first maxima occurs when the path difference is equal to one wavelength, which is 6320 A∘6320\,\mathop {\text{A}}\limits^ \circ.
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