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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Mixed difficulty

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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 271-300 of 418 questions

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271 Medium
271 of 418
Understanding +4 | -1
The sky appears blue, because
View answer and explanation
Correct option: A
Blue light is scattered the most
Blue light is scattered the most, which is why the sky appears blue. This phenomenon is known as Rayleigh scattering.
272 Medium
272 of 418
Application +4 | -1
In Young's double slit experiment if the separation between coherent sources is doubled and distance of screen from slit plane is halved, then fringe width becomes
View answer and explanation
Correct option: C
One-fourth
Option C (One-fourth) is correct because the fringe width is directly proportional to the distance to the screen and inversely proportional to the separation between the slits.
273 Medium
273 of 418
Application +4 | -1
nth{n^{th}} Bright fringe if red light (λ1=7500A∘)\left( {{\lambda _1} = 7500\mathop \mathrm{A}\limits^ \circ } \right) coincides with (n+1)th{\left( {n + 1} \right)^{th}} bright fringe of green light (λ2=6000A∘)\left( {{\lambda _2} = 6000\mathop \mathrm{A}\limits^ \circ } \right). The value of n=?n = ?
View answer and explanation
Correct option: A
4
Option A (4) is correct because the condition for the coincidence of bright fringes is nλ1=(n+1)λ2n\lambda_1 = (n+1)\lambda_2, which can be solved for n.
274 Medium
274 of 418
Understanding +4 | -1
Wavefront of a wave has direction with wave motion
View answer and explanation
Correct option: B
Perpendicular
Option B (Perpendicular) is correct because the wavefront is defined as the locus of points with the same phase, and the wave motion is perpendicular to it.
275 Medium
275 of 418
Understanding +4 | -1
An optically active compound
View answer and explanation
Correct option: A
Rotates the plane polarized light
Option A (Rotates the plane polarized light) is correct because an optically active compound has the ability to rotate the plane of polarization of plane-polarized light.
276 Medium
276 of 418
Understanding +4 | -1
VO  and  VE{V_O}\;and\;{V_E} represent the velocities, μO  and  μE{\mu _O}\;and\;{\mu _E} the refractive indices of ordinary and extraordinary rays for a doubly refracting crystal. Then
View answer and explanation
Correct option: C
Vo≤VE,μo≥μE{V_o} \le {V_E},{\mu _o} \ge {\mu _E} if the crystal is calcite
Option C (Vo≤VE,μo≥μE{V_o} \le {V_E},{\mu _o} \ge {\mu _E} if the crystal is calcite) is correct because for calcite, the refractive index of the ordinary ray is greater than or equal to that of the extraordinary ray, and the velocity is inversely related.
277 Medium
277 of 418
Application +4 | -1
In Young’s double slit experiment, distance between two slits is 0.28  mm0.28\;\mathrm{mm} and distance between slits and screen is 1.4  m1.4\;\mathrm{m}. Distance between central bright fringe and third bright fringe is 0.9  cm0.9\;\mathrm{cm}, what is the wavelength of light used?
View answer and explanation
Correct option: B
6000A∘6000\mathop \mathrm{A}\limits^ \circ
Option B (6000A∘6000\mathop \mathrm{A}\limits^ \circ) is correct because the wavelength is calculated using the formula for fringe width in Young's double-slit experiment.
278 Medium
278 of 418
Understanding +4 | -1
Plane polarised 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
View answer and explanation
Correct option: C
There is no change in intensity
Option C (There is no change in intensity) is correct because the intensity of plane-polarized light passing through a polaroid remains constant when the polaroid is rotated about the direction of light.
279 Medium
279 of 418
Understanding +4 | -1
In single slit diffraction pattern
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because in single-slit diffraction, the central fringe is the widest and brightest, and other fringes have decreasing width and intensity.
280 Medium
280 of 418
Understanding +4 | -1
Assertion : Polaroid glasses reduce glare from road or water etc.

Reason : Reflected waves are partially polarised.
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 (If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion) is correct because reflected light is partially polarized, and Polaroid glasses reduce glare by blocking this polarized light.
281 Easy
281 of 418
Application +4 | -1
Monochromatic light of wavelength 600 nm600 \mathrm{~nm} is incident from air on a glass surface (refractive index of glass is 1.51.5). The frequency of the refracted light will be
View answer and explanation
Correct option: A
5×1014 Hz5 \times 10^{14} \mathrm{~Hz}
Option A (5×1014 Hz5 \times 10^{14} \mathrm{~Hz}) is correct because the frequency of light remains the same when it is refracted.
282 Easy
282 of 418
Analyzing +4 | -1
Statement-1: In Young's double slit experiment the two slits are at distance dd apart. Interference pattern is observed on a screen at distance DD 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.
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 because the wavelength is proportional to the square of the distance between the slits, and intensity is indeed zero for a dark fringe.
283 Medium
283 of 418
Application +4 | -1
A ray of light falls on a transparent glass slab with refractive index (relative to air) of 1.62\mathrm{1}\mathrm{.62}. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is
View answer and explanation
Correct option: A
tan⁡−1  (1.62){\tan ^{ - 1}}\,\,\left( {1.62} \right)
Option A (tan⁡−1  (1.62){\tan ^{ - 1}}\,\,\left( {1.62} \right)) is correct because it is derived from the relationship between refractive index and the angle of incidence when reflected and refracted rays are perpendicular.
284 Medium
284 of 418
Understanding +4 | -1
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
View answer and explanation
Correct option: C
Partially polarised
Option C (Partially polarised) is correct because the light exhibits a maximum and minimum intensity, but not complete darkness, indicating a partially polarized state.
285 Easy
285 of 418
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 the angle of incidence at which the reflected light is completely polarized is Brewster's angle, given by μ=tan⁡θP\mu = \tan \theta_P, where μ\mu is the refractive index.
286 Medium
286 of 418
Remembering +4 | -1
Wavefront means
View answer and explanation
Correct option: A
All particles in it have same phase
Option A (All particles in it have same phase) is correct because a wavefront is defined as a surface containing points affected by a wave that are all in the same phase.
287 Hard
287 of 418
Application +4 | -1
Two coherent light sources having intensity in the ratio 2 ×2\, \times produce an interference pattern. The ratio lmax⁡−lmin⁡lmax⁡+lmin⁡\frac{{\begin{array}{cccccccccccccccccccc} {{l_{\max }} - }&{{l_{\min }}} \end{array}}}{{\begin{array}{cccccccccccccccccccc} {{l_{\max }} + }&{{l_{\min }}} \end{array}}} will be
View answer and explanation
Correct option: D
22x2x+1\frac{{2\sqrt {2x} }}{{2x + 1}}
Option D is correct. The ratio is found using the formula for maximum and minimum intensities in interference.
288 Medium
288 of 418
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 fringe width is inversely proportional to the wavelength, and the field of view remains the same.
289 Medium
289 of 418
Application +4 | -1
In Young's double slit experiment, slits are separated by 2 mm2 \mathrm{~mm} and the screen is placed at a distance of 1.2 m1.2 \mathrm{~m} from the slits. Light consisting of two wavelengths 6500  A∘6500\,\,\mathop {\text{A}}\limits^ \circ and 5200  A∘5200\,\,\mathop {\text{A}}\limits^ \circ are used to obtain interference fringes. Then, the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is
View answer and explanation
Correct option: A
0.312 mm0.312 \mathrm{~mm}
Option A (0.312 mm0.312 \mathrm{~mm}) is correct because it is the calculated separation between the fourth bright fringes of the two wavelengths.
290 Medium
290 of 418
Understanding +4 | -1
Oil floating on water looks coloured due to interference of light. What should be the order of magnitude of thickness of oil layer in order that this effect may be observed?
View answer and explanation
Correct option: A
10,000A∘10,000\mathop \mathrm{A}\limits^ \circ
Option A (10,000A∘10,000\mathop \mathrm{A}\limits^ \circ) is correct because the thickness of the oil layer must be on the order of the wavelength of visible light for interference to be observed.
291 Easy
291 of 418
Understanding +4 | -1
Assertion: Incoming light reflected by earth is partially polarized.
Reason: Atmospheric particle polarize the light.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A (If both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because atmospheric particles scatter sunlight, causing it to become partially polarized.
292 Medium
292 of 418
Application +4 | -1
A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through 30∘{30^\circ } with respct to the preceding one. The percentage of incident intensity that passes through the system will be
View answer and explanation
Correct option: D
12%12\%
Option D (12%12\%) is correct because the intensity of light passing through multiple polaroids is calculated using Malus's law, considering the angle between each successive polaroid.
293 Medium
293 of 418
Understanding +4 | -1
A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
View answer and explanation
Correct option: C
Hyperbola
Option C (Hyperbola) is correct because the interference fringes formed in Young's double-slit experiment are hyperbolic in shape.
294 Medium
294 of 418
Application +4 | -1
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
View answer and explanation
Correct option: A
2
Option A (2) is correct because the ratio of intensities is calculated using the formula for intensity in interference and the phase difference.
295 Medium
295 of 418
Remembering +4 | -1
Light waves can be polarized as they are
View answer and explanation
Correct option: A
Transverse
Option A (Transverse) is correct because polarization is a phenomenon exhibited by transverse waves, where the oscillations occur in a single plane.
296 Medium
296 of 418
Application +4 | -1
Two parallel slits 0.6 mm0.6\,mm apart are illuminated by light source of wave length 6000 A∘6000\,\mathop \mathrm{A}\limits^ \circ. The distance between two consecutive dark fringes on a screen 1  m1\,\,m away from the slits is
View answer and explanation
Correct option: A
1 mm1\,mm
Option A (1 mm) is correct because it uses the formula for fringe width in Young's double-slit experiment to calculate the distance between consecutive dark fringes.
297 Medium
297 of 418
Remembering +4 | -1
Resolving power of microscope depends upon
View answer and explanation
Correct option: B
wavelength of light used (inversely proportional)
Option B (wavelength of light used (inversely proportional)) is correct because the resolving power of a microscope is inversely proportional to the wavelength of the light used.
298 Easy
298 of 418
Understanding +4 | -1
Assertion: Distance between position of bright and dark fringe remain same in YDSE
Reason: Fringe width β=λDd\beta=\frac{\lambda D}{d}
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
Option A is correct because the fringe width in Young's double-slit experiment is constant and depends on wavelength, distance to the screen, and slit separation.
299 Medium
299 of 418
Application +4 | -1
A beam of light consisting of two wavelengths 650  nm650\;nm and 520  nm520\;nm is used to illuminate the slit of a Young’s double slit experiment. Then the order of the bright fringe of the longer wavelength that coincide with a bright fringe of the shorter wavelength at the least distance from the central maximum is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the condition for coinciding bright fringes is n1λ1=n2λ2n_1 \lambda_1 = n_2 \lambda_2, and solving for the smallest integer nn gives n=4n=4.
300 Medium
300 of 418
Analyzing +4 | -1
Assertion: We cannot observe diffraction pattern from a wide slit iluminated by monochromatic light.

Reason: In diffraction pattern, all the bright bands are not of the same intensity.
View answer and explanation
Correct option: B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
Option B is correct. While diffraction does occur, it's less noticeable with wide slits, and the intensity variation is a key characteristic of diffraction patterns.
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