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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 271-300 of 396 questions

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271 Medium
271 of 396
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.
272 Medium
272 of 396
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.
273 Medium
273 of 396
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.
274 Medium
274 of 396
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.
275 Medium
275 of 396
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.
276 Medium
276 of 396
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.
277 Medium
277 of 396
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.
278 Medium
278 of 396
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.
279 Medium
279 of 396
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.
280 Medium
280 of 396
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.
281 Medium
281 of 396
Application +4 | -1
In an experiment to measure the speed of light by Fizeau's apparatus, following data are used : Distance between the mirrors =12.0 km=12.0 \mathrm{~km}, Number of teeth in the wheel =180=180. Find the minimum angular speed of the wheel for which the image is not seen.
View answer and explanation
Correct option: A
1.25×104 deg/sec1.25 \times 10^4\,\mathrm{deg/sec}
Option A (1.25×104 deg/sec1.25 \times 10^4\,\mathrm{deg/sec}) is correct because it calculates the minimum angular speed of the wheel using the Fizeau's apparatus formula.
282 Medium
282 of 396
Understanding +4 | -1
Assertion: The soap film in sun light is colourful.

Reason: Thin films produce interfernce of light.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the colorful appearance of soap films is due to the interference of light waves reflecting off the two surfaces of the thin film.
283 Medium
283 of 396
Application +4 | -1
In the Young’s double slit experiment, the distance between the slits varies in time as d(t)=  d0+a0  sin⁡ωtd(t) = \,\,{d_0} + {a_0}\,\,\sin \omega t
where d0  ,  ω{d_0}\,\,,\,\,\omega and a0{a_0} are constants. The difference between the largest fringe width and the smallest fringe
width obtained over time is given as
View answer and explanation
Correct option: B
2λDa0(d02−a02)\frac{{2\lambda D{a_0}}}{{\left( {d_0^2 - a_0^2} \right)}}
Option B is correct because the difference between the largest and smallest fringe widths is derived by considering the maximum and minimum values of the time-varying slit separation in Young's double-slit experiment.
284 Medium
284 of 396
Remembering +4 | -1
The solar spectrum during a complete solar eclipse is
View answer and explanation
Correct option: A
Continuous
Option A (Continuous) is correct because during a complete solar eclipse, the solar spectrum is continuous due to the light received from the chromosphere.
285 Medium
285 of 396
Remembering +4 | -1
Electromagnetic radiation of highest frequency is
View answer and explanation
Correct option: D
γ−rays\gamma - rays
Option D (γ−rays\gamma - rays) is correct because gamma rays have the highest frequency in the electromagnetic spectrum.
286 Medium
286 of 396
Understanding +4 | -1
In the phenomena of diffraction of light, when blue light is used in the experiment in spite of red light, then
View answer and explanation
Correct option: A
fringes will become narrower
Option A (fringes will become narrower) is correct because the fringe width in diffraction is directly proportional to the wavelength, and blue light has a shorter wavelength than red light.
287 Medium
287 of 396
Analyzing +4 | -1
Assertion: Diffraction is common in sound but not common in light waves.

Reason: Wavelength of light is more than the wavelength of sound.
View answer and explanation
Correct option: C
Assertion is true and Reason is false
Option C is correct. Assertion is true because diffraction is more common with sound waves due to their longer wavelengths. Reason is false because the wavelength of light is much shorter than the wavelength of sound.
288 Medium
288 of 396
Understanding +4 | -1
Large aperture of telescope are used for
View answer and explanation
Correct option: B
Greater resolution
Option B (Greater resolution) is correct because a larger aperture increases the resolving power of a telescope.
289 Medium
289 of 396
Understanding +4 | -1
A Young’s double slit set up for interference is shifted from air to within water, then the fringe width
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because the fringe width is inversely proportional to the refractive index of the medium.
290 Medium
290 of 396
Understanding +4 | -1
Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will see
View answer and explanation
Correct option: D
Interference in which width of the fringe will be slightly increased
Option D is correct because evacuating the chamber reduces the refractive index, increasing the wavelength and thus slightly increasing the fringe width.
291 Medium
291 of 396
Understanding +4 | -1
Assertion(A): It is not possible to have interference between the waves produced by two violins.

Reason(R): For interference of two waves, the phase difference between the waves must remain constant.
View answer and explanation
Correct option: A
Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
Option A (Both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because interference requires a constant phase difference, which is not maintained by two violins.
292 Medium
292 of 396
Application +4 | -1
Fringe width observed in the Young's double slit experiment is β\beta. If the frequency of the source is doubled, the fringe width will
View answer and explanation
Correct option: B
become β/2\beta / 2
Option B (become β/2\beta / 2) is correct because the fringe width is inversely proportional to the frequency of the source.
293 Medium
293 of 396
Analyzing +4 | -1
Assertion : In YDSE the fringe separation doesn’t contain any information about intensity of individual source.

Reason : Fringe width depends on wavelength of source.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion
Option B (If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the wavelength affecting fringe width doesn't explain why intensity doesn't affect fringe separation.
294 Medium
294 of 396
Application +4 | -1
In YDSE, the distance from central position where the intensity fall to half of maxima.
View answer and explanation
Correct option: B
λD4d\frac{{\lambda D}}{{4d}}
Option B (λD4d\frac{{\lambda D}}{{4d}}) is correct because it is the distance from the central position where the intensity falls to half of the maximum intensity in YDSE.
295 Medium
295 of 396
Understanding +4 | -1
Wave that cannot be polarised are
View answer and explanation
Correct option: A
Longitudinal
Option A is correct because longitudinal waves, such as sound waves, oscillate parallel to the direction of propagation and cannot be polarized.
296 Medium
296 of 396
Remembering +4 | -1
Soap bubble appears coloured due to the phenomenon of
View answer and explanation
Correct option: A
Interference
Option A (Interference) is correct because the colored appearance of soap bubbles is due to the interference of light waves reflected from the inner and outer surfaces of the bubble.
297 Medium
297 of 396
Remembering +4 | -1
Fringe pattern in YDSE is formed due to the phenomenon of
View answer and explanation
Correct option: C
Interference of light
Option C (Interference of light) is correct because the fringe pattern in Young's Double Slit Experiment (YDSE) is a direct result of the interference of light waves.
298 Medium
298 of 396
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. Covering one slit with a transparent film that reduces light intensity will cause the fringe pattern to shift *towards* the covered slit, not away from it. The other statements are correct.
299 Medium
299 of 396
Remembering +4 | -1
In Young’s double slit experiment, if D,d{D,d} and λ\lambda represent, the distance between the slits, the distance of the screen from the slits and wavelength of light used respectively, then the band width is inversely proportional to
View answer and explanation
Correct option: B
dd
Option B (dd) is correct because the fringe width (bandwidth) in Young's double-slit experiment is inversely proportional to the distance between the slits (dd).
300 Medium
300 of 396
Understanding +4 | -1
Sunlight is passed through a polariser. Now the polariser is rotated through 360∘360^{\circ}. If the light passing through the polariser is observed then its intensity
View answer and explanation
Correct option: D
Varies between maximum to minimum twice in a rotation
Option D is correct because sunlight is partially polarized, so rotating the polarizer will cause the intensity to vary between a maximum and a minimum twice in a full rotation.
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