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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Mixed difficulty

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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 301-330 of 418 questions

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301 Medium
301 of 418
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.
302 Medium
302 of 418
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.
303 Medium
303 of 418
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.
304 Medium
304 of 418
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.
305 Medium
305 of 418
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.
306 Medium
306 of 418
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.
307 Medium
307 of 418
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.
308 Medium
308 of 418
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.
309 Medium
309 of 418
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.
310 Medium
310 of 418
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.
311 Medium
311 of 418
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.
312 Medium
312 of 418
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.
313 Medium
313 of 418
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.
314 Medium
314 of 418
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.
315 Medium
315 of 418
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.
316 Medium
316 of 418
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.
317 Medium
317 of 418
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.
318 Medium
318 of 418
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.
319 Medium
319 of 418
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).
320 Medium
320 of 418
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.
321 Medium
321 of 418
Remembering +4 | -1
The electromagnetic theory of light failed to explain
View answer and explanation
Correct option: A
Photoelectric effect
Option A (Photoelectric effect) is correct because the electromagnetic theory of light could not explain the discrete nature of energy exchange in the photoelectric effect, which required the concept of photons.
322 Medium
322 of 418
Application +4 | -1
In young's experiment find the number of maximas formed if distance between the slits is 1.5λ1.5 \lambda. Where λ\lambda is wavelength of light used:
View answer and explanation
Correct option: B
3
Option B (3) is correct because only three integer values of 'n' satisfy the condition for maxima formation in Young's experiment.
323 Medium
323 of 418
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 when the screen is perpendicular to the line joining the sources, the fringes formed are circular.
324 Medium
324 of 418
Application +4 | -1
In YDSE, a thin mica sheet of thickness 2×10−6 m2 \times 10^{-6} \mathrm{~m} and refractive index (μ=1.5)(\mu=1.5) is introduced in the path of the wave in front of the slit above the central line of the screen. The wavelength of the wave used is 5000A∘5000 \mathop {\text{A}}\limits^ \circ. The central bright maximum will shift
View answer and explanation
Correct option: A
2 fringes upward
Option A (2 fringes upward) is correct because the central bright maximum shifts upward by a distance equal to 2 fringe widths when the mica sheet is introduced.
325 Medium
325 of 418
Application +4 | -1
In a double slit experiment, two parallel slits are illuminated first by light of wavelength 400 nm400 \mathrm{~nm} and then by light of unknown wavelength. The fourth order dark fringe resulting from the known wavelength of light falls in the same place on the screen as the second order bright fringe from the unknown wavelength. The value of unknown wavelength of the light is
View answer and explanation
Correct option: B
700 nm700 \mathrm{~nm}
Option B (700 nm700 \mathrm{~nm}) is correct because the condition for dark fringes in a double-slit experiment is (n+12)Dλ1d=mDλ2d(n + \frac{1}{2})\frac{{D\lambda_1}}{{d}} = m \frac{{D\lambda_2}}{{d}}, where nn and mm are integers.
326 Medium
326 of 418
Understanding +4 | -1
When a compact disc is illuminated by small source of white light, coloured bands are observed. This is due to
View answer and explanation
Correct option: B
Diffraction
Option B (Diffraction) is correct because the colored bands are formed due to the diffraction of light by the closely spaced tracks on the CD surface.
327 Medium
327 of 418
Understanding +4 | -1
Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are βG,βR  and  βB{\beta _G},{\beta _R}\;and\;{\beta _B}, respectively. Then
View answer and explanation
Correct option: D
βR>βG>βB{\beta _R} > {\beta _G} > {\beta _B}
Option D is correct. Fringe width is directly proportional to wavelength, and red light has the longest wavelength, followed by green, and then blue, resulting in the order βR>βG>βB{\beta _R} > {\beta _G} > {\beta _B}.
328 Medium
328 of 418
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
The correct answer is A because closing one slit eliminates the interference pattern, leaving only a single-slit diffraction pattern, which appears as a bright slit.
329 Medium
329 of 418
Understanding +4 | -1
In a Young’s double slit experiment, the central point on the screen is
View answer and explanation
Correct option: A
Bright
Option A (Bright) is correct because in Young's double-slit experiment, the central point has zero path difference, leading to constructive interference and a bright fringe.
330 Medium
330 of 418
Understanding +4 | -1
Keeping all parameter fixed if distance between two slits in Young's double slit experiment is decreased, then fringe width will
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because the fringe width in Young's double-slit experiment is inversely proportional to the distance between the slits.
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