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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 1-30 of 396 questions

1 Medium

1 of 396

Understanding +4 | -1

Which of the following properties of electromagnetic waves does not support particle nature of radiation (electromagnetic wave)?

View answer and explanation

Correct option: C

Polarization

Option C (Polarization) is correct because polarization demonstrates the wave nature of electromagnetic waves, not the particle nature.

2 Medium

2 of 396

Understanding +4 | -1

Spherical wavefronts, emanating from a point source, strike a plane reflecting surface. What will happen to these wave fronts, immediately after reflection?

View answer and explanation

Correct option: C

They will remain spherical, with the same curvature, but sign of curvature reversed.

Option C is correct because reflection from a plane surface reverses the direction of the wavefronts, but their curvature remains the same in magnitude, only the sign changes.

3 Medium

3 of 396

Application +4 | -1

An astronomical telescope is able to resolve two stars at an angular separation of

10^{-3}

degree. If wavelength of light used is

500 {~nm}

, then the diameter of the objective of the telescope is

View answer and explanation

Correct option: A

3.5 {~cm}

Option A (

3.5 {~cm}

) is correct because it is calculated using the formula for the limit of resolution of a telescope, considering the given wavelength and angular separation.

4 Medium

4 of 396

Understanding +4 | -1

An interference pattern was made by using red light. If the red light changes with blue light, the fringes will become

View answer and explanation

Correct option: B

Narrower

Option B is correct because the fringe width is directly proportional to the wavelength of light, and blue light has a shorter wavelength than red light.

5 Medium

5 of 396

Remembering +4 | -1

H-polaroid is prepared by

View answer and explanation

Correct option: D

Stretching polyvinyl alcohol and then impregnation with iodine

Option D (Stretching polyvinyl alcohol and then impregnation with iodine) is correct because H-polaroids are manufactured by stretching polyvinyl alcohol films and then impregnating them with iodine, which aligns the polymer chains and creates a polarizing effect.

6 Medium

6 of 396

Application +4 | -1

Specific rotation of sugar solution is

0.01\;SI

units.

200\;kg{m^{ - 3}}

of impure sugar solution is taken in a polarimeter tube of length

0.25\;m

and an optical rotation of

0.4\;rad

is observed. The percentage of purity of sugar is the sample is

View answer and explanation

Correct option: A

80\%

Option A (

80\%

) is correct because the percentage of purity is calculated by comparing the concentration of pure sugar to the concentration of the impure solution.

7 Medium

7 of 396

Understanding +4 | -1

Two waves are represented by the equations

{y_1} = a\sin \omega t,\;and\;{y_2} = a\cos \omega t

. The first wave

View answer and explanation

Correct option: D

Lags the second by

{\frac{\pi }{2}}

Option D (Lags the second by

{\frac{\pi }{2}}

) is correct because

{y_1}

can be expressed as a sine function, while

{y_2}

is a cosine function, which is a sine function leading by

{\frac{\pi }{2}}

8 Medium

8 of 396

Understanding +4 | -1

The angle of incidence at which reflected light it totally polarized for reflection from air to glass (refractive index n) is

View answer and explanation

Correct option: D

{\tan ^{ - 1}}\left( n \right)

Option D (

{\tan ^{ - 1}}( n )

) is correct because the angle of incidence for total polarization (Brewster's angle) is given by the arctangent of the refractive index.

9 Medium

9 of 396

Application +4 | -1

Two coherent sources of intensity ratio

\beta

interfere. Then the value of

\left(I_{\max }-I_{\min }\right) /\left(I_{\max }+I_{\min }\right)

is :-

View answer and explanation

Correct option: D

\frac{2 \sqrt{\beta}}{1+\beta}

Option D (

\frac{2 \sqrt{\beta}}{1+\beta}

) is correct because it represents the correct ratio of the difference and sum of maximum and minimum intensities in terms of the intensity ratio

\beta

10 Medium

10 of 396

Application +4 | -1

In young’s double slit experiment, 5th dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7th bright fringe is obtained at the same point. The thickness of the film in terms of wavelength

\lambda

and refractive index

\mu

will be

View answer and explanation

Correct option: D

\frac{{2.5\lambda }}{{\left( {\mu - 1} \right)}}

Option D (

\frac{{2.5\lambda }}{{{{\left( {\mu - 1} \right)}}}}

) is correct because it is derived from equating the path difference due to the film with the change in fringe position.

11 Medium

11 of 396

Application +4 | -1

Red light of wavelength

625\;nm

is incident normally on an optical diffraction grating with

2 \times {10^5}\;lines/m

. Including central principal maxima, how many maxima may be observed on a screen which is for from the grating?

View answer and explanation

Correct option: B

Option B (17) is correct because the number of maxima is determined by the grating equation and the condition that

\sin \theta

must be less than or equal to 1.

12 Medium

12 of 396

Understanding +4 | -1

In an interference experiment, phase difference for points where the intensity is minimum is

(n = 1,2,3, \ldots \ldots )

View answer and explanation

Correct option: C

\left( {2n - 1} \right)\pi

Option C (

\left( {2n - 1} \right)\pi

) is correct because the phase difference for minimum intensity (destructive interference) is an odd multiple of

\pi

13 Medium

13 of 396

Understanding +4 | -1

Two waves originating from source

{S_1}\;and\;{S_2}

having zero phase difference and common wavelength

\lambda

will show complete destructive interference at a point

P

, is

({S_1}P - {S_2}P) =

View answer and explanation

Correct option: D

\frac{{11\lambda }}{2}

Option D (

\frac{{11\lambda }}{2}

) is correct because for destructive interference, the path difference must be an odd multiple of

\lambda/2

14 Medium

14 of 396

Application +4 | -1

The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length

1\;m

. The width of slit is

0.3\;mm

. If third minimum is formed at a distance of

5\;mm

from central maximum, then wavelength of light will be

View answer and explanation

Correct option: A

5000\mathop \mathrm{A}\limits^ \circ

Option A (

5000\mathop \mathrm{A}\limits^ \circ

) is correct because it is calculated using the formula for the position of minima in Fraunhofer diffraction.

15 Medium

15 of 396

Understanding +4 | -1

Two light sources are said to be coherent if they are obtained from

View answer and explanation

Correct option: B

A single point source

Option B (A single point source) is correct because coherent sources are derived from a single source to maintain a constant phase relationship.

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16 Medium

16 of 396

Application +4 | -1

Two coherent sources of intensity ratio

\beta

interfere. Then the value of

\frac{{{a_{\max }}}}{{{a_{\min }}}}

View answer and explanation

Correct option: A

\frac{{\sqrt \beta + 1}}{{\sqrt \beta - 1}}

Option A (

\frac{{\sqrt \beta + 1}}{{\sqrt \beta - 1}}

) is correct because it represents the ratio of maximum to minimum amplitudes in terms of the intensity ratio β.

17 Medium

17 of 396

Remembering +4 | -1

The penetration of light into the region of geometrical shadow is called

View answer and explanation

Correct option: C

Diffraction

Option C (Diffraction) is correct because diffraction is the phenomenon where light bends around corners, penetrating the region of geometrical shadow.

18 Medium

18 of 396

Analyzing +4 | -1

Assertion : When light ray is incident at polarizing angle on glass, reflected light is partially polarized.

Reason : The intensity of light decreases in polarization.

View answer and explanation

Correct option: D

If both Assertion and reason are false

Option D (If both Assertion and reason are false) is correct because when light is incident at the polarizing angle, the reflected light is completely polarized, not partially, and the intensity of light does decrease in polarization.

19 Medium

19 of 396

Remembering +4 | -1

Brewster’s angle in terms of refractive index (n) of the medium

View answer and explanation

Correct option: D

{\tan ^{ - 1}}\left[ n \right]

Option D (

{\tan ^{ - 1}}\left[ n \right]

) is correct because Brewster's angle is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

20 Medium

20 of 396

Understanding +4 | -1

In YDSE, a laser light is used to illuminate the closely spaced slits, as a result, a clear interference pattern is obtained on screen. If we place smoke particles between the double slit and the screen, then the effects of interference will be observed:

View answer and explanation

Correct option: B

In between the space, between the screen and slits

Option B (In between the space, between the screen and slits) is correct because interference occurs throughout the space between the slits and the screen, and smoke particles make the interference pattern visible in that space.

21 Medium

21 of 396

Understanding +4 | -1

An unpolarized beam of light is incident on a glass surface at an angle of incidence equal to the polarizing angle of the glass. Read the following statements:
(i) The reflected beam is completely polarized
(ii) The refracted beam is partially polarized
(iii) The angle between the reflected and the refracted beam is

90^{\circ}

.
Which of the above statements is/are true?

View answer and explanation

Correct option: C

(i) and (iii)

Option C ((i) and (iii)) is correct because at the polarizing angle, the reflected light is completely polarized and the angle between reflected and refracted rays is 90 degrees.

22 Medium

22 of 396

Application +4 | -1

In Young’s experiment, the distance between the slits is reduced to half and the distance between the slit and screen is doubled, then the fringe width

View answer and explanation

Correct option: D

Will become four times

Option D (Will become four times) is correct because the fringe width is directly proportional to the distance between the slit and screen and inversely proportional to the distance between the slits.

23 Medium

23 of 396

Remembering +4 | -1

If for a calcite crystal

{\mu _0}\;\mathrm{and}\;{\mu _e}

are the refractive indices of the crystal for

O

- ray and

E

-ray respectively, then along the optic axis of the crystal

View answer and explanation

Correct option: A

{\mu _0} = {\mu _e}

Option A (

{\mu _0} = {\mu _e}

) is correct because along the optic axis, the ordinary and extraordinary rays experience the same refractive index.

24 Medium

24 of 396

Application +4 | -1

The first diffraction minimum due to single slit diffraction is

\theta

, for a light of wavelength

5000 \mathop {\text{A}}\limits^{\text{o}}

. If the width of slit is

1 \times 10^{-4} \mathrm{~cm}

. Then, the value of

\theta

View answer and explanation

Correct option: A

30^{\circ}

Option A (

30^{\circ}

) is correct because it is the angle calculated using the single-slit diffraction formula

d \sin \theta = \lambda

for the first minimum.

25 Medium

25 of 396

Application +4 | -1

In Young’s double slit experiment we get 60 fringes in the field of view of monochromatic light of wavelength

{4000\;\mathop \mathrm{A}\limits^ \circ }

. If we use monochromatic light of wavelength

{6000\;\mathop \mathrm{A}\limits^ \circ }

, then the number of fringes obtained in the same field of view are

View answer and explanation

Correct option: C

Option C (40) is correct because the number of fringes is inversely proportional to the wavelength of light used.

26 Medium

26 of 396

Application +4 | -1

In a Young’s double slit experiment, the two slits act as coherent sources of waves of equal amplitude

A

and wavelength

\lambda

. In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is

{I_1}

and in the second case

{I_2}

, then the ratio

\frac{{{I_1}}}{{{I_2}}}

View answer and explanation

Correct option: B

Option B (2) is correct because the ratio of intensities for coherent and incoherent sources is 2, given equal amplitudes and wavelengths.

27 Medium

27 of 396

Understanding +4 | -1

Assertion: When tiny circular obstacles is placed in the path of light from some distnce, a bright spot is sen at the centre ofthe shadow of the obstacle.

Reason: Destructive interference occuts at the centre of the shadow.

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 bright spot is due to constructive interference, not destructive interference.

28 Medium

28 of 396

Application +4 | -1

The time period of rotation of the sun is 25 days and its radius is

7 \times {10^8}\;m

. The Doppler shift for the light of wavelength

6000\mathop \mathrm{A}\limits^ \circ

emitted from the surface of the sun will be

View answer and explanation

Correct option: A

0.04\mathop \mathrm{A}\limits^ \circ

Option A (

0.04\mathop \mathrm{A}\limits^ \circ

) is correct because it is calculated using the Doppler shift formula and the sun's rotational velocity.

29 Medium

29 of 396

Understanding +4 | -1

Assertion: Thin films such as soap bubble or a thin layer of oil on water show beautiful colours when illuminated by sunlight.

Reason: The colors are obtained by dispersion of light only.

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 colors in thin films are due to interference, not just dispersion.

30 Medium

30 of 396

Remembering +4 | -1

Which source is associated with a line emission spectrum

View answer and explanation

Correct option: B

Neon street sign

Option B (Neon street sign) is correct because excited neon atoms emit light at specific wavelengths, creating a line spectrum.

Revision notes and key takeaways

Review WAVE OPTICS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 396 single-correct-answer MCQs on WAVE OPTICS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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