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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Hard

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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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1 Hard

1 of 9

Application +4 | -1

Two coherent light sources having intensity in the ratio

2\, \times

produce an interference pattern. The ratio

\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

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

Option D (

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

) is correct because it is the result of applying the formula for the ratio of the difference and sum of maximum and minimum intensities in an interference pattern.

2 Hard

2 of 9

Application +4 | -1

In an experiment similar to Young's experiment, interference is observed using waves associated with electrons. The electrons are being produced in an electron gun. In order to increase the fringe width-

View answer and explanation

Correct option: B

Electron gun voltage be decreased.

Option B (Electron gun voltage be decreased) is correct because decreasing the voltage increases the de Broglie wavelength, thus increasing the fringe width.

3 Hard

3 of 9

Remembering +4 | -1

In Young’s double slit experiment, a minimum is obtained when the phase difference of superimposing waves is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because a minimum (destructive interference) in Young's double-slit experiment occurs when the phase difference is an odd multiple of

\pi

4 Hard

4 of 9

Application +4 | -1

In young’s single slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

View answer and explanation

Correct option: C

{\sin ^{ - 1}}\left[ {0.001} \right]

Option C (

{\sin ^{ - 1}}\left[ {0.001} \right]

) is correct because the half angular width of the central maximum in single-slit diffraction is given by

\sin \theta = \frac{\lambda}{d}

5 Hard

5 of 9

Application +4 | -1

In a double-slit experiment, green light

(5303\mathop \mathrm{A}\limits^\mathrm{o} )\mathrm{ }

falls on a double slit having a separation of

19.44\mu m

and a width of

4.05\mu m

. The number of bright fringes between the first and the second diffraction minima is

View answer and explanation

Correct option: A

Option A (5) is correct because the number of bright fringes between the first and second diffraction minima is approximately the ratio of the slit separation to the slit width.

6 Hard

6 of 9

Application +4 | -1

In Young's double slit experiment

\frac{\mathrm{d}}{\mathrm{D}}=10^{-4}

(

\mathrm{d}=

distance between slits,

\mathrm{D}=

distance of screen from the slits). At a point

\mathrm{P}

on the screen the resulting intensity is equal to the intensity due to individual slit

\mathrm{I}_0

. Then the minimum distance of point

\mathrm{P}

from the central maximum is

(\lambda=6000 \mathop {\text{A}}\limits^ \circ)

View answer and explanation

Correct option: A

2 \mathrm{~mm}

The correct answer is A (2 mm) because it is the minimum distance calculated using the condition for intensity equal to individual slit intensity.

7 Hard

7 of 9

Application +4 | -1

In YDSE of equal width slits, if intensity at the centre of screen is

I_0

, then intensity at a distance of

\beta / 4

from the central maxima is (

\beta

is the fringe width):

View answer and explanation

Correct option: B

\frac{I_0}{2}

Option B (

\frac{I_0}{2}

) is correct because at a distance of β/4 from the central maxima, the phase difference is π/2, and the resulting intensity is half of the maximum intensity.

8 Hard

8 of 9

Application +4 | -1

A slit of width ‘a’ is illuminated by white light. For red light

\left( {\lambda = 6500\,{A^ \circ }} \right)

, the first minima is obtained at

\left( {\lambda = 6500\,{A^ \circ }} \right)

, then the value of ‘a’ will be

View answer and explanation

Correct option: C

1.24 microns

Option C (1.24 microns) is correct because it is calculated using the formula for the first minima in single-slit diffraction,

a \sin \theta = \lambda

9 Hard

9 of 9

Application +4 | -1

Two coherent light sources having intensity in the ratio

2\, \times

produce an interference pattern. The ratio

\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

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

Option D is correct. The ratio is found using the formula for maximum and minimum intensities in interference.

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 9 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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