Chapter: Physics· JEE

In Young’s double-slit experiment, the width of one of the slits is three times the other slit.

In Young’s double-slit experiment, the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

a. 4: 1

b. 2: 1

c. 3: 1

d. 1: 4

Doubtly Asked 5 Apr 2021 Student doubt jee-2021 Save WhatsApp Report

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The correct answer is a) 4:1

Explaination ::

A₂=3A₁

\( Amplitude ∝ width \space of \space slit\)

I_max = (A₁ + 3A₁)² ..............1

and I_min = (A₁ – 3A₁)² .................2

\({I_{max} \over I_{min }}= {(A_1+3A_1)^2\over (A_1-3A_1)^2}\) \(={16A^2_1\over 4A^2_1}={4\over1}\)

4:1

Doubtly · 5 Apr 2021 Report

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In Young’s double-slit experiment, the width of one of the slits is three times the other slit.

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