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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
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The correct answer is  a) 4:1

Explaination ::

A2=3A1

$$Amplitude ∝ width \space of \space slit$$

Imax = (A1 + 3A1)    ..............1

and     Imin = (A1 – 3A1)          .................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

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