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Wave Optics Part 1

Section: Physics

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Question : 14 of 100

Marks: +1, -0

Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of

9:16

. (Assuming intensity of light is directly proportional to the width of slits)

[30-Jun-2022-Shift-1]

$3:4$
$4:3$
$7:1$
$49:1$

Solution:

Given

w_1:w_2 = 9:16

. Since intensity is directly proportional to slit width,

I_1:I_2 = 9:16

. Amplitudes are proportional to the square root of intensity, so

A_1:A_2 = \sqrt{9}:\sqrt{16} = 3:4

. In Young's double-slit interference,

I_{\max}/I_{\min} = (A_1 + A_2)^2/(A_1 - A_2)^2

. Thus,

I_{\max}/I_{\min} = ((3+4)/(4-3))^2 = (7/1)^2 = 49:1

. Therefore, the correct option is

49:1

.

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