JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Wave Optics

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The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd . If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:4 then what is the value of x ?
(Assume that the field strength varies according to the slit width.)

(A)2
(B)3
(C)5✓ Correct
(D)4

Answer: (C)

Solution

Taking the field amplitude from each slit as proportional to its width (as stated),

a_{1} \propto d

and

a_{2} \propto x d

.

The maximum and minimum intensities are

I_{ma x} \propto (a_{1} + a_{2})^{2}, I_{min} \propto (a_{1} - a_{2})^{2}

\frac{I _{ma x}}{I _{min}} = (\frac{a _{1} + a _{2}}{a _{1} - a _{2}})^{2} = \frac{9}{4}

\frac{a _{1} + a _{2}}{a _{1} - a _{2}} = \frac{3}{2}

With

a_{1} = d

a_{2} = x d

and

x > 1

\frac{d + x d}{x d - d} = \frac{x + 1}{x - 1} = \frac{3}{2}

2 (x + 1) = 3 (x - 1)

2 x + 2 = 3 x - 3

x = 5

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Wave Optics

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