JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Wave Optics

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In a Young's double slit experiment, the width of the one of the slit 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) $1 : 4$
(B) $4 : 1$ ✓ Correct
(C) $3 : 1$
(D) $2 : 1$

Answer: (B)

Solution

Let the amplitude from the narrower slit be

A

. Then the amplitude from the wider slit is

3 A

(since amplitude is proportional to slit width).

Maximum intensity:

I_{ma x} = (A + 3 A)^{2} = 16 A^{2}

Minimum intensity:

I_{min} = (3 A - A)^{2} = 4 A^{2}

Ratio:

\frac{I _{ma x}}{I _{min}} = \frac{16 A ^{2}}{4 A ^{2}} = \frac{4}{1} = 4 : 1

Answer: (B)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

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