JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Wave Optics

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Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

(A) $(22 + 1) : (22 - 1)$
(B) $(3 + 22) : (3 - 22)$ ✓ Correct
(C) $9 : 1$
(D) $3 : 1$

Answer: (B)

Solution

Brightness from a slit is proportional to its width, so the narrower slit gives intensity

I_{0}

and the wider one

2 I_{0}

. The interference extremes are

I_{m a x} = (I_{0} + 2 I_{0})^{2} = I_{0} (1 + 2)^{2} = I_{0} (3 + 22)

and

I_{m i n} = (2 I_{0} - I_{0})^{2} = I_{0} (3 - 22)

. The ratio of maximum to minimum intensity is

(3 + 22) : (3 - 22)

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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