Physics · Wave Optics

JEE Main 2025 — 3 April, Evening Shift — Question 55

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

Option A:
$(2 \sqrt{2}+1):(2 \sqrt{2}-1)$
Option B:
$3: 1$
Option C:
$9: 1$
Correct
Option D:
$(3+2 \sqrt{2}):(3-2 \sqrt{2})$

Answer: C

Step-by-step solution

Amplitude $\propto$ width $\Rightarrow$ Intensity $\propto$ width $^{2}$

$l_{1}=I, l_{2}=4 l$ ; ratio $\left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}$

Ratio $=\left(\frac{1+2}{1-2}\right)^{2}=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications