Physics · Wave Optics

JEE Main 2025 — 3 April, Morning Shift — Question 66

Two coherent monochromatic light beams of intensities $4 /$ and $9 /$ are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is $x$ I. The value of $x$ is

Answer: 24

Numerical answer — enter this value.

Step-by-step solution

$I_{\text {max }}=(\sqrt{4 I}+\sqrt{9 I})^{2}=25 I$ $I_{\text {min }}=(\sqrt{4 I}-\sqrt{9 I})^{2}=I$ $\Delta l=24 l$

Answer key and solution verified before publishing.

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