Physics · Wave Optics

JEE Main 2024 — 6 April, Shift 2 — Question 58

Two coherent monochromatic light beams of intensities $I$ and $4I$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $xI$ . The value of $x$ is $\qquad$ .

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{I}_{\max }=(\sqrt{\mathrm{I}}+\sqrt{4 \mathrm{I}})^{2}=9 \mathrm{I}$

$\mathrm{I}_{\min }=(\sqrt{4 \mathrm{I}}-\sqrt{\mathrm{I}})^{2}=\mathrm{I}$

$\therefore \mathrm{I}_{\max }-\mathrm{I}_{\min }=8 \mathrm{I}$

$\implies x = 8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications