Physics · Wave Optics

JEE Main 2024 — 9 April, Shift 1 — Question 53

In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, the minimum distance

of the point from the central maximum is $\qquad$ $\mu \mathrm{m}$ . (Given: $\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$ )

Answer: 200

Numerical answer — enter this value.

Step-by-step solution

$I=I_{0} \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

$\frac{\mathrm{I}_{0}}{4}=\cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

$\Delta \phi=\frac{2 \pi}{3}$

$\frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3}$

$\mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications