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Physics · Wave Optics

JEE Main 2024 — 5 April, Shift 1 — Question 49

In Young's double slit experiment, carried out with light of wavelength 5000A5000 A, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x=0 cmx=0 \mathrm{~cm}. The value of xx for third maxima is \qquad mm

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

β=λDd=5×10−7×23×10−4=10×10−33 m\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}=\frac{10 \times 10^{-3}}{3} \mathrm{~m}

For 3rd 3^{\text {rd }} maxima y3=3β=10×10−3 m=10 mmy_{3}=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}

Answer key and solution verified before publishing.

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

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