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Physics · Wave Optics

JEE Main 2025 — 4 April, Evening Shift — Question 65

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm . If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is x×10−3 cmx \times 10^{-3} \mathrm{~cm}, where xx-value is \qquad -.

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

d=1.5 mmd=1.5 \mathrm{~mm} D=2 mD=2 \mathrm{~m}

λ=400 nm\lambda=400 \mathrm{~nm}

20λDd=2λa\frac{20 \lambda D}{d}=\frac{2 \lambda}{a}

a=d10D=1.510 mma=\frac{d}{10 D}=\frac{1.5}{10} \mathrm{~mm}

=150×10−3 cm10=\frac{150 \times 10^{-3} \mathrm{~cm}}{10}

=15×10−3 cm=15 \times 10^{-3} \mathrm{~cm}

Answer key and solution verified before publishing.

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

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