Physics · Wave Optics

JEE Main 2024 — 8 April, Shift 2 — Question 53

Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light wavelength 500 nm is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is $\ldots \ldots \ldots . \times 10^{-4} \mathrm{~m}$ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{d}=1 \mathrm{~mm}, \mathrm{D}=1 \mathrm{~m}, \lambda=500 \mathrm{~nm}$

\begin{aligned} & 10\left(\frac{\lambda D}{d}\right)=\frac{2 \lambda D}{a} & \mathbf{a}=\frac{d}{5} &=\frac{10 \times 10^{-4} \mathrm{~m}}{5} &=2 \times 10^{-4} \end{aligned}

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications