Physics · Wave Optics

JEE Main 2025 — 2 April, Morning Shift — Question 68

If the measured angular separation between the second minimum to the left to the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is $\qquad$ $\mu \mathrm{m}$ .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$2^{\text {nd }}$ minima $\Rightarrow \theta_{1}=\left(\frac{2 \lambda}{a}\right)$

$3^{\text {rd }}$ minima $\Rightarrow \theta_{2}=\frac{3 \lambda}{a}$

$\theta_{1}+\theta_{2}=\frac{\pi}{6}$

$\frac{5 \lambda}{a}=\frac{\pi}{6}$

$a=\frac{30 \lambda}{\pi}=\frac{30 \times 0.628}{3.14}=6 \mu \mathrm{~m}$

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Wave Optics
Topic: Diffraction of Light Waves

A light wave is propagating with plane wave fronts of the type x+y+z= constant. The angle made by the direction of wave propagation with…

Answer: 6

Step-by-step solution

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