In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm , the angular width of central maximum is α × 10 -2 degrees. The value of α is .

Physics · Wave Optics

JEE Main 2026 — 2 April, Morning Shift — Question 15

In single slit diffraction pattern, the wavelength of light used is $628\mathrm{nm}$ and slit width is $0.2\mathrm{mm}$ , the angular width of central maximum is $\alpha \times 10^{-2}$ degrees. The value of $\alpha$ is ______.

Answer: 36

Numerical answer — enter this value.

Step-by-step solution

$\theta_{cm} = \frac{2\lambda}{d} = \frac{2\times628\times10^{-9}}{0.2\times10^{-3}} rad = 628\times10^{-5} rad \times \frac{180}{\pi} \approx 36\times10^{-2} degrees$

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Wave Optics
Topic: Diffraction of Light Waves