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JEE Main 2020 (7 Jan, Shift 1) · Diffraction at a Single Slit

Diffraction at a Single Slit question from JEE Main 2020 (7 Jan, Shift 1)

Q66JEE Main 2020 · 7 JAN · SHIFT 1Medium

Diffraction at a Single Slit

Visible light of wavelength 6000 $\times$ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at $\theta$ 1, then $\theta$ 1, is close to:

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Answer: (c)

Minima: $a\sin\theta_n = n\lambda$ .
Second minimum: $\sin 60^\circ = \dfrac{2\lambda}{a} \Rightarrow \dfrac{\lambda}{a} = \dfrac{\sqrt3}{4} \approx 0.433$ .
First minimum: $\sin\theta_1 = 0.433 \Rightarrow \theta_1 \approx 25.7^\circ$ .
So the answer is (c) $25^\circ$ .

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