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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 θ\theta1, then θ\theta1, is close to:

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Answer: (c)

  1. Minima: asin⁡θn=nλa\sin\theta_n = n\lambda.
  2. Second minimum: sin⁡60∘=2λa⇒λa=34≈0.433\sin 60^\circ = \dfrac{2\lambda}{a} \Rightarrow \dfrac{\lambda}{a} = \dfrac{\sqrt3}{4} \approx 0.433.
  3. First minimum: sin⁡θ1=0.433⇒θ1≈25.7∘\sin\theta_1 = 0.433 \Rightarrow \theta_1 \approx 25.7^\circ.
  4. So the answer is (c) 25∘25^\circ.
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