JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Wave Optics

Visible light of wavelength

6000 \times 1 0^{- 8}

cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at

6 0^{\circ}

from the central maximum. If the first minimum is produced at

θ_{1}

then

θ_{1}

is close to:

(A) $2 0^{\circ}$
(B) $3 0^{\circ}$
(C) $2 5^{\circ}$ ✓ Correct
(D) $4 5^{\circ}$

Answer: (C)

Solution

For single-slit diffraction the minima satisfy

d sin θ = nλ

. The second minimum at

6 0^{\circ}

gives

d sin 6 0^{\circ} = 2 λ

, so

d = \frac{2 λ}{3 /2} = \frac{4 λ}{3}

. The first minimum then obeys

sin θ_{1} = \frac{λ}{d} = \frac{3}{4} \approx 0.433

, which gives

θ_{1} \approx 2 5^{\circ}

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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