JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Wave Optics

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In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:

(A) $1. 7^{\circ}$
(B) $0.0 7^{\circ}$
(C) $0.5 7^{\circ}$ ✓ Correct
(D) $0.1 7^{\circ}$

Answer: (C)

Solution

In Young's experiment the angular separation between consecutive fringes is

θ = \frac{λ}{d}

. With

λ = 500

= 500 \times 1 0^{- 9}

m and

d = 0.05

= 5 \times 1 0^{- 5}

θ = \frac{500 \times 1 0 ^{- 9}}{5 \times 1 0 ^{- 5}} = 1 0^{- 2}

rad. Converting,

1 0^{- 2} \times \frac{180}{π} \approx 0.5 7^{\circ}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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