JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Wave Optics

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In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm used, the fringe separation will be :

(A)1 mm
(B)0.75 mm
(C)0.25 mm✓ Correct
(D)0.50 mm

Answer: (C)

Solution

The fringe width in Young's double slit experiment is given by

β = \frac{λ D}{d}

, where

λ = 500

= 500 \times 1 0^{- 9}

D = 1

m, and

d = 2

= 2 \times 1 0^{- 3}

m. Substituting:

β = \frac{500 \times 1 0 ^{- 9} \times 1}{2 \times 1 0 ^{- 3}} = \frac{500 \times 1 0 ^{- 9}}{2 \times 1 0 ^{- 3}} = 250 \times 1 0^{- 6}

= 0.25

mm.

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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