JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Wave Optics

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Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be :

(A)0.23 mm✓ Correct
(B)0.33 mm
(C)0.63 mm
(D)0.46 mm

Answer: (A)

Solution

When the Young's double slit apparatus is immersed in a liquid of refractive index

μ

, the wavelength of light shortens to

λ = \frac{λ _{0}}{μ}

, where

λ_{0}

is the wavelength in air. The fringe width in the liquid becomes

β = \frac{λ D}{d} = \frac{λ _{0}}{μ} \cdot \frac{D}{d} .

Substituting

λ_{0} = 690 \times 1 0^{- 9} m

μ = 1.44

D = 0.72 m

d = 1.5 \times 1 0^{- 3} m

β = \frac{690 \times 1 0 ^{- 9} \times 0.72}{1.44 \times 1.5 \times 1 0 ^{- 3}} = \frac{496.8 \times 1 0 ^{- 9}}{2.16 \times 1 0 ^{- 3}} = 0.23 \times 1 0^{- 3} m = 0.23 mm .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Wave Optics

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