JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Wave Optics

In Young's double slit experiment, monochromatic light of wavelength

5000 \overset{˚}{A}

is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______

\times 1 0^{- 6} m

Answer: 125

Solution

The resultant intensity on the screen is

I = 2 I_{0} (1 + cos ϕ)

, with maximum intensity

I_{ma x} = 4 I_{0}

ϕ = 0

. Setting

I = I_{ma x} /2 = 2 I_{0}

gives

cos ϕ = 0

, i.e.

ϕ = π /2

. The phase difference relates to the path difference by

ϕ = \frac{2 π}{λ} d sin θ

, and for small angles

sin θ \approx y / D

. So

\frac{2 π}{λ} \cdot \frac{d y}{D} = \frac{π}{2}

, giving

y = \frac{λ D}{4 d} = \frac{5 \times 1 0 ^{- 7} \times 1}{4 \times 1 0 ^{- 3}} = 125 \times 1 0^{- 6}

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Wave Optics

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