JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Wave Optics

In a Young's double slit experiment, the intensity at a point is

(\frac{1}{4})^{th}

of the maximum intensity, the minimum distance of the point from the central maximum is ____

μ m

.
(Given :

λ = 600 nm, d = 1.0 mm, D = 1.0 m

)

Answer: 200

Solution

I = I_{0} cos^{2} (Δ ϕ /2) = I_{0} /4 \Rightarrow Δ ϕ = 2 π /3

. Path difference

\frac{y d}{D} = \frac{λ}{2 π} Δ ϕ = \frac{λ}{3}

, so

y = \frac{λ D}{3 d} = \frac{600 \times 1 0 ^{- 9} \times 1}{3 \times 1 0 ^{- 3}} = 2 \times 1 0^{- 4}

= 200 μ

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Integer answer

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