JEE Main 2024 Physics question
In a Young's double slit experiment, the intensity at a point is (1/4)ᵗʰ of the maximum intensity, the minimum distance of the point from the central maximum is .......... μm. (Given : λ = 600 nm, d = 1.0 mm, D = 1.0 m)
Answer: 200
Solution
I = I₀cos²(Δφ/2) = I₀/4 gives Δφ/2 = π/3, so the minimum phase difference is Δφ = 2π/3. With Δφ = (2π/λ)(yd/D): y = λD/(3d)
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