In Young's double slit experiment, intensity at a point is (1/4) of the maximum intensity. The angular position of this point is (a) sin ⁻¹((λ)/(D)) (b) sin ⁻¹((λ)/(2d)) (c) sin ⁻¹((λ)/(3d)) (d) sin ⁻¹((λ)/(4d))

Answer: (c). I = Imaxcos ²φ/2 = Imax4 ⇒ cos φ/2 = 1/2 ⇒ φ = 2π/3 Path difference Δ x = (λ)/(2π)φ = (λ)/(3) dsin θ = (λ)/(3) ⇒ θ = sin ⁻¹((λ)/(3d)) So the answer is (c).

In Young's double slit experiment, intensity at a point is (1/4) of… | MHT CET 2024 (16 May, Shift 1)

Q53MHT CET 2024 · 16 MAY · SHIFT 1Difficult

Superposition, Coherence & Intensity

In Young's double slit experiment, intensity at a point is $\left(\frac{1}{4}\right)$ of the maximum intensity. The angular position of this point is

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Answer: (c)

$I = I_{max}\cos^2\frac{\phi}{2} = \frac{I_{max}}{4} \Rightarrow \cos\frac{\phi}{2} = \frac{1}{2} \Rightarrow \phi = \frac{2\pi}{3}$
Path difference $\Delta x = \frac{\lambda}{2\pi}\phi = \frac{\lambda}{3}$
$d\sin\theta = \frac{\lambda}{3} \Rightarrow \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right)$
So the answer is (c).

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