In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is λ(λ = wavelength) is β. The intensity at a point where the path difference is λ/3, will be cos π/3 = 1/2] (a) β (b) β/2 (c) (β)/(4) (d) β/8

Answer: (c). Path difference λ is a bright fringe, so β = Imax. For λ/3: φ = 2π/3, I = βcos ² 60° = (β)/(4). So the answer is (c) (β)/(4).

In the Young's double slit experiment, the intensity at a point on… | MHT CET 2024 (9 May, Shift 2)

Q48MHT CET 2024 · 9 MAY · SHIFT 2Medium

Superposition, Coherence & Intensity

In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is $\lambda(\lambda =$ wavelength $)$ is $\beta$ . The intensity at a point where the path difference is $\lambda/3$ , will be $\cos\frac{\pi}{3} = 1/2]$

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Answer: (c)

Path difference $\lambda$ is a bright fringe, so $\beta = I_{max}$ .
For $\lambda/3$ : $\phi = 2\pi/3$ , $I = \beta\cos^2 60^\circ = \dfrac{\beta}{4}$ .
So the answer is (c) $\dfrac{\beta}{4}$ .

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