In biprism experiment, the maximum intensity is I₀. If the path difference between the two interfering waves is (λ)/(3), then intensity at the point on the screen is [ sin 30° = cos 60° = 0.5, sin 60° = cos 30° = √(3)/2 ] (a) I₀/4 (b) I₀/3 (c) I₀/2 (d) I₀

Answer: (a). Phase difference φ = (2π)/(λ)·(λ)/(3) = 2π/3. I = I₀cos ²φ/2 = I₀cos ² 60° = I₀/4. So the answer is (a) I₀/4.

In biprism experiment, the maximum intensity is I₀. If the path… | MHT CET 2026 (19 Apr, Shift 2)

Q5MHT CET 2026 · 19 APR · SHIFT 2Medium

Biprism, Thin Films & Other Devices

In biprism experiment, the maximum intensity is $I_0$ . If the path difference between the two interfering waves is $\frac{\lambda}{3}$ , then intensity at the point on the screen is

[ $\sin 30^\circ = \cos 60^\circ = 0.5,\ \sin 60^\circ = \cos 30^\circ = \sqrt{3}/2$ ]

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Answer: (a)

Phase difference $\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{3} = \frac{2\pi}{3}$ .
$I = I_0\cos^2\frac{\phi}{2} = I_0\cos^2 60^\circ = \frac{I_0}{4}$ .
So the answer is (a) $\frac{I_0}{4}$ .

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