JEE Main 2026 Apr 2 Shift 2, Physics Q21: Young’s Double Slit Experiment

JEE Main2026Apr 2, Shift 2Physics

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{λ}{x}$ where $λ$ is wavelength of the incident light. The value of $x$ is _____.

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Solution

For Young's double slit: $I = I_{m a x} cos^{2} (ϕ /2)$ , where $ϕ$ is the phase difference.

Setting $I = \frac{3}{4} I_{m a x}$ : $cos^{2} (ϕ /2) = \frac{3}{4}$ , so $cos (ϕ /2) = \frac{3}{2}$ , giving $ϕ /2 = π /6$ and $ϕ = π /3$ .

Path difference $Δ x = \frac{ϕ}{2 π} λ = \frac{π /3}{2 π} λ = \frac{λ}{6}$ .

$∴ x = 6$ .

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JEE Main 2026 Apr 2 Shift 2, Physics Q21: Young’s Double Slit Experiment

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