The maximum intensity in a Young's double slit experiment is I₀. Distance between the slits (d) is 5λ, where λ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D = 10d is ______. (a) I₀/4 (b) I₀/2 (c) I₀ (d) 3I₀/4

Answer: (b). In front of a slit, Δ = d²/2D = d/20 = (λ)/(4), so φ = π/2. I = I₀cos ²(φ/2) = I₀cos ²π/4 = I₀/2 So the answer is (b) I₀/2.

The maximum intensity in a Young's double slit experiment is I₀… | JEE Main 2026 (5 Apr, Shift 2)

Q21JEE Main 2026 · 5 APR · SHIFT 2Medium

YDSE: Film, Medium & Two Wavelengths

The maximum intensity in a Young's double slit experiment is $I_0$ . Distance between the slits $(d)$ is $5\lambda$ , where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D = 10d$ is ______.

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Answer: (b)

In front of a slit, $\Delta = \dfrac{d^2}{2D} = \dfrac{d}{20} = \dfrac{\lambda}{4}$ , so $\phi = \dfrac{\pi}{2}$ .
$I = I_0\cos^2(\phi/2) = I_0\cos^2\tfrac{\pi}{4} = \dfrac{I_0}{2}$
So the answer is (b) $\frac{I_0}{2}$ .

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