JEE Main 2026 Apr 5 Shift 2, Physics Q17: Young’s Double Slit Experiment

JEE Main2026Apr 5, Shift 2Physics

The maximum intensity in a Young's double slit experiment is $I_{0}$ . Distance between the slits ( $d$ ) is $5 λ$ , 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 $D = 10 d$ is _____.

A
$\frac{I _{0}}{4}$
B
$\frac{I _{0}}{2}$
C
$I_{0}$
D
$\frac{3 I _{0}}{4}$

Solution

The point directly opposite one slit lies at $y = d /2$ from the central axis. The phase difference at this point (for small angles) is:

$Δ ϕ = \frac{2 π}{λ} \cdot \frac{y d}{D} = \frac{2 π}{λ} \cdot \frac{d ^{2}}{2 D} = \frac{π d ^{2}}{λ D}$ .

Substituting $d = 5 λ$ and $D = 10 d = 50 λ$ :

$Δ ϕ = \frac{π ( 5 λ ) ^{2}}{λ \cdot 50 λ} = \frac{π}{2}$ .

$I = I_{0} cos^{2} (\frac{Δ ϕ}{2}) = I_{0} cos^{2} (π /4) = \frac{I _{0}}{2}$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q17: Young’s Double Slit Experiment

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