The maximum intensity in a Young's double slit experiment is $\mathrm{I}_{0}$. 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 $\mathrm{D}=10 \mathrm{~d}$ is ______

(D) $\frac{3 \mathrm{I}_{0}}{4}$

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Wave Optics

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 ______

Answer: (B)

The point directly opposite one slit is displaced from the central axis by

y = \frac{d}{2}

. The path difference there is

Δ = \frac{y d}{D} = \frac{d ^{2}}{2 D}

. With

d = 5 λ

and

D = 10 d = 50 λ

Δ = \frac{( 5 λ ) ^{2}}{2 ( 50 λ )} = \frac{λ}{4}

. The phase difference is

δ = \frac{2 π}{λ} Δ = \frac{π}{2}

, and the intensity is

I = I_{0} cos^{2} \frac{δ}{2} = I_{0} cos^{2} \frac{π}{4} = \frac{I _{0}}{2}

. The intensity at the point opposite one of the slits is

\frac{I _{0}}{2}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option