The maximum intensity in a Young's double slit experiment is I 0 . 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

Physics · Wave Optics

JEE Main 2026 — 5 April, Evening Shift — Question 16

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

Option A:
$\dfrac{I_0}{4}$
Option B:
$\dfrac{I_0}{2}$
Correct
Option C:
$I_0$
Option D:
$\dfrac{3I_0}{4}$

Answer: B

Step-by-step solution

At point opposite one slit, path difference $\Delta = \frac{y d}{D}$ with $y = d/2$ .

So $\Delta = \frac{d^2}{2D} = \frac{25\lambda^2}{2\times10d} = \frac{25\lambda}{100} = \lambda/4$ .

Phase difference $\phi = 2\pi\Delta/\lambda = \pi/2$ .

Intensity $I = I_0 \cos^2(\phi/2) = I_0 \cos^2(\pi/4) = I_0/2$ .

The correct option is (2) corresponding to $I_0/2$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications