Physics · Wave Optics

JEE Main 2026 — 24 January, Evening Shift — Question 42

In the Young's double slit experiment the intensity produced by each one of the individual slits is $I_{0}$ . The distance between two slits is 2 mm . The distance of screen from slits is 10 m . The wavelength of light is $6000 \AA$ . The intensity of light on the screen in front of one of the slits is $\_\_\_\_$ .

Option A:
$2 I_{0}$
Option B:
$I_{0}$
Correct
Option C:
$\frac{I_{0}}{2}$
Option D:
$4 I_{0}$

Answer: B

Step-by-step solution

$\mathrm{d}=2 \mathrm{~mm}$ $\mathrm{D}=10 \mathrm{~m}$ $\lambda=6000 \AA$ $y=\frac{d}{2}($ in front of one slit $)$ $I=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{\lambda} \cdot \frac{y}{D} d\right)$ $\Rightarrow \mathrm{I}=\mathrm{I}_{0}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications