Physics · Wave Optics

JEE Main 2026 — 2 April, Evening Shift — Question 18

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times the maximum intensity. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength. The value of $x$ is _____

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$I = I_{max}\cos^2(\phi/2) = \frac34 I_{max}$ → $\cos(\phi/2)=\sqrt{3}/2$ → $\phi/2=30^\circ$ → $\phi=\pi/3$ path difference = $\frac{\phi}{2\pi}\lambda = \frac{\lambda}{6}$ → $x=6$

Answer key and solution verified before publishing.

Practise Wave Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications