Q 12-10-038JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
In a Young's double slit experiment, the intensity at some point on the screen is found to be $\dfrac34$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\dfrac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 6
$I=I_{max}\cos^2\dfrac{\phi}{2}=\dfrac34I_{max}\Rightarrow\cos\dfrac\phi2=\dfrac{\sqrt3}{2}\Rightarrow\phi=\dfrac\pi3$
Path difference $=\dfrac{\lambda}{2\pi}\phi=\dfrac{\lambda}{6}$, so $x=6$.
Solution by Sreeraj P, M.Sc Physics