Q 12-10-033JEE MainJEE Main 2026 (5 Apr, Shift 2)Medium
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 ______.
Answer: (B) $\dfrac{I_0}{2}$
Concept. Opposite a slit, $y=\dfrac d2$, and path difference $=\dfrac{yd}{D}$.
Step 1. Path. $\dfrac{(d/2)d}{10d}=\dfrac{d}{20}=\dfrac{5\lambda}{20}=\dfrac\lambda4$.
Step 2. Intensity. $\phi=\dfrac\pi2$: $I_0\cos^2\dfrac\pi4=\dfrac{I_0}{2}$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
