JEE Main 2026 — Wave Optics Question with Solution

The position of the point on the screen exactly opposite to one of the slits is at a distance $y=\frac{d}{2}$ from the central maximum.

The path difference $\Delta x$ at this point is given by:
$\Delta x=\frac{yd}{D}$

Substituting $y=\frac{d}{2}$ and $D=10d$:
$\Delta x=\frac{\left(\frac{d}{2}\right)d}{10d}=\frac{d^2}{20d}=\frac{d}{20}$

Given that the distance between the slits is $d=5\lambda$, we have:
$\Delta x=\frac{5\lambda }{20}=\frac{\lambda }{4}$

The corresponding phase difference $\Delta \varphi$ is:
$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\left(\frac{\lambda }{4}\right)=\frac{\pi }{2}$

The intensity at this point is given by:
$I=I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

Substituting $\Delta \varphi =\frac{\pi }{2}$:
$I=I_0{\cos }^2\left(\frac{\pi }{4}\right)=I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{2}$

Answer: $\frac{I_0}{2}$