In Young's double slit experiment, the intensity of light at a point on the screen is \({ }^{\prime} \mathrm{K}^{\prime}\) unit for path difference \({ }^{\prime} \lambda^{\prime}\). What would be the intensity at a point if path
difference is \(\frac{\lambda^{\prime}}{4}\) ?

When path difference is \(\lambda\), phase difference \(\phi\) is \(2 \pi\) and when path difference is \(\frac{\lambda}{4}\), the phase difference is \(\frac{\pi}{2}\). If \(\mathrm{I}_{0}\) is the intensity of the two waves then the resultant intensity is given by
\(
\begin{aligned}
I &=4 I_{0} \cos ^{2} \frac{\phi}{2} \\
\therefore K &=4 I_{0} \cos ^{2} \frac{2 \pi}{2}=4 I \cos ^{2} \pi=4 I_{0} \quad \because \cos \pi=-1
\end{aligned}
\)
\(\text { When } \phi=\frac{\pi}{2} \)
\(\mathrm{I}=4 \mathrm{I}_{0} \cos ^{2} \cdot \frac{\pi}{4} \)
\(=4 \mathrm{I}_{0} \frac{1}{2}=\frac{\mathrm{K}}{2} \quad \because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)