In Young's double slit experiment, intensity at a point is \(\left(\frac{1}{4}\right)\) of the maximum intensity. The angular position of this point is

For any point in interference pattern,
\(I=I_{\max } \cos ^2 \frac{\phi}{2} \)
\( \therefore \frac{I_{\max }}{4}=I_{\max } \cos ^2 \frac{\phi}{2} \)
\( \therefore \cos ^2 \frac{\phi}{2}=\frac{1}{4} \)
\( \therefore \cos \frac{\phi}{2}=\frac{1}{2} \)
\( \therefore \frac{\phi}{2}=60^{\circ}=\frac{\pi}{3} \)
\( \therefore \phi=\frac{2 \pi}{3}\)
We know that, \(\phi=\left(\frac{2 \pi}{\lambda}\right) \Delta \mathrm{x}\), where \(\Delta \mathrm{x}\) is path difference. and \(\Delta \mathrm{x}=\mathrm{d} \sin \theta\)
\(\therefore \frac{2 \pi}{3}=\frac{2 \pi}{\lambda}(\mathrm{~d} \sin \theta) \)
\( \therefore \frac{\lambda}{3 \mathrm{~d}}=\sin \theta \)
\( \therefore \theta=\sin ^{-1}\left(\frac{\lambda}{3 \mathrm{~d}}\right)\)