In Young's double slit experiment, the intensity at a point is (1 /4) of the maximum intensity. The angular position of this point is 

 Let P be the point at angular position $\theta$ where the intensity is (1 /4 ) of &e maximum intensity as shown in fig. (a). We know that 
 

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_{max}{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right) \\ \frac{{\mathrm{I}}_{max}}{4}={\mathrm{I}}_{max}{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)\text{ or }4{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)=1 \\ {\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)=\frac{1}{4}\text{ or }\cos \left(\frac{\mathrm{\varphi }}{2}\right)=\frac{1}{2} \\ \frac{\mathrm{\varphi }}{2}={\cos }^{-1}\left(\frac{1}{2}\right)={60}^{\circ }\text{ or }\mathrm{\varphi }={120}^{\circ }=\left(\frac{2\mathrm{\pi }}{3}\right) \\ \frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times \mathrm{dsin}\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}\text{ or }\mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{\lambda }}{3\mathrm{d}}\right) \end{gathered}$$