In Young's double slit experiment, the intensity at a point P on the screen is ${\left(\frac{3}{4}\right)}^{\mathrm{th}}$ of the maximum intensity in the interference pattern. If d is the separation between the slits and $\lambda$ is the wavelength of light, the angular separation between point P and the center of the screen is:

In Young's double-slit experiment,
$\mathrm{I}=4{\mathrm{I}}_0{\cos }^2(\mathrm{\varphi }/2)$
where $\mathrm{\varphi }$ is the phase difference between the interfering beams.
${\mathrm{I}}_{\max }=4{\mathrm{I}}_0$
$\text{ Given }=\frac{3}{4}{\mathrm{I}}_{\max }$
$$\begin{gathered} \text{ or } 4{\mathrm{I}}_0·{\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)=\frac{3}{4}{\mathrm{I}}_{\max } \\ \Rightarrow {\cos }^2\left(\frac{\mathrm{\varphi }}{2}\right)=\frac{3}{4} \\ \Rightarrow \cos \left(\frac{\mathrm{\varphi }}{2}\right)=\frac{\sqrt{3}}{2} \\ \frac{\mathrm{\varphi }}{2}=\frac{\mathrm{\pi }}{6} \text{ or }\mathrm{\varphi }=\frac{\mathrm{\pi }}{3} \end{gathered}$$
Now refer to Fig

If $\mathrm{\theta }$ is the angular separation between P and ${\mathrm{P}}_0$, then $\mathrm{tan\theta }=\frac{\mathrm{y}}{\mathrm{D}}$
path difference between two waves at P,

$∆\mathrm{x}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}\mathrm{\varphi }=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{\mathrm{\lambda }}{6}$

again,

 $∆\mathrm{x}=\mathrm{d} \sin \left(\mathrm{\theta }\right)\simeq \mathrm{d} \tan \left(\mathrm{\theta }\right)$

$$\begin{gathered} \Rightarrow d \tan \left(\theta \right)=\frac{\mathrm{\lambda }}{6} \\ \Rightarrow \tan \left(\theta \right)=\frac{\mathrm{\lambda }}{6\mathrm{d}} \\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{\lambda }}{6\mathrm{d}}\right) \end{gathered}$$