Unpolarised light from air incidents on the surface of a...

Figure showing the ray diagram of refraction at a plane surface. Given, $i_p=$ angle of polarisation
So, $i_p=\tan ^{-1} \mu \Rightarrow i_p=\tan ^{-1} \sqrt{2}=\sin ^{-1} \sqrt{\frac{2}{3}}$ From Snell's law, $\sin i_p=\mu \sin r$ Here, $\mu=1.414=\sqrt{2} \Rightarrow \sin r=\frac{\sin i_p}{\mu}$$\begin{aligned} \sin r & =\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{3}} \\ r & =\sin ^{-1} \frac{1}{\sqrt{3}} \end{aligned}$$$angle of refraction Now, the angle of deviation,$$\begin{aligned} & \delta=i_p-r \\ & \delta=\sin ^{-1} \sqrt{\frac{2}{3}}-\sin ^{-1} \frac{1}{\sqrt{3}} \end{aligned}Angle of reflection, $\begin{gathered} \theta=90^{\circ}-r \Rightarrow r=90^{\circ}-\theta \\ \sin ^{-1} \frac{1}{\sqrt{3}}=90^{\circ}-\theta \Rightarrow \frac{1}{\sqrt{3}}=\sin \left(90^{\circ}-\theta\right) \\ \cos \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1} \frac{1}{\sqrt{3}} \end{gathered}$ Angle between incident and completely polarised light is given by\begin{aligned} & \phi=i_p+\theta \\ & \phi=\sin ^{-1} \sqrt{\frac{2}{3}}+\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \sin ^{-1} \sqrt{\frac{2}{3}} \end{aligned}$$ Hence, the correct option is (d).