\(n=\frac{\sin 90^{\circ}}{\sin C}=\frac{1}{\sin C}\)

\(\therefore \sin C=\frac{1}{n}=\frac{\sqrt{3}}{2}\)

\(\therefore \mathrm{C}=60^{\circ}\)

Applying Snell's Law at \(P\)

\(n=\frac{\sin \theta}{\sin (90-C)} \Rightarrow \sin \theta=n \times \sin (90-C) ;\) from \((1)\)

\(\Rightarrow \sin \theta=n \cos\)

\(\therefore \theta=\sin ^{-1}\left[\frac{2}{\sqrt{3}} \times \cos 60^{\circ}\right]\)

or \(\quad \theta=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)