Explanation:
A: Given, \(\mathrm{i}=0^{\circ}, \mu=\sqrt{2}\)
Then, \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}_1}=\mu\)
\(\begin{aligned}
\frac{\sin 0^{\circ}}{\sin r_1}=\sqrt{2} \\
\sin r_1=\frac{0}{\sqrt{2}}=0=\sin 0^{\circ} \\
r_1=0^{\circ}
\end{aligned}\)
And, \(\quad \frac{\sin \mathrm{r}_2}{\sin \mathrm{e}}=\frac{1}{\mu}\)
\(\sin \mathrm{e}=\sqrt{2} \sin \mathrm{r}_2\)
Since,
\(\begin{aligned}
r_1+r_2=60^{\circ} \Rightarrow 0+r_2=60^{\circ} \\
r_2=60^{\circ}
\end{aligned}\)
So,
\(\sin \mathrm{e}=\sqrt{2} \sin 60^{\circ}\)
\(\begin{aligned}
\sin \mathrm{e}=\sqrt{2} \times \frac{\sqrt{3}}{2} \\
\mathrm{e}=\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)>1
\end{aligned}\)
So, light is incident at more than critical angle and totally internally reflected.
So, Deviation angle \((\delta)=\mathrm{i}-\left(\mathrm{r}_1+\mathrm{r}_2\right)\)
\(\begin{aligned}
\delta=0-60^{\circ} \\
\delta=60^{\circ} \quad \text { (magnitude only) }
\end{aligned}\)
