For a situation shown in figure, find the refrective index of glass...

Clearly $r+i_c=90^{\circ}$$\begin{aligned} & r=90-i_c \\ & \mu=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ}}{\sin \left(90-i_c\right)}\end{aligned}$$$From eqns (i) and (ii)$$\begin{aligned} & \sin i_c=\frac{2 \cos i_c}{\sqrt{3}} \Rightarrow \tan i_c=\frac{2}{\sqrt{3}} \\ \text{ or } & i_c \approx 49^{\circ} 11^{\prime} \approx 50^{\circ} \\ \Rightarrow & \mu=\frac{1}{\sin i_c}=\frac{1}{\sin 50^{\circ}}=1.31\end{aligned}$$