The refracting angle of a prism $60^{\mathrm{o}}$.The refractive index of the material of the prism is $\sqrt{\frac{7}{3}}$. The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

$\begin{array}{l}\text{ Given }A=60^{\circ },\mu =\sqrt{\frac{7}{3}}\\ \text{ we know that }{\gamma }_1+r_2=A\\ \Rightarrow r_1+r_2=60^{\circ }\end{array}$

$\begin{array}{l}\text{ Snell’s law at first refraction }\\ \begin{array}{l}\sin i=\mu \sin {\gamma }_1\\ \mathrm{Sin}{\gamma }_1=\sqrt{\frac{3}{7}}\sin i\end{array}\\ \text{ Snell’s law at second refraction }\end{array}$

$\begin{array}{rl}\mu \sin r_2& =1\\ & \Rightarrow \sin r_2=\sqrt{\frac{3}{7}}\\ r_2={\sin }^{-1}\left(\sqrt{\frac{3}{7}}\right)& \\ r_2=& 40.89^{\circ }\Rightarrow r_1=19.1066^{\circ }\end{array}$

$\begin{array}{rl}\Rightarrow \mathrm{Sin}{r}_1& =0.327\\ \sin i& =\sqrt{\frac{7}{3}}\sin r_1\\ \sin i& =1/2\\ i& =30^{\circ }\end{array}$