A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be ${40}^{\circ }$. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Given,
Minimum angle of deviation $\left(D_m\right)={40}^{\circ }$
Refracting angle of prism $\left(A\right)={60}^{\circ }$
For a prism of the angle 𝐴, refractive index $n_g$ placed in a medium of refractive index $n_a$ ,
$n_{ga}=\frac{n_g}{n_a}=\frac{\sin \left(\frac{A+Dm}{2}\right)}{\sin \frac{A}{2}}$
$n_g=\frac{\sin \left(\frac{{60}^{\circ }+{40}^{\circ }}{2}\right)}{\sin \frac{{60}^{\circ }}{2}}$
$n_g=\frac{\sin \left({50}^{\circ }\right)}{\sin {30}^{\circ }}=1.532$

After Prism is submerged in water $(n_w=1.33)$
$\frac{n_g}{n_w}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \frac{A}{2}}$
Putting values $\frac{1.53}{1.33}=\frac{\sin \left(\frac{{60}^{\circ }+D_m}{2}\right)}{\sin \frac{{60}^{\circ }}{2}}$
$\sin \left(\frac{{60}^{\circ }+D_m}{2}\right)=\frac{1.53}{1.33\times 2}=0.575$
$\left(\frac{{60}^{\circ }+D_m}{2}\right)={\sin }^{-1}\left(0.575\right)$
$\left(\frac{{60}^{\circ }+D_m}{2}\right)=33.10$
$D_m={10.20}^{\circ }\cong {10}^{\circ }$
Final Anser : $n\cong 1.53$ and $D_m$ for prism in water $\cong {10}^{\circ }$