A prism is made of glass of unknown refractive index.

Solution:

Angle of minimum deviation, $\delta_{\mathrm{m}}=40^{\circ}$

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism $=\mu^{\prime}$

The angle of deviation is related to refractive index $\left(\mu^{\prime}\right)$ as:

$\mu^{\prime}=\frac{\sin \frac{\left(A+\delta_{m}\right)}{2}}{\sin \frac{A}{2}}$

$=\frac{\sin \frac{\left(60^{\circ}+40^{\circ}\right)}{2}}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=1.532$

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let $\delta_{\mathrm{m}}^{\prime}$ be the new angle of minimum deviation for the same prism.

 

The refractive index of glass with respect to water is given by the relation:

$\mu_{\mathrm{g}}^{\mathrm{w}}=\frac{\mu^{\prime}}{\mu}=\frac{\sin \frac{\left(A+\delta_{\mathrm{m}}^{\prime}\right)}{2}}{\sin \frac{A}{2}}$

$\sin \frac{\left(A+\delta_{\mathrm{m}}^{\prime}\right)}{2}=\frac{\mu^{\prime}}{\mu} \sin \frac{A}{2}$

$\sin \frac{\left(A+\delta_{\mathrm{m}}^{\prime}\right)}{2}=\frac{1.532}{1.33} \times \sin \frac{60^{\circ}}{2}=0.5759$

$\frac{\left(A+\delta_{\mathrm{m}}^{\prime}\right)}{2}=\sin ^{-1} 0.5759=35.16^{\circ}$

$60^{\circ}+\delta_{\mathrm{m}}^{\prime}=70.32^{\circ}$

$\therefore \delta_{\mathrm{m}}{ }^{\prime}=70.32^{\circ}-60^{\circ}=10.32^{\circ}$

Hence, the new minimum angle of deviation is 10.32°.