$A$ 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 $60^{\circ}$. If the prism is placed in water (refractive index $1.33$),predict the new angle of minimum deviation of a parallel beam of light.

(N/A) Given: Angle of minimum deviation $\delta_{m} = 40^{\circ}$,Angle of prism $A = 60^{\circ}$,Refractive index of water $\mu_{w} = 1.33$.
$1$. Finding the refractive index of the prism material $(\mu_{g})$:
Using the prism formula: $\mu_{g} = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$
$\mu_{g} = \frac{\sin((60^{\circ} + 40^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin 50^{\circ}}{\sin 30^{\circ}} = \frac{0.766}{0.5} = 1.532$.
$2$. Finding the new angle of minimum deviation $(\delta'_{m})$ in water:
The relative refractive index of glass with respect to water is $\mu_{g/w} = \frac{\mu_{g}}{\mu_{w}} = \frac{1.532}{1.33} \approx 1.1519$.
Using the formula: $\mu_{g/w} = \frac{\sin((A + \delta'_{m})/2)}{\sin(A/2)}$
$\sin((60^{\circ} + \delta'_{m})/2) = 1.1519 \times \sin(30^{\circ}) = 1.1519 \times 0.5 = 0.57595$.
$(60^{\circ} + \delta'_{m})/2 = \sin^{-1}(0.57595) \approx 35.17^{\circ}$.
$60^{\circ} + \delta'_{m} = 70.34^{\circ}$.
$\delta'_{m} = 70.34^{\circ} - 60^{\circ} = 10.34^{\circ}$.
Thus,the new angle of minimum deviation is $10.34^{\circ}$.