For a glass prism of refractive index $\displaystyle \sqrt{3}$, the value of angle of minimum deviation $\displaystyle \delta$ is equal to the angle of prism A. Find the angle of prism A. How will the angle $\displaystyle \delta$ be affected if the prism were immersed in water ?

Calculation of Angle of prism & $\displaystyle 2$ Effect of water on prism & $\displaystyle 1$ \begin{gathered} \mu=\sqrt{3} \mu=\frac{\sin \dfrac{\left(A+\delta_{m}\right)}{2}}{\sin (A / 2)} \end{gathered}$\displaystyle \delta=A$ $\displaystyle \sqrt{3}=\frac{\sin \dfrac{(A+A)}{2}}{\sin (A / 2)}=\frac{\sin A}{\sin \dfrac{A}{2}}$ $\displaystyle \sqrt{3}=\frac{2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}=2 \cos \frac{A}{2}$ $\displaystyle \frac{\sqrt{3}}{2}=\cos \frac{A}{2}$ $\displaystyle \therefore A=60^{\circ}$ With increase in refractive index of surrounding medium, angle of deviation $\displaystyle \left(\delta_{\mathrm{m}}\right)$ will decrease