For a glass prism of refractive index √ 3, the value of angle of minimum deviation δ is equal to the angle of prism A. Find the angle of prism A. How…

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