The refractive index of the material of an equilateral prism is  $\sqrt{3}$. What is the angle of minimum deviation in degrees?

$$\begin{gathered} \mathrm{Refractive} \mathrm{index}, \mathrm{\mu } =\frac{\sin \left({\displaystyle \frac{\mathrm{A}+{\mathrm{\delta }}_{\mathrm{m}}}{2}}\right)}{\sin {\displaystyle \frac{\mathrm{A}}{2}}} \\ \mathrm{Now} \mathrm{A} = 60° \mathrm{and} \mathrm{\mu } = \sqrt{3} . \mathrm{Therefore}, \\ \sqrt{3}=\frac{\sin \left({\displaystyle 30°+\frac{{\mathrm{\delta }}_{\mathrm{m}}}{2}}\right)}{\sin 30°} \\ \mathrm{or} \sin \left(30°+\frac{{\mathrm{\delta }}_{\mathrm{m}}}{2}\right)=\sqrt{3} \sin 30°=\frac{\sqrt{3}}{2} \\ \mathrm{or} 30°+\frac{{\mathrm{\delta }}_{\mathrm{m}}}{2}=60° \mathrm{which} \mathrm{gives} {\mathrm{\delta }}_{\mathrm{m}}= 60° \end{gathered}$$