Given: angle of prism \(A =60^{\circ}\)
refractive index \(\eta=1.5\)
required, angle of minimum deviation \(D_{m}\)
$$
\begin{aligned}
&\eta=\frac{\frac{\sin \left(D_{m}+A\right)}{2}}{\frac{\sin A}{2}} \\
&1.5=\frac{\frac{\sin \left(D_{m}+60\right)}{2}}{\frac{\sin 60}{2}} \\
&\left.1.5 \times \sin 30=\frac{\sin \left(D_{m}\right.}{2}+60\right) \\
&\frac{3}{4}=\frac{\sin \left(D_{m}+60\right)}{2} \\
&\sin { }^{-1}\left(\frac{3}{4}\right)=\frac{D_{m}+60}{2} \\
&48.59=\frac{D_{m}+60}{2} \\
&48.59 \times 2=D_{m}+60 \\
&D_{m}=97.18-60^{\circ}=37.2^{\circ}
\end{aligned}
$$