Problem (IIT JEE 2008): Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60 degree). In the position of minimum deviation, the angle of refraction will be,
Solution: At the angle of minimum deviation ($\delta_m$), angle of incidence is equal to the angle of emergence, the angle of refraction ($r$) is equal to half of the prism angle ($A$) and ray inside the prism is parallel to the prism base. Further, refractive index is given by, \begin{align} \label{aga:eqn:1} \mu=\frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}, \end{align} and the angle of incidence by, \begin{align} \label{aga:eqn:2} i=\frac{A+\delta_m}{2}. \end{align} From above equations, $\delta_m$ and $i$ depend on $\mu$ (colours). However, for the given prism, $r=A/2={30}\;\mathrm{degree}$ is independent of $\mu$. Readers are encouraged to find $\delta_m$ and $i$ for red and violet colours if $\mu_\text{red}=1.514$ and $\mu_\text{violet}=1.523$.