IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Deviation and Dispersion of Light by a Prism

## Problems from IIT JEE

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,

1. 30 degree for both the colours.
2. greater for the violet colour.
3. greater for the red colour.
4. equal but not 30 degree for both the colours.

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 $\mu_\text{violet}=1.523$.