Snell's equation tells us that
$\begin{align*}\sin \theta = n(\lambda) \sin \theta'(\lambda)\end{align*}$
This implies that
$\begin{align*}n(\lambda) = \frac{\sin \theta}{\sin \theta'(\lambda)}\end{align*}$
We differentiate this equation with respect to $\lambda$ :
$\begin{align*}\frac{d n }{d \lambda} = \frac{- \cos \theta' \sin \theta}{\sin^2 \theta'} \cdot \frac{d \theta'}{d \lambda} = ...$
So, if the spread in the wavelength is $\delta \lambda$ , then the spread in the angle of refraction is given by
$\begin{align*}\delta \theta' = \delta \lambda \left|\frac{d \theta'}{d \lambda} \right| = \delta \lambda \left|\frac{\tan \th...$
Therefore, answer (E) is correct.

Solution to 2001 Problem 97