Solution to 2001 Problem 97

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.

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