Past exam of the physics 2 paper of the University of Oxford / 2015 / 11 / b / Solution

Regard the refraction angle as a function of the angle of incidence, $\beta =\arcsin (\sin \alpha /n)$. Implicit differentiation of Snell's law gives ${\beta }^{\prime }=\cos \alpha /(n\cos \beta )$. Differentiating the deviation angle gives ${\delta }^{\prime }=2-4{\beta }^{\prime }$. An interior stationary point therefore satisfiesThe stationary deviation of a primary rainbow ray is a minimum: differentiating $\cos \alpha =n\cos \beta {\beta }^{\prime }$ again and using Snell's law gives ${\beta }^{\prime \prime }=\tan \beta (({\beta }^{\prime }{)}^2-1)$. At the stationary point, ${\beta }^{\prime }=1/2$, and hence ${\delta }^{\prime \prime }=-4{\beta }^{\prime \prime }=3\tan \beta >0$ for a positive interior refraction angle. This minimum is responsible for the concentration of emerging light rays near the primary rainbow direction.