 Solution

Let the two internal angles be $r={\theta }_2$ and $\varphi -r={\theta }_3$, since the optical prism geometry gives ${\theta }_2+{\theta }_3=\varphi$. By Snell's law, the corresponding external angles are $F(r)$ and $F(\varphi -r)$, where $F(r)=\arcsin (n\sin r)$. The deviation isOn the transmitted branch with $n>1$ and $0<r<\arcsin (1/n)$,Thus $F^{\prime }$ is strictly increasing. The stationary condition ${\gamma }^{\prime }(r)=F^{\prime }(r)-F^{\prime }(\varphi -r)=0$ forces $r=\varphi -r$, and the positive second derivative makes this the unique minimum deviation. ConsequentlyThe argument assumes a transmitted ray is possible; the symmetric internal angles must lie below the critical angle.