 Minimum deviation of a prism

Consider a homogeneous optical prism with apex angle $0<A<\pi$, in the same exterior medium on both sides. Let $n>1$ be the relative refractive index, the inside refractive index divided by the exterior refractive index. For a ray crossing both faces in the usual principal section, the internal angles obey $r_1+r_2=A$. Snell's law gives external incidence and emergence angles $i=f(r_1)$ and $e=f(r_2)$, whereThe deviation is $\Delta =i+e-A$. On the interior of this interval,Thus convexity gives $f(r_1)+f(r_2)\ge 2f(A/2)$. The minimum occurs for the symmetric path $r_1=r_2=A/2$, yieldingA path with nongrazing entry and exit requires $n\sin (A/2)<1$. Equality describes only a grazing limit; a larger value prevents transmission across both faces in this section. The condition $n>1$ matters: if $n<1$, the sign of $f^{\prime \prime }$ reverses, so the symmetric path maximizes this signed deviation instead. Measuring the minimum deviation therefore determines $n=\sin ((A+{\Delta }_{\min })/2)/\sin (A/2)$ for an ordinary glass-in-air optical prism.