= Minimum deviation of a prism
{title2=$\Delta_{\min}=2\arcsin(n\sin(A/2))-A$}
Consider a homogeneous <optical prism> with apex <angle> $0<A<\pi$, in the same exterior medium on both sides. Let $n>1$ be the \b[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 $e=f(r_2)$, where
$$
f(r)=\arcsin(n\sin r),\qquad 0\leq r\leq\arcsin(1/n).
$$
The deviation is $\Delta=i+e-A$. On the interior of this interval,
$$
f''(r)=\frac{n(n^2-1)\sin r}{(1-n^2\sin^2r)^{3/2}}\geq0.
$$
Thus <convexity> gives $f(r_1)+f(r_2)\geq2f(A/2)$. The minimum occurs for the symmetric path $r_1=r_2=A/2$, yielding
$$
\Delta_{\min}=2\arcsin\!\bigl(n\sin(A/2)\bigr)-A.
$$
A 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 $f''$ 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>.
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