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At minimum deviation, ${\theta }_2={\theta }_3=\varphi /2$ and ${\theta }_1={\theta }_4=({\gamma }_{\min }+\varphi )/2$. Substituting in Snell's law givesFor a known prism apex angle, measuring the minimum deviation with monochromatic light determines its refractive index. Repeating the measurement at several wavelengths measures optical dispersion. If the surrounding medium is not air, the measured ratio is relative to that medium's refractive index.

The reversibility of an optical ray interchanges entry and exit while preserving the total deviation. Away from the turning point, a ray and its reversed configuration give the same deviation with the entry and exit angles exchanged. At minimum deviation the two configurations merge: the path is symmetric and the entry and exit angles are equal.

Physically, distributing the refraction symmetrically between the two faces avoids making one face contribute disproportionately large bending. The strict convexity established in the preceding part proves that this stationary symmetric configuration is the minimum, rather than merely following from reversibility alone.

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.