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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.

Let $r_1,r_2$ be the internal angle of refraction values measured from the two face normals, and let $i,e$ be the external incidence and emergence angles. Geometry of an optical prism gives $r_1+r_2=A$ and total deviation $\delta =(i-r_1)+(e-r_2)=i+e-A$.

For an isosceles optical prism whose internal beam is parallel to its base, reflection symmetry gives $r_1=r_2=A/2$. Applying Snell's law at both faces, with air refractive index approximated by one, gives $\sin i=n\sin (A/2)=\sin e$. For transmitted forward rays, $i=e$, so $i=(\delta +A)/2$. HenceThis is the symmetric minimum deviation geometry at a single wavelength. Transmission requires $n\sin (A/2)\le 1$; otherwise the proposed symmetric path cannot enter and leave through the two faces.

A prism minimum-deviation index measurement infers a transparent sample's relative refractive index from two angles: its known apex angle $A$ and the smallest deviation ${\delta }_{\min }$ of a selected spectral line. A spectrometer combines a source and slit, a collimator, a rotating optical prism table, and a viewing optical telescope with an angular readout. The collimator supplies a nearly parallel beam; the optical telescope locates its emergent direction. This turns an otherwise difficult measurement of the phase velocity of light into a measurement of angles.

For the symmetric path through a uniform optical prism, with the same external medium at both faces and positive refractive indices,The angles $i,e,r_1,r_2$ are measured from face normals. This is the minimum deviation relation, not a formula for an arbitrary prism orientation. An absolute refractive index additionally requires $n_{\mathrm{ext}}$.

For example, stipulate $A=60^{\circ }$ and $n_{\mathrm{rel}}=3/2$. Then $i=e=\arcsin (3/4)\simeq 48.5904^{\circ }$ and ${\delta }_{\min }\simeq 37.1808^{\circ }$. Substitution recovers $n_{\mathrm{rel}}=1.5$. These synthetic values illustrate the geometry; they are not readings for the prism in the laboratory protocol below.

A Michigan State University teaching experiment uses a mercury lamp, slit, collimator, prism table and optical telescope, with a subsidiary angular scale read to one arcminute. First focus the optical telescope on a distant object; with the prism removed, align its crosshair to a narrow vertical slit image, focus the collimator, and establish the direct-beam zero. Insert and centre the prism. Select the green mercury spectral line, tabulated at $546.1\mathrm{nm}$, then rotate the prism while tracking its image with the optical telescope. Locate the turning point where continued prism rotation reverses the image motion, and read the deviation. Re-establish the minimum separately for each other spectral line. The protocol assumes $A=60^{\circ }$; that assumed apex angle is not an independently measured result. ^[ref]

For an uncertainty-aware implementation, repeat the turning-point search from both prism-rotation directions, recheck the direct-beam reference, and record individual readings before averaging. Independently determine $A$ or include its calibration uncertainty; record the chosen spectral line, sample identity, temperature and surrounding-air conditions. These are proposed controls deduced from the inference, not claimed additional instructions or results of that manual. Finite slit width, unresolved spectral lines, imperfect collimation, table tilt, angular calibration and sample nonuniformity can limit the inferred refractive index. One arcminute readout resolution alone does not establish one arcminute total measurement uncertainty. A single selected wavelength gives one index, not a measured optical dispersion curve.

A short geometric derivation uses $r_1+r_2=A$ and $\delta =i+e-A$. Reversibility exchanges entrance and exit. At the symmetric stationary path, $i=e$ and $r_1=r_2$; applying Snell's law gives the displayed index relation. The plotted calculation checks that this stationary point is a minimum on the chosen transmitting branch. To assess local sensitivity, differentiate the inferred index using angles in radians:This first-order relation exposes both angular contributions; it is not an instruction to add signed systematic errors as independent random errors. In the synthetic example, a deviation error of $+1$ arcminute at fixed $A$ changes the inferred relative refractive index by approximately $+1.92\times 10^{-4}$.