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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 $+1.92\times 10^{-4}$.

Refraction changes the direction of a transmitted wave when it crosses between media with different phase velocities. For light at a planar interface, Snell's law relates the angles measured from the normal vector through the refractive indices. At normal incidence the direction remains unchanged even though the wavelength and phase velocity can change. The worked example under geometrical optics distinguishes transmission from total internal reflection.

For light, refraction explains why a submerged object looks nearer the surface and why an optical lens can form an image. A wavefront crossing a boundary must retain a consistent wave phase along that boundary: its normal spacing changes with phase velocity, and oblique light rays change direction. Normal incidence separates these two effects especially clearly.

Take stationary, transparent, homogeneous media with isotropy, linearity of their optical response, and real positive refractive indices, at one chosen frequency $f$. With $c$ the vacuum speed of light and ${\lambda }_0=c/f$ the vacuum wavelength, medium $j$ hasThe frequency is unchanged at a stationary boundary; the phase velocity, wavelength and wavenumber can change. Here $n_j$ is a phase refractive index, not the group refractive index. If there is optical dispersion, evaluate $n_j$ at this same frequency on both sides.^[ref]

For an original idealized air–water example, stipulate $n_{\mathrm{a}}=1$ and $n_{\mathrm{w}}=4/3$, with vacuum wavelength $600\mathrm{nm}$. Then $f=c/(600\mathrm{nm})\approx 4.99654\times 10^{14}\mathrm{Hz}$, the water phase velocity is $3c/4$, and its wavelength is $450\mathrm{nm}$. At an air angle of incidence $30^{\circ }$, Snell's law gives a water angle of refraction $\arcsin (3/8)\approx 22.0243^{\circ }$. These stipulated values illustrate the relations; actual refractive indices depend on material conditions and frequency. By reversibility of an optical ray, the same path emerging from water has outgoing air angle $30^{\circ }$.

With incident light travelling to the right, use positive object distance $u$ to the left, image distance $d$ to the right, and curvature radius $R$ toward a right-hand centre. For a ray hitting at small height $Y$, its incoming inclination, outgoing downward inclination and inward-normal downward inclination are $\alpha \simeq Y/u$, $\beta \simeq Y/d$ and $\theta \simeq Y/R$. The paraxial approximation to Snell's law gives $n_1(\alpha +\theta )=n_2(\theta -\beta )$. Cancelling $Y$ proves the title equation. Virtual conjugates and other centre orientations require corresponding signed distances. Applying the relation at both faces of a thin lens proves the lensmaker's equation.

Let a plane incident wavefront first meet a flat interface at $A$, and let another point $B$ on that front reach $C$ on the interface after $\Delta t$. Its path has length $BC=v_1\Delta t$. The secondary wavelet from $A$ has radius $AD=v_2\Delta t$ in the second medium; the new wavefront through $C$ is tangent to it at $D$. Right triangles with common hypotenuse $AC$ give $\sin {\theta }_1=BC/AC$ and $\sin {\theta }_2=AD/AC$, where both ray angles are measured from the interface normal. Consequently $\sin {\theta }_1/\sin {\theta }_2=v_1/v_2=n_2/n_1$, which is Snell's law. The construction uses phase velocities in homogeneous isotropic media.

The refractive index compares the phase velocity of light in a medium with the vacuum speed of light $c$. For a transparent, uniform medium with isotropy and a positive-index propagation branch,Here $\omega >0$ is the angular frequency, $k>0$ is the wavenumber, and $v_{\mathrm{ph}}=\omega /k$ is the speed of a surface of constant wave phase. The refractive index is a dimensionless quantity; it is a property of a specified material response at a specified frequency, rather than a universal number attached to the word “glass”. More general absorbing or negative-index media require the conventions discussed in refractive index and electromagnetic response. ^[ref]

For an original model example, choose $n=3/2$ at vacuum wavelength ${\lambda }_0=600\mathrm{nm}$. The phase velocity is $v_{\mathrm{ph}}=2c/3\simeq 1.999\times 10^{8}m/s$, the frequency is $f=c/{\lambda }_0\simeq 4.997\times 10^{14}\mathrm{Hz}$, and the medium wavelength is $400\mathrm{nm}$. These are stipulated model values, not measured properties of glass or water. A value at this single frequency does not determine a pulse's group velocity.

The usefulness of refractive index is that one measured response connects several different observations: Snell's law predicts refraction, optical path length predicts accumulated wave phase, and changes with frequency describe optical dispersion. Even at normal incidence, where a light ray does not bend, a changed refractive index changes wavelength and wave phase. The following comparison isolates that effect; the oblique geometry is developed in wavefront.

The relative refractive index of medium two with respect to medium one is $n_{21}=n_2/n_1$, using their absolute refractive indices at the same frequency. For transparent media with isotropy and positive refractive indices, Snell's law gives $n_{21}=\sin i/\sin r$, with the angles measured from the interface normal. Thus a refraction measurement in air directly determines an index relative to air. Converting it to an absolute refractive index requires the refractive index of that air; replacing it by one is an approximation.

In a stationary reciprocal optical medium, reversing the direction of a ray produces a ray along the same path. Snell's law is unchanged when its two media and ray directions are interchanged.

A thin lens models an optical lens by collapsing its refracting surfaces into one plane, neglecting thickness relative to the relevant radius of curvature values and object and image distances. Here use an ideal converging thin lens in air and the paraxial approximation: light rays make small angles with the optical axis, so Snell's law can be linearized using the small-angle approximation. Equal surrounding refractive indices on both sides give the familiar symmetric equation.^[ref]

Put the thin lens at $x=0$, with incident light travelling to the right. An object at $x=-u$ has $u>0$. The image coordinate is $x=v$: $v>0$ for a real image and $v<0$ for a virtual image. The converging thin lens has focal length $f>0$, and heights $h_o,h_i$ are signed relative to the optical axis. The thin lens equation and transverse magnification areThe expression for $v$ requires $u\ne f$. Negative transverse magnification means inverted; positive transverse magnification means upright. Remember that the positive distance $u$ differs in sign from the object's coordinate $-u$. ^[ref].

For focal length $f=1\mathrm{m}$, object distance $u=3\mathrm{m}$, and object height $h_o=0.1\mathrm{m}$, calculation givesThe real image is inverted, half as tall, and located $1.5\mathrm{m}$ to the right of the thin lens. A screen at that position receives the image because the outgoing light rays actually meet there.

For the same focal length $f=1\mathrm{m}$ and object height $h_o=0.1\mathrm{m}$, move the object inside the front optical focus, to $u=0.5\mathrm{m}$. ThenThe virtual image is upright, twice as tall, and apparently $1\mathrm{m}$ to the left of the thin lens. Outgoing light rays still travel to the right; only their backward extensions intersect at the virtual image. A screen there cannot collect that image from the outgoing bundle. This geometry explains why a converging optical lens can act as a magnifier when held close to an object. ^[ref].

Both diagrams use two rules for ray tracing. A light ray from the object's tip parallel to the optical axis reaches the thin lens at $(0,h_o)$ and exits along the straight line through the rear optical focus $(f,0)$. Another light ray through the centre $(0,0)$ continues undeviated in the ideal thin lens model. Their intersection, or the intersection of their backward extensions, locates the image tip. ^[ref].

A short geometric proof outline makes the signs transparent. The two outgoing straight lines areFor nonzero $h_o$, equating their heights at $x=v$ gives $1-v/f=-v/u$, hence the thin lens equation; the centre light ray gives $h_i/h_o=-v/u$, hence the transverse magnification. This derives the image equations from the ideal ray tracing rules, whose physical justification comes from Snell's law in the paraxial approximation.

At $u=f$, the two outgoing straight lines are parallel: there is no finite image position. For $u>f$ the real image has $m<0$; for $0<u<f$ the virtual image has $m>0$. These cases let an optical camera designer choose a receiving position and let a user understand a magnifier. An actual optical lens has finite apertures, which cause diffraction and impose diffraction-limited resolution; optical aberrations can further blur the image. The diagrams are ideal geometrical optics, with exact coordinates within that model, rather than exact predictions of Maxwell's equations. ^[ref].

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]