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An Abbé refractometer places a small sample between an illuminating prism and a measuring prism, then locates the boundary between bright and dark parts of an angular field. That limiting ray determines the sample's refractive index through Snell's law and the known measuring-prism geometry.

The prism surfaces must be clean and the sample must make good optical contact: a trapped air layer introduces a different interface. Because refractive index varies with wavelength and temperature, a useful reading specifies both; an instrument using white illumination needs compensation for the colored boundary caused by dispersion. A reference liquid checks the reading before an unknown sample is measured.

The angle of refraction is the angle between the transmitted light ray and the surface normal vector, measured on the transmitted side. With positive refractive indices, a propagating transmitted light ray satisfies $n_i\sin {\theta }_i=n_t\sin {\theta }_t$ in Snell's law. The incident and transmitted angles are measured from the same normal line, not from the surface and not from the emergent light ray at a second surface.

For an interface between two lossless, isotropic, nonmagnetic media with unequal refractive index values $n_1,n_2$, Brewster's angle is the incidence angle at which the reflected amplitude in transverse magnetic polarization vanishes. The Fresnel equations and Snell's law give $\tan {\theta }_B=n_2/n_1$; the reflected and transmitted rays are perpendicular there. The usual formula assumes equal magnetic permeabilities.

A photon emitted at angle ${\theta }_C$ to a normally incident particle can cross a planar boundary from index $n_1$ to $n_2$ only if Snell's law permits a real transmitted angle. Combining $\cos {\theta }_C=1/(n_1\beta )$ with that condition gives $n_1^2-{\beta }^{-2}\le n_2^2$. Otherwise the photon undergoes total internal reflection. For a relativistic particle in glass of index $1.55$ followed by air, the condition fails despite emission inside the glass.

For a positive refractive index depending only on $x$, Fermat principle applied to $\int \mu (x)\sqrt{1+y^{\prime 2}}dx$ has $y$ as a cyclic coordinate. The Euler-Lagrange equation therefore makes the displayed quantity constant along a stationary ray. When ${\mu }^2>p^2$, its slope is $y^{\prime }=p/\sqrt{{\mu }^2-p^2}$. If $\theta$ is the angle to the $x$ axis, this is conservation of $\mu \sin \theta$, the stratified-medium form of Snell's law. For fixed endpoint values, the integrand has second derivative $\mu (1+y^{\prime 2}{)}^{-3/2}>0$ with respect to $y^{\prime }$. Its strict convexity and the constant momentum show, by the tangent-line inequality and integration, that any such ray is the unique minimum among competing graph paths with the same endpoints.

For two homogeneous inviscid fluids, specify displacement amplitudes along one fixed interface normal. At a real positive frequency, their ratios equal those of the corresponding normal velocities. For propagating plane waves with incidence angles ${\theta }_j$ from the normal, letContinuity of normal velocity and pressure givesThe condition $|r_n|=|t_n|$ therefore becomes $|Z_{1,n}-Z_{2,n}|=2Z_{1,n}$. Positive impedances make this equivalent to $Z_{2,n}=3Z_{1,n}$, giving $r_n=-1/2$, $t_n=1/2$.

At normal incidence this is simply ${\rho }_2{c}_2=3{\rho }_1{c}_1$. At oblique incidence, Snell's law also requires $\sin {\theta }_1/c_1=\sin {\theta }_2/c_2$, so both angular factors enter. Equal normal displacement amplitudes then correspond to reflected and transmitted power fractions$1/4$ and$3/4$, because transmitted power has an impedance factor. Full displacement magnitudes along each ray use different cosine factors; the measured component must be stated.

At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

Fermat's principle characterizes a light ray between fixed endpoints by stationary travel time, or stationary optical path length in a stationary isotropic medium. With refractive index $n(\mathbf{x})$ and path element $ds$, the travel time is $T=c^{-1}\int nds$, so an allowed ray has vanishing first variation of this integral. Stationary does not always mean the globally shortest time.

For a planar interface between two uniform media, vary the point at which a broken straight path crosses the interface. Setting the travel-time derivative to zero gives $n_1\sin {\theta }_1=n_2\sin {\theta }_2$, Snell's law, with angles measured from the normal. This makes the link between a variational statement and an observable direction of refraction explicit. The geometrical optics description requires scales large enough compared with the wavelength; diffraction supplies effects that a single ray cannot describe.

To isolate the readout mechanism, replace the processed photographic emulsion by $N$ equally spaced, identical weak scattering planes, with separation $d$. Take the readout refractive index $n$ to be real and independent of wavelength, and let $\theta$ be the internal angle from the plane normal. Adjacent specularly backscattered contributions differ in wave phase by $2k_{z}d=4\pi nd\cos \theta /{\lambda }_0$. Their geometric series givesHere $a$ is a small plane reflection wave amplitude, assumed constant in this illustrative band; $F_N=1$ at $\delta =0$ by continuity. Since incident and reflected waves occupy the same external medium, $|r_N{|}^2$ is the model reflectance when transmission factors are absorbed into $a$. Require $N|a|\ll 1$ for the weak-scattering sum to remain self-consistent. Normalizing to $F_N$ discards absolute brightness; it does not predict a highly reflecting historical plate.

The first-order peak obeysThis is Bragg's law with an angle measured from the normal, rather than the glancing Bragg angle. For external viewing angle $\alpha$ in air approximated by refractive index one, Snell's law gives $n\sin \theta =\sin \alpha$, henceThe expression describes specular readout: illumination must supply the matching incident direction. Tilting a plate under fixed narrow illumination can make the signal disappear as well as change its selected colour.

For the chosen $n=1.5$ and $d=200\mathrm{nm}$, calculated peaks are $600\mathrm{nm}$ at $\alpha =0$, $565.69\mathrm{nm}$ at $30^{\circ }$, and $529.15\mathrm{nm}$ at $45^{\circ }$. Greater obliquity selects a shorter wavelength. Uniform shrinkage to $d^{\prime }=sd$, with unchanged $n$, multiplies every peak wavelength by $s$: a chosen $s=0.95$ moves the normal peak to $570\mathrm{nm}$. More generally, ${\lambda }_B^{\prime }/{\lambda }_B=(n^{\prime }/n)s$ at normal incidence. A change of thickness alone is insufficient if processing also changes the refractive index.

Finite $N$ gives a finite spectral response. The nearest zeros satisfy $\delta =\pm 2\pi /N$, giving ${\lambda }_{\pm }={\lambda }_B/(1\pm 1/N)$. At the normal $600\mathrm{nm}$ peak, $N=30$ gives neighbouring zeros at approximately $580.65$ and $620.69\mathrm{nm}$; $N=10$ gives $545.45$ and $666.67\mathrm{nm}$. Thus greater sampled depth narrows the central band in this model. The small-detuning distance from peak to a nearest zero is approximately ${\lambda }_B/N$; it is not the full width at half maximum.

Real processed photographic emulsions have a continuous, irregular depth response rather than identical planes. Absorption, wavelength-dependent refractive index, unequal scattering strengths, a nonideal exposure reflector, changes during photographic processing and multiple scattering of light can alter both peak positions and shapes. A transfer calculation with measured material parameters would be needed for quantitative reflectance. The present model explains the phase-selection mechanism and the tradeoff between depth and spectral width; it cannot establish faithful reproduction of an arbitrary incident optical spectrum. Its useful lesson is that a photograph can record colour through sub-wavelength structure, while its eventual appearance still depends on illumination, viewing geometry and processing.

Geometrical optics describes propagation by rays and Snell's law, neglecting diffraction on scales much larger than the wavelength.

A light ray is a path in an approximation, not a physical thread of zero width. In a transparent, homogeneous, isotropic medium it is straight and perpendicular to a wavefront; boundaries redirect it by the law of reflection or Snell's law. The approximation is useful when the wavelength is small compared with relevant apertures, surface curvature scales, and distances over which the refractive index changes. For an illustrative wavelength of $500\mathrm{nm}$ and a $10\mathrm{mm}$ aperture, their ratio is $5\times 10^{-5}$. That small ratio supports ray tracing for image location, but does not eliminate diffraction when calculating the fine structure of a focused spot. At a wave caustic, geometrical optics can fail despite large optical elements. ^[ref].

The mechanism of refraction is a change in phase velocity. For transparent isotropic media at the chosen frequency, the refractive index iswhere $c$ is the vacuum speed of light and the angles are measured from the interface normal vector. At a stationary planar interface the frequency and tangential wavevector match across the boundary. Since the wavenumber is $k=n\omega /c$, that matching gives Snell's law. The transmitted wavelength changes while the frequency stays fixed. Here $n$ uses phase velocity, which need not equal group velocity. ^[ref]. ^[ref].

Choose model air with $n_1=1$ and model glass with $n_2=1.5$, without claiming these are exact values for every sample or wavelength. At ${\theta }_1=45^{\circ }$, Snell's law givesThe transmitted light ray bends toward the normal vector; its phase velocity is $c/1.5$, and its wavelength is two thirds of the air value. By reversibility of an optical ray, incidence from glass at $28.1255^{\circ }$ returns the outgoing air angle $45^{\circ }$.

Incidence from glass at $45^{\circ }$ is a different experiment. It would require $\sin {\theta }_2=1.5\sin 45^{\circ }\approx 1.06066$, so there is no real transmitted angle. The critical angle for total internal reflection isAbove this angle, an ideal lossless glass-air boundary produces total internal reflection. The physical optics description still permits an evanescent wave in the air, although no propagating transmitted light ray crosses the boundary. ^[ref]. ^[ref].

The usefulness of geometrical optics is that complicated instruments can be assembled from these local rules. Applying Snell's law at successive curved surfaces changes a spreading bundle into a converging one. The Fermat principle unifies the construction through stationary optical path length; stationarity does not require a global minimum. ^[ref]. Fang's MIT notes explicitly connect this principle to specular reflection, refraction, and thin lens imaging, providing a route from individual paths to complete instruments. ^[ref].

Grazing-incidence X-ray optics redirect X-rays by shallow specular reflections along mirror surfaces. Riccardo Giacconi's lecture describes coaxial paraboloidal and hyperboloidal mirrors using two reflections, and the later Einstein observatory's imaging capability. Focusing separates collecting area from detector area: the collected source signal can remain large while the detector background is drawn from a smaller region.^[ref]

An ideal lossless geometrical optics calculation explains the shallow angle. Take vacuum refractive index one and a material with real refractive index $n=1-\delta$, where $0<\delta \ll 1$. Let $\theta$ be the grazing angle measured from the surface. Snell's law written for angles from the normal requires $\cos \theta =n\sin {\theta }_t$. A propagating transmitted ray exists only if $\cos \theta \le n$. Thus the ideal threshold isThe last step uses the Taylor series $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$. For a chosen $\delta =2\times 10^{-5}$, ${\theta }_c\simeq 0.006325$ radians, or $0.3624^{\circ }$. Absorption makes the refractive index complex and softens this ideal threshold; surface roughness and alignment errors also reduce throughput or worsen the point spread function. These are reasons to calibrate X-ray effective area and angular resolution, rather than assume that reflection is perfect.

The benefit can be quantified without specifying a historical telescope. Suppose a source retains fraction $f$ of its counts inside an extraction region, while a uniform detector background contributes $qat$ counts over detector area $a$. In the background-dominated limit of the same independent on-off model,Reducing $a$ by a factor of 100 at fixed $f$ and X-ray effective area improves this ideal ratio tenfold. A collimator restricts admitted directions; focusing optics bring selected rays together. An image also separates nearby sources and identifies clean background regions, connecting optical design directly to a better constrained inverse problem.

For vacuum incidence at fixed angle ${\theta }_i$ into a transparent isotropic medium, Snell's law gives $n({\lambda }_0)=\sin {\theta }_i/\sin {\theta }_t({\lambda }_0)$. Thus $dn/d{\lambda }_0=-n\cot {\theta }_{t}d{\theta }_t/d{\lambda }_0$, andAngles must be differentiated in radians. This determines the local group velocity from a first derivative of measured refraction angles; pulse broadening needs further information about group velocity dispersion and absorption.

At a stationary, uniform planar boundary, incident and transmitted plane waves have the same variation of wave phase along the interface and in time. The figure marks places where equal-wave phase wavefronts meet the boundary. Although their perpendicular spacing differs, their intersections must advance along the boundary together. This is the local mechanism behind Snell's law.^[ref]

For a short derivation, let $x$ run along the interface and $z$ increase into the transmitted medium. Write each wave phase as ${\Phi }_j=k_{j,x}x+k_{j,z}z-{\omega }_{j}t+{\varphi }_j$. The electromagnetic boundary conditions must hold at every $x,t$ on $z=0$; their matching requires ${\omega }_i={\omega }_t$ and $k_{i,x}=k_{t,x}$. Constant offsets ${\varphi }_j$ and wave amplitudes are determined separately. Since $k_{j,x}=k_j\sin {\theta }_j$, this givesFor nonzero tangential wavenumber, the spacing of equal-wave phase intersections along the interface is ${\lambda }_i/\sin {\theta }_i={\lambda }_t/\sin {\theta }_t$. It is not the perpendicular wavelength. At normal incidence, the tangential wavenumber vanishes and the interface is uniform in wave phase.^[ref]

The lensmaker's equation connects focal length to the refractive index and shape of a thin lens. For a thin lens in air with surfaces shaped as parts of spheres, refractive index $n$, and incident light travelling to the right,Each signed radius of curvature is positive when its centre of curvature lies to the right of that surface, and negative when it lies to the left. A symmetric biconvex thin lens with $n=1.5$, $R_a=1\mathrm{m}$, and $R_b=-1\mathrm{m}$ therefore has $1/f=0.5(1+1){\mathrm{m}}^{-1}=1{\mathrm{m}}^{-1}$, giving $f=1\mathrm{m}$ and optical power $+1\mathrm{D}$. This is an ideal shape example consistent with the thin lens diagrams, not a specification of their finite thickness.

In a uniform surrounding medium of refractive index $n_s$ on both sides, replace $n-1$ by $n/n_s-1$. Thus the same shape reverses its focusing sign if its refractive index is lower than its surroundings: for $n=1.5$ and $n_s=2$, the same radii give $1/f=-0.5{\mathrm{m}}^{-1}$, or $f=-2\mathrm{m}$. Positive focal length therefore cannot be inferred from convex shape in every environment. A proof outline applies Snell's law with the small-angle approximation at each surface, adds their direction changes, and neglects the distance between them. ^[ref].

An optical lens is a transparent element, or assembly of elements, that redirects a bundle of light rays to form images or change how the bundle spreads. In an ordinary glass optical lens, refraction at the surfaces provides the redirection: their changing normal vectors make Snell's law bend different light rays by different amounts. A converging optical lens brings incoming light rays parallel to its optical axis toward an optical focus; a diverging optical lens makes them spread as if from an optical focus on the incoming side. ^[ref].

For a usual glass thin lens in air, a form thicker at the centre than at the edges normally converges, with positive focal length; one thinner at the centre normally diverges. Shape alone is insufficient: the refractive index relative to the surroundings also matters, as the lensmaker's equation shows. An aperture selects which light rays pass; an optical lens also changes their directions. The distinction explains how an optical camera can gather light from one object point over many incoming directions and deliver it to one image location.

The beautiful simplification is that an entire bundle can be characterized by a focal length, and two constructed light rays can locate an image. The thin lens examples below turn that geometry into a prediction of image position, orientation and transverse magnification. A real optical lens retains optical aberrations and diffraction, so the construction is a useful leading model rather than an exact solution of Maxwell's equations.^[ref]

Early mathematical treatments include Johannes Kepler's 1611 Dioptrice, whose title explicitly concerns vision through newly invented instruments using transparent glass, and René Descartes's 1637 La Dioptrique, whose seventh discourse discusses improving vision with shaped glass. These are landmarks in explaining and designing optical instruments, rather than dates for the invention of the optical lens. ^[ref]; ^[ref].

Use a slit light source, a rectangular transparent block on plain paper, pencil, ruler and protractor. Trace the block outline, keep it fixed, and mark widely separated points on incident and emergent light rays. Remove the block and join the entry and exit points to reconstruct its internal light ray. Draw the surface normal vector at entry; measure $i$ outside and $r$ inside from that normal line. Choose a path exiting the opposite parallel face rather than a side edge. Repeat for a range of $i$ with the same block and source colour; record $i,r,\sin i,\sin r$ and repeat-reading spread. Plot $\sin i$ vertically against $\sin r$ horizontally: Snell's law predicts slope $n_{\mathrm{block}}/n_{\mathrm{air}}$, approximately the block's refractive index. The emergent light ray should be parallel to the incident one. Beam width, pencil marks, protractor resolution, block movement and colour spread limit precision; a narrow-colour source reduces spread from optical dispersion. These are measured angular readings, not a direct timing of phase velocity.^[ref]

A normal incidence trial checks the drawn normal and the absence of bending, but the ratio $\sin i/\sin r$ is then $0/0$. Small nonzero angles also make that ratio sensitive to reading errors. Compare several trials through the slope rather than relying on one nearly axial pair. This rectangular-block measurement complements the existing total internal reflection experiment, which probes a different boundary condition.

For a homogeneous optical prism in air, let its apex angle be $\varphi$ and refractive index $n>1$. A transmitted ray through both refracting faces exists only iffor a nongrazing symmetric path. At minimum deviation, both internal angles are $\varphi /2$ and both external angles are $({\delta }_{\min }+\varphi )/2$. Snell's law gives

To see the feasibility condition, the internal angles sum to $\varphi$ and each must be below the escape critical angle $\arcsin (1/n)$. The symmetric path satisfies both restrictions exactly when the displayed inequality holds. Under the ordinary transmitted-ray geometry, symmetry also minimizes the sum of the two external angles.

Measuring ${\delta }_{\min }$ therefore determines $n$ at the chosen wavelength. For $\varphi =60^{\circ }$ and $n=1.5$, the deviation is approximately $37.18^{\circ }$. The formula does not describe an emerging ray when the required refraction is forbidden.

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.

At normal incidence, the incident light ray follows the surface normal vector. Snell's law gives ${\theta }_t=0$ for positive refractive indices, so the transmitted light ray does not bend. Nevertheless, its phase velocity and wavelength change if the refractive index changes. A single axial light ray therefore cannot measure apparent depth: neighboring light rays and their directions supply that information. A reflected component can still be present; no bending of the transmitted light ray does not imply complete transmission.