Geometrical optics

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].

The angle through which an optical element changes a light ray's propagation direction, with the chosen branch and convention specified. For the one-internal-reflection branch through a spherical drop, entry refraction, internal specular reflection, and exit refraction give $\delta =\pi +2\alpha -4\beta$, with $\sin \alpha =n\sin \beta$. The corresponding primary rainbow radius about the antisolar direction is $\pi -\delta$.

For a light ray in a spherically stratified refractive index $n(r)$, $p=n(r)r\sin \psi$ is constant, where $\psi$ is the angle to the radius. Approximate the medium by concentric homogeneous shells. Straight propagation within each shell conserves $r\sin \psi$, the line's distance from the centre; Snell's law at each shell boundary conserves $nr\sin \psi$. Passing to the smooth limit proves the result. In an exterior region with $n=1$, $p$ is the distance of the asymptotic straight light ray from the centre. Consequently a surface-tangent light ray at radius $R$ and refractive index $n_s$ has vacuum asymptote at altitude $(n_s-1)R$, provided the light ray reaches the exterior without trapping.

Ray transfer matrix analysis propagates paraxial ray height and slope with a two-by-two matrix. Free propagation through $L$ uses $\left(\begin{array}{cc}1& L\\ 0& 1\end{array}\right)$, a thin optical lens uses $\left(\begin{array}{cc}1& 0\\ -1/f& 1\end{array}\right)$, and an unfolded concave-mirror reflection uses $\left(\begin{array}{cc}1& 0\\ -2/R& 1\end{array}\right)$. Multiply matrices in propagation order, with the first operation at the right.

For forward paraxial slopes and a lossless same-index optical system with $AD-BC=1$ and $B\ne 0$, Fermat's principle gives endpoint derivatives ${\partial }_{x}l=-\theta$ and ${\partial }_{x^{\prime }}l={\theta }^{\prime }$. Substituting $\theta =(x^{\prime }-Ax)/B$ and ${\theta }^{\prime }=(D{x}^{\prime }-x)/B$ and integrating proves the displayed optical path. Free space has $B=L>0$ and increases path length by $(x^{\prime }-x{)}^2/(2L)$, fixing the sign. Its quadratic phase generates the associated linear canonical transform; $B=0$ is a separate imaging limit.

The rapidly varying real phase of a short-wavelength field written $a(\mathbf{r},t)e^{iS(\mathbf{r},t)}$. Its derivatives define the local wavevector $\mathbf{k}=\nabla S$ and angular frequency $\omega =-{\partial }_{t}S$. The WKB approximation assumes that $a$ and background parameters vary little across one wavelength.

The accumulated phase along a one-dimensional ray. In an oscillatory WKB approximation, the two travelling waves have phases $\pm \int kdx$. In an evanescent region, the positive local decay rate $\kappa =\sqrt{-k^2}$ replaces $k$, producing exponents $\pm \int \kappa dx$.

For the refractive index $\mu =1/y$ on the upper half-plane, Fermat principle and the Beltrami identity give $1/[y\sqrt{1+(dy/dx{)}^2}]=1/R$. The nonvertical rays are arcs of circles centered on $y=0$; vertical lines are the remaining rays. These are also geodesics of the Poincaré half-plane model. A ray entering the strip $0<y\le 1$ toward decreasing $y$ at angle $\theta$ to the vertical has limiting horizontal displacement $\tan (\theta /2)$, at most one. The line $y=0$ is approached only after infinite optical path length because the refractive index diverges there.

For a light ray parametrized by arc length $s$, with unit tangent vector $t$ in a smooth positive refractive index $\mu$, the optical momentum is $p=\mu t$. This is the momentum for optical path-length normalization. The Euler-Lagrange equation for Fermat principle, $\int \mu (r)|r^{\prime }(\tau )|d\tau$, gives $d(\mu t)/ds=\nabla \mu$. Therefore an index independent of $z$ conserves $p_z=\mu dz/ds$. In coordinates $r(z),\theta (z)$ this is $\mu /\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$, the first integral supplied by the Beltrami identity for several dependent variables.

For a positive refractive index $\mu (r)$ in cylindrical coordinates, a light ray with constant radius $R>0$ and varying axial coordinate has the form $\theta (z)={\theta }_0+\omega z$ when it exists. Fermat principle makes $L=\mu (r)\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$ stationary. Conservation of the momentum conjugate to the cyclic angle makes ${\theta }^{\prime }$ constant along a constant-radius path. The radial Euler-Lagrange equation is thengiving the displayed formula. A nonzero real helix requires $0<-R{\mu }^{\prime }(R)/\mu (R)<1$. If ${\mu }^{\prime }(R)=0$, $\omega =0$ gives a straight axial light ray; a generic radial profile need not support a constant-radius helix. Its helical pitch is $2\pi /|\omega |$. This ray model is useful for a graded-index optical fiber when the refractive index varies smoothly and geometrical optics applies; it does not by itself impose wave-mode quantization or boundary confinement.

A mirage is an apparent displaced image caused by refraction through a spatially varying refractive index. Above a hot surface, low-index air near the ground can bend a light ray back upwards before it reaches the surface. An observer tracing the arriving ray backwards along its local tangent can then infer an apparent image below the turning region. This explains the appearance of a distant reflective surface without requiring reflection at the road itself.

For height $x$ and horizontal position $y$, Fermat principle in a positive index profile $n(x)=1+\beta x$, $\beta >0$, conserves $n(x)y^{\prime }/\sqrt{1+(y^{\prime }{)}^2}$ on a graph $y(x)$. For positive magnitude $c$ of the horizontal optical momentum, integration gives the displayed catenary. Its minimum height is $x_0=(c-1)/\beta$, at $y=y_0$. Thus $c>1$ produces a smooth turning ray wholly above a road at $x=0$. The two graph branches have opposite signs of $dy/dx$; $x(y)$ continues smoothly through their common horizontal tangent. For a signed negative graph constant, use $|c|$ in the prefactor; $c=0$ gives a vertical ray separately.

For speed $c(y)=1/y$ in $y>0$, Fermat principle gives the elapsed-time integrand $y\sqrt{1+(y^{\prime }{)}^2}$. Its symmetric stationary light rays through $(\pm d,H)$ are the displayed catenaries. Set $t=d/a$. The endpoint equation is $\cosh t/t=H/d$. This function decreases and then increases, with unique minimum where $t\tanh t=1$. Thus above the minimum there are two stationary rays, at the minimum one, and below it none. The travel time on either ray is $T=ad+(a^2/2)\sinh (2d/a)$, and its minimum height is $a$.

For $d=1$ and $H=\cosh 1$, one solution is $a=1$, with $T=1+\sinh 2/2$. A second has $a\simeq 0.7021026952$, with $T\simeq 2.8224491447$. Hence these endpoints do not uniquely specify the ray. Stationarity is distinct from global minimization: paths descending almost to $y=0$ have limiting travel time $H^2$, below either stationary value in this example.

An angular readout using a reflected light ray: rotating a mirror by $\theta$ rotates the reflected light ray by $2\theta$, giving spot displacement $x\simeq 2D\theta$ at distance $D$ for small angles.

A wave caustic is a locus where the projection of a family of rays or wavefronts becomes singular. Several ray paths may coalesce there, and the geometrical approximation can predict divergent intensity. Finite-wavelength diffraction replaces that ideal divergence by a resolved wave pattern. The construction applies to acoustic as well as optical waves.

A path of light propagation in the geometrical optics approximation. In a smoothly varying propagation speed, Fermat principle makes its travel time stationary.

For speed $c(y)=y+c_0>0$, Fermat principle and the Beltrami identity yield circular light rays whose centres lie on $y=-c_0$. The ray between $(\pm a,0)$ in the upper half-plane is $y=\sqrt{a^2+c_0^2-x^2}-c_0$.

A light ray whose path remains in a plane containing the optical axis. In a straight step-index optical fiber, this geometry gives the complementary axis and interface-normal angles used to derive the optical fiber acceptance angle.

For the wave equation ${\phi }_{tt}-\nabla \cdot (c^2\nabla \phi )=0$ with slow variables and phase $e^{i\theta /\epsilon }$, define $\omega =-{\theta }_T$ and $\mathbf{k}=\nabla \theta$. The leading balance is ${\omega }^2=c^2|\mathbf{k}{|}^2$. Its positive-frequency dispersion relation is $\Omega =c|\mathbf{k}|$, and the group velocity is $c\mathbf{k}/|\mathbf{k}|$.

The first-order imaginary balance of the wave equation in geometrical optics gives ${\partial }_T(\omega A^2)+\nabla \cdot (c^2\mathbf{k}{A}^2)=0$ for a real leading amplitude. The action density $\omega A^2$ travels with the group velocity $c^2\mathbf{k}/\omega$. For a complex amplitude replace $A^2$ by $|A{|}^2$.

For an individual monochromatic component in $c=c(Y)$, conserved $\omega ,k$ give $\ell =\sqrt{{\omega }^2/c^2-k^2}$. With amplitude independent of $X,T$, wave-action transport for a variable-speed wave conserves $c^2\ell A^2$ on a positive-$\ell$ branch, hence $A=\mu /(c\sqrt{\ell })$. The divergence at $\ell =0$ signals failure of the outer WKB approximation.

A smooth phase gives ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ and ${\mathbf{k}}_T=-\nabla \omega$. These mixed-derivative identities allow the spatially differentiated dispersion relation to be rewritten as Hamiltonian ray equations for a local dispersion relation. Away from phase singularities they express the curl-free character of a local wavevector.

For a smooth phase $\Theta$, define local wavenumber $k={\Theta }_x$ and local angular frequency $\omega =-{\Theta }_t$. Equality of mixed derivatives gives $k_t+{\omega }_x=0$. Dividing by $2\pi$ expresses conservation of signed crest count with density $k/(2\pi )$ and flux $\omega /(2\pi )$. When a homogeneous local dispersion relation gives $\omega =\omega (k)$, the equation becomes $k_t+v_g(k)k_x=0$; its characteristic curves move at the group velocity. This phase compatibility is distinct from energy or wave-action conservation.

For paraxial rays, the Lagrange optical invariant is $H=n(y_1{u}_2-y_2{u}_1)$. Lossless paraxial ray transfer preserves this quantity: in coordinates $(y,nu)$, free propagation and a thin lens act by matrices with determinant one, which preserve the oriented area of two ray vectors. Consequently a full angular optical slit width $\theta$ at an optical pupil diameter $D$ satisfies $D\theta =A_{\mathrm{cam}}p/f_{\mathrm{cam}}$ for its image width $p$ and optical camera optical pupil width $A_{\mathrm{cam}}$ in the same meridional plane.

Etendue measures how much position-and-direction space a bundle of light rays occupies. A small spot accepting a wide range of directions can have the same etendue as a large spot accepting a narrow range. This makes etendue useful for deciding whether an optical instrument can deliver an extended source's entire admitted radiant flux to a small receiver. For forward crossings of a surface, the geometrical optics definition iswhere $\mathcal{B}$ is the occupied set of surface positions and directions, $n$ is the local refractive index, $dA$ is actual surface area, $d\Omega$ is solid angle, and $\theta$ is the angle to the forward unit normal. The cosine supplies the projected area. Opposite crossing directions must be counted separately with positive projected area, rather than cancelling them in a signed integral. ^[ref]

The shorthand $G\simeq n^{2}A\Omega$ assumes constant refractive index, a narrow nearly normal bundle, and the same admitted directions across the whole area $A$. At larger angles, use the cosine-weighted integral; a position-dependent angular acceptance cannot generally be replaced by the product of two independent extents.

Lossless reversible propagation preserves the etendue of the same bundle of light rays. Here the model uses stationary isotropic media, specular reflection and refraction, with every selected light ray reaching the output and with no optical gain or scattering of light. An aperture that blocks some incident light rays changes which bundle is counted; the full incident and transmitted etendue need not be equal. ^[ref]

The underlying phase space has four coordinates. In a planar section with constant refractive index, let $(x,y)$ locate a light ray and setThe Jacobian determinant from spherical coordinates givesThus etendue is a four-dimensional volume, preserved by the optical counterpart of Liouville theorem in Hamiltonian mechanics. Individual position and direction extents may change, and correlations between them matter. The Lagrange optical invariant is a related paraxial approximation invariant for pairs of light rays, rather than the full four-dimensional measure.

Basic radiance is radiance divided by the square of the refractive index. For radiance $L$ at a fixed wavelength,Lossless propagation conserves reduced radiance along each light ray; $L$ itself is conserved only when the endpoint refractive indices agree. A larger irradiance at a smaller image can result from admitting more directions without increasing reduced radiance. Uniform wave absorption can lower radiant flux while leaving the occupied geometric etendue unchanged; an opaque aperture removes part of that occupied set. ^[ref]

Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

Insert a centred opaque circular stop of diameter $d=6\mathrm{mm}$ directly in the plane of the $D=10\mathrm{mm}$ bundle. It transmits the same $30^{\circ }$ directions at every surviving position, but discards the surrounding annulus. ThereforeWith the uniform radiance of etendue of a circular aperture and cone, the transmitted radiant flux is $2.2207\mathrm{mW}$ and the blocked fraction is $64\%$. Surviving light rays retain their radiance; the stop has not concentrated them. Lossless propagation after the stop preserves the etendue of this surviving subset, not that of the original incident bundle. A stop placed elsewhere can select positions and directions together, so its physical area ratio alone need not give the transmitted radiant flux fraction.

An optical concentrator redirects light from an entrance area $A_{\mathrm{in}}$ to a smaller receiver area $A_{\mathrm{out}}$. Its geometric concentration ratio is $C=A_{\mathrm{in}}/A_{\mathrm{out}}$. Specify the admitted angular distribution and losses before interpreting $C$ as an irradiance gain: if a fraction $\eta$ of incident radiant flux reaches the receiver, the ratio of mean receiver to mean entrance irradiance is $\eta C$.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

A proposed optical bench test uses a steady illuminated diffuse patch, a relay made from two optical lens elements, interchangeable centred circular stops, and a calibrated optical power meter large enough to collect the whole output bundle. Start with small cone angles so an ordinary relay approximates the area-angle tradeoff. Measure the patch's image diameter and the accepted angular extent at both conjugate planes; the transverse magnification gives the spatial scale, but does not by itself measure etendue.

Map the local radiance with a baffled photodiode on translation and rotation stages, using two small aligned holes to select a narrow known solid angle. Keep the sampled position fixed while scanning direction, calibrate the sampling footprint and angular response, and repeat at several positions. Numerically integrate $n^2\cos \theta dAd\Omega$ over the occupied samples, and integrate $(L/n^2)dG$ to predict the collected radiant flux. This checks the filled-cone assumption instead of inferring it from two outer dimensions.

Compare the relay without a stop and with a stop of $0.6$ times the illuminated diameter in the same plane. The uniform-bundle prediction is a radiant flux ratio of $0.36$, with unchanged local radiance in the surviving samples. Cover the input for a dark reading, monitor source drift, repeat the open-stop reading, check detector linearity with calibrated optical attenuators, and enlarge the collecting detector or scan it to check for missed light rays. Lens losses, angular response, illumination gradients, alignment errors and diffraction can spoil the ideal prediction; their effects must be measured separately. This is a procedure to try, with predicted readouts, not purported experimental data.

An optical axis is the reference line through the centres of rotational symmetry of an optical system. A folded system can have successive axis segments related by reflection.

The incident ray, reflected ray and surface normal lie in one plane; incidence and reflection angles measured from the normal are equal. For a unit incident direction $\mathbf{d}$ and unit normal $\mathbf{n}$, reflection gives $\mathbf{d}-2(\mathbf{d}\cdot \mathbf{n})\mathbf{n}$.

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.

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].

An optical focus is a point where a specified bundle of light rays converges, or where its backward extensions meet. For a converging thin lens receiving light rays parallel to the optical axis, the rear optical focus is $(f,0)$ when the thin lens is at $x=0$; its front optical focus is $(-f,0)$. An object's image is generally elsewhere: the optical focus describes the parallel incoming bundle, while the thin lens equation locates the image for a finite object distance.

The focal length $f$ specifies the signed distance from an ideal thin lens to its optical focus for incident light rays parallel to the optical axis. A converging thin lens has $f>0$, with a rear optical focus beyond the thin lens; a diverging thin lens has $f<0$, with a virtual optical focus on the incoming side. The focal length is a property of the optical lens in its specified surroundings, not the image distance for every object. For example, a converging thin lens with $f=1\mathrm{m}$ images an object $3\mathrm{m}$ away at $1.5\mathrm{m}$, as computed below. ^[ref].

The optical power is reciprocal signed focal length, $\Phi =1/f$. With $f$ expressed in metres, optical power is expressed in dioptres: $f=1\mathrm{m}$ gives $\Phi =+1\mathrm{D}$, while $f=-0.5\mathrm{m}$ gives $\Phi =-2\mathrm{D}$. The sign distinguishes converging and diverging optical lens elements; the magnitude measures how strongly they change the direction of a light ray at a given height. ^[ref].

For two ideal thin lens elements in contact in air, optical powers add: ${\Phi }_{\mathrm{total}}={\Phi }_1+{\Phi }_2$. Thus $+2\mathrm{D}$ and $-1\mathrm{D}$ combine to give $+1\mathrm{D}$, or $f_{\mathrm{total}}=1\mathrm{m}$. Separation changes the rule: for two ideal thin lens elements separated by $d$ in air, their equivalent optical power is ${\Phi }_1+{\Phi }_2-d{\Phi }_1{\Phi }_2$. The contact rule is the $d=0$ case.^[ref]

A dioptre, also spelled diopter, is the unit of optical power equal to one reciprocal metre. In $\Phi =1/f$, convert focal length to metres before taking the reciprocal: $f=25\mathrm{cm}=0.25\mathrm{m}$ gives $\Phi =4\mathrm{D}$, not $0.04\mathrm{D}$. A dioptre measures focusing strength, distinct from the transverse magnification of an image.

A thick lens retains the separation of its refracting surfaces when predicting image formation. In the paraxial approximation with air on both sides, object and image distances measured from the respective principal planes satisfy the familiar imaging equation. Measuring instead from one mechanical centre generally introduces an error.^[ref]

The two principal planes are reference planes that represent a centred imaging system in the paraxial approximation. In air on both sides, measure object distance from the front principal plane, image distance from the rear principal plane, and focal length from each principal plane to its corresponding optical focus. They need not coincide with a surface or mechanical centre. Their signed separation matters in the Bessel lens displacement method.^[ref]

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].

Use positive distances for a real incident object and a real outgoing image. The first thin lens makes an image at $v_1=f_1{d}_1/(d_1-f_1)$. The second has object distance $L-v_1$, possibly negative, and the thin lens equation gives the displayed expression. At $L=0$ it becomes $d_2=F{d}_1/(d_1-F)$ with $F^{-1}=f_1^{-1}+f_2^{-1}$. For $d_1>0$, a real final image requires both $F>0$ and $d_1>F$; the inequality alone is insufficient when $F<0$.

For two thin converging lenses, the first lens may form its would-be image beyond the second lens. The rays entering lens 2 then converge, so its Cartesian real-object distance is negative. Apply $1/f=1/u+1/v$ separately at each lens rather than treating that intermediate focus as an object upstream. In the example $u_1=2f$ with separation $d<f$, $u_2=-(2f-d)$ and $v_2=f(2f-d)/(3f-d)$. Multiplying the two signed transverse magnifications determines final inversion; a positive outgoing image distance means a real image.

For an optical element, a virtual object is a point toward which incident rays already converge, so they would meet beyond the element if it were removed. It is distinct from a virtual image, which is an apparent origin of outgoing rays. In the real-is-positive thin-lens convention, a virtual object beyond the lens has $u<0$, and $1/u+1/v=1/f$. Thus $u\to -\infty$ gives a real image in the rear focal plane of a converging lens.

For a converging thin lens forming a real image of a real object, let $u,v$ be positive distances from the lens and $f$ the focal length. Set $u^{\prime }=u-f$ and $v^{\prime }=v-f$, measuring from the neighboring focal points. Substituting into $1/u+1/v=1/f$ gives $(f+u^{\prime })(f+v^{\prime })=f(2f+u^{\prime }+v^{\prime })$, so $u^{\prime }{v}^{\prime }=f^2$. Other object-image geometries require an explicitly consistent signed-distance convention.

The Bessel lens displacement method measures a converging thin lens's focal length by moving only the optical lens between a fixed object and screen. If their separation is $L>4f$, two settings give sharp real images: one enlarged, the other reduced. Read their separation $d$ on an optical bench, then calculateThe two settings exchange object and image distances, so their signed transverse magnifications are reciprocals. This makes the experiment a visible test of conjugate optical planes as well as a focal length measurement.^[ref]

For an original stipulated example, put the object at $x=0$, the screen at $x=L=45\mathrm{cm}$, and take an ideal thin lens with $f=10\mathrm{cm}$. Its two settings are $x_{-}=15\mathrm{cm}$ and $x_{+}=30\mathrm{cm}$, with $d=15\mathrm{cm}$. The transverse magnifications are $m_{-}=-2$ and $m_{+}=-1/2$: both real images are inverted. An object $0.50\mathrm{cm}$ tall would give signed image heights $-1.00\mathrm{cm}$ and $-0.25\mathrm{cm}$. Conversely, stipulated readings $L=45\mathrm{cm}$ and $d=15\mathrm{cm}$ give $f=(45^2-15^2)/(4\times 45)=10\mathrm{cm}$. These are model calculations, not measured data or a manufacturer's calibration.

A fixed offset between a carrier's reference mark and an ideal thin lens plane cancels when subtracting the two carrier readings. That useful cancellation does not make a thick lens equivalent to one unknown optical centre: its two principal planes can be separated. The Bessel lens displacement experiment below makes this limitation explicit.

Place a converging thin lens at coordinate $x$ between the object at $0$ and the screen at $L$. In air and the paraxial approximation, the required object and image distances are $u=x$ and $v=L-x$. The thin lens equation gives a quadratic equation:Its discriminant is $L^2-4Lf$. For $L>4f>0$, the two roots and their separation areBoth roots lie inside $(0,L)$, and both object and image distances exceed $f$. The two roots therefore describe finite real images on the screen. Exchanging the roots exchanges $u,v$; from transverse magnification $m=-v/u$,The image-height ratio between the enlarged and reduced settings is $|m_{-}|/|m_{+}|=|m_{-}{|}^2$, rather than $|m_{-}|$.

At $L=4f$, the roots coincide: $d=0$, $x=L/2=2f$, $u=v=2f$ and $m=-1$. There is one setting, not two separately measurable settings. For $0<L<4f$ the discriminant is negative, so no position between the fixed object and screen gives a sharp real image there. This does not forbid a real image elsewhere if the screen is moved. Another short proof observes that $x(L-x)$ has maximum $L^2/4$ at $x=L/2$; the focusing requirement $x(L-x)=Lf$ is possible exactly when $L\ge 4f$.

KIT's teaching apparatus uses a glass spoke-wheel or test negative as the object, a halogen lamp with an illuminating optical lens, an imaging optical lens, a screen and a triangular rail with a position scale. The illuminating optical lens concentrates the lamp filament's image on the imaging optical lens's centre; translucent paper checks that illumination. This reduces the contribution of peripheral light rays and their spherical aberration. The instructions require fixed object-screen separation exceeding $4f$, at least ten displacement measurements, and separate red- and blue-filter measurements to test chromatic aberration.^[ref]

The following measurement protocol elaborates that apparatus rather than reporting additional KIT results. Align the target centre, optical lens and screen normal along one optical axis; measure $L$ between the target pattern plane and receiving screen face. Lock the target and screen. Slide the imaging optical lens to obtain the enlarged real image, then the reduced real image, reading the same carrier reference mark at each setting. Record $r_{-},r_{+}$ and $d=r_{+}-r_{-}$ for each pair. Record corresponding target and image intervals to test the reciprocal transverse magnifications independently; merely observing different image sizes is a weaker test.

Use the same central target detail to judge best optical focus. Bracket each setting by moving through both visibly blurred sides, repeat from both travel directions, and record the accepted position range as well as repeat scatter. Rail divisions set a readout limit but do not alone specify focusing uncertainty. Check scale calibration, carrier play and drift; a common additive carrier offset cancels in $d$, whereas scale-factor error remains. Keep illumination, aperture, target, screen and optical lens orientation fixed within a series. Change the aperture and colour filter in separate control series: spherical aberration and chromatic aberration can shift the apparent best optical focus, while a very small aperture introduces stronger diffraction and makes focus discrimination harder. Record these settings alongside the inferred focal length, rather than interpreting all disagreement as random measurement error.

For a rigid thick lens in air, let $h=x_{H^{\prime }}-x_H$ be the signed separation of its rear and front principal planes along the direction of propagation. Then the positive object and image distances measured from the principal planes obey $u+v=L-h$. Translating the whole assembly without reversing it moves both principal planes equally, so the measured carrier displacement still equals the difference of the two solutions for $u$. The same algebra therefore givesHere $f$ is the effective focal length measured from the principal planes. This deduction uses the principal-plane imaging equation, not the uncorrected thin lens formula.^[ref] Mechanical centre thickness is not generally $h$. Record thickness and optical lens orientation, but use a justified bound or model for $h$ if it matters at the claimed precision. The two readings $L,d$ alone cannot determine both $f$ and unknown $h$; cancelling a carrier offset does not cancel this limitation.

For small measurement errors, write $f=L/4-d^2/(4L)$. Its partial derivatives and first-order propagation of uncertainty areHere the $\sigma$ values are standard deviations used to describe uncertainty, and $\mathrm{Cov}(L,d)$ is the covariance. Independent carrier-reading errors give ${\sigma }_d^2={\sigma }_{r_{-}}^2+{\sigma }_{r_{+}}^2$; correlated readings require the term $-2\mathrm{Cov}(r_{-},r_{+})$. Averaging repeated independent pairs reduces the random component of the uncertainty of the mean, but does not remove calibration or model errors.

As a stipulated uncertainty calculation, let $L=45\mathrm{cm}$, $d=15\mathrm{cm}$, ${\sigma }_L=0.10\mathrm{cm}$ and ${\sigma }_{r_{-}}={\sigma }_{r_{+}}=0.05\mathrm{cm}$, with all three input errors independent. Then ${\sigma }_d=0.0707\mathrm{cm}$, $A=5/18$, $B=-1/6$ and ${\sigma }_f\simeq 0.0302\mathrm{cm}$. The result would be $f=10.000\mathrm{cm}$ with this standard uncertainty in the stipulated thin lens model; these numbers are not a measured uncertainty budget. A small neglected principal-plane separation gives an additional first-order bias $f_{\mathrm{thin}}-f\simeq Ah$, distinct from that random budget.

The two kinds of sensitivity must be distinguished. At fixed focal length,These derivatives diverge as $L$ approaches $4f$ from above: a small change in object-screen spacing strongly changes the separation of the two settings. Yet the inferred focal length's displacement sensitivity $|B|=d/(2L)$ tends to zero. That does not make coalescing settings easy to measure. For fixed $L$, define the axial mismatch between the predicted real image and the screen as $g(x)=fx/(x-f)-(L-x)$, for $x>f$. At a focused setting,Near $u=v=2f$, moving the optical lens changes this mismatch only weakly, making visual discrimination of the two settings difficult. At the example's settings $g^{\prime }(15\mathrm{cm})=-3$ and $g^{\prime }(30\mathrm{cm})=3/4$, so equal carrier motion produces different axial defocus responses. This is an axial mismatch calculation, not a full prediction of perceived image sharpness.

Choose $L$ sufficiently above $4f$ to separate the two accepted focusing ranges, while retaining both real images on the available screen and rail. There is no universal best $L$: focusing uncertainty itself depends on aperture, transverse magnification, optical aberrations and the target. Close to $d=0$, the nonnegative measured separation and the quadratic term in $d$ also make first-order propagation of uncertainty unreliable if its error is comparable to its value. Repeatability, controls and the principal plane correction determine whether the elegant two-distance formula is an adequate measurement model.

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.