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An achromatic lens combines optical lens elements with different optical dispersion so that two specified wavelengths have the same focal length in the paraxial approximation. This brings their parallel incoming light rays to a common optical focus, helping an optical camera form a sharper image under mixed-colour illumination. A common implementation is an achromatic doublet: a positive crown glass element paired with a negative flint glass element. ^[ref].

The achromatic doublet does not require either material to have zero optical dispersion. Instead, the positive element gains positive optical power toward blue, while the negative element becomes more negative. Suitable Abbe numbers and signed optical powers let these changes cancel even though their reference optical powers do not cancel. For example, the achromatic doublet power balance calculation below combines $+10\mathrm{D}$ with $-5\mathrm{D}$ to obtain $+5\mathrm{D}$, or focal length $0.2\mathrm{m}$, with zero primary F–C optical power difference. The smaller Abbe number of the negative element makes its weaker reference optical power sufficient to oppose the positive element's colour change.

An achromatic lens is corrected at two chosen wavelengths, not necessarily at every wavelength between them. The remaining focal length variation is the secondary spectrum. Correcting axial chromatic aberration also does not automatically remove lateral chromatic aberration throughout an imaging system, spherical aberration, or diffraction. Actual designs therefore need ray tracing with the intended wavelengths, element thicknesses, separations and apertures. ^[ref]; ^[ref].

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.

Coma makes an off-axis optical point source appear asymmetric, often with a compact head and a flaring tail like a comet. Different parts of the optical pupil send its light rays to different transverse image positions. A sharp central star image therefore does not guarantee sharp images across an optical telescope's field of view. In an idealized, aligned system with rotational symmetry, coma vanishes on the optical axis; misalignment can introduce coma even at the nominal field centre.^[ref]

Coma differs from spherical aberration, whose dependence on optical pupil radius can blur an on-axis point without selecting a transverse direction. Astigmatism gives different optical foci in two perpendicular sections of an off-axis bundle; a through-focus sequence can change its elongated image's orientation. Diffraction persists when these optical aberrations vanish. A real asymmetric point spread function may mix several effects, so its appearance alone is not a unique diagnosis of coma.

The classical primary description belongs to the nineteenth-century Seidel aberration classification of five primary monochromatic image errors.^[ref] It remains useful in modern design: OpticStudio's Seidel Coefficients by ANSYS, Inc. documentation lists coma as $W_{131}$ and tabulates each surface's contribution as well as the system sum (Discussion). Cancellation of contributions can reduce coma without making every surface individually free of it. The same documentation limits this calculation to specified systems with axially symmetric surfaces; higher-order terms and broken rotational symmetry require more general ray tracing.

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 institution-issued AY 105 Lab Experiment 3: Optical aberrations by California Institute of Technology (Spring 2011 offering, Part A, pp. 2–3, 7–8) specifies a concrete optical bench: a lamp, opal diffuser and condensing optical lens illuminate a $25$ or $30\mu \mathrm{m}$ optical pinhole; a photographic telephoto optical lens acts as a collimator, producing an approximately $75\mathrm{mm}$ beam. A test optical lens sits on a rotation stage, followed by an adjustable optical microscope. The procedure establishes the on-axis image, then rotates the test optical lens a few degrees and examines the off-axis image through optical focus. It compares this readout with through-focus ray tracing at the corresponding field angles and offers a Ronchi ruling to calibrate the measuring eyepiece.

Part B (pp. 9–10) substitutes a spherical mirror of $1\mathrm{m}$ radius of curvature, with the optical microscope beside the incoming beam so the image is necessarily off-axis. It calls for measuring the field angle, calibrating the eyepiece and recording the image's length and width at minimum size. These are specified apparatus dimensions and instructions, not supplied measurement results or dimensions of the normalized figures.

A controlled extension would record images at zero and equal positive and negative field angles, repeat zero afterwards, and scan through optical focus at every setting. Keep the illuminated optical pupil and illumination fixed, recenter the optical microscope without clipping the bundle, and use its calibrated scale to separate a mere image displacement from a change in blur shape. The pure Seidel coma pupil mapping predicts reversal and linear growth; a stationary asymmetric central image instead calls for checking alignment. A focus-dependent change between perpendicular elongations suggests an astigmatism contribution. These comparisons test the model rather than assuming every off-axis flare is pure coma.

The optical pinhole has finite angular size, approximately its diameter divided by the collimator's focal length; the collimator and observing optical microscope also contribute their own optical aberrations. A broadband lamp can mix chromatic aberration with the geometric blur, and diffraction limits the smallest visible structure. A narrow spectral band, repeated field settings and a reference image help separate these effects. Consequently a measured length and width are useful checks on off-axis image formation, but they cannot by themselves determine a unique coma coefficient or recover the full point spread function.

Let the rapidly varying phase be $\Theta$, so $\mathbf{k}=\nabla \Theta$ and $\omega =-{\Theta }_t$. The dispersion law is the Hamilton-Jacobi equation ${\Theta }_t+\Omega (\nabla \Theta ,\mathbf{x},t)=0$. Differentiating it and following a path with group velocity gives the ray tracing equationsThe final relation is obtained by the chain rule along that path; its two spatial terms cancel.

The medium is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $k_2=K$ are conserved. For deep water far from shore, ${\kappa }_{\infty }={\omega }_{\infty }^2/g$. Choose the incident branch with $k_1<0$ and $K=({\omega }_{\infty }^2/g)\sin {\theta }_{\infty }>0$; choosing the opposite alongshore direction changes its sign. Since the dispersion relation depends on direction only through $\kappa$, the group-velocity component ratio is $k_2/k_1$. Hence the required implicit ray equation isNear shore, $\kappa h\to 0$ and $\kappa \sim {\omega }_{\infty }/\sqrt{g\alpha }{x}^{-p/2}$. Thus $y^{\prime }\sim -K\sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$, whose integral givesThe rays approach the shoreline normally. A crest is a level set of $\int k_{1}dx+Ky-\omega t$. Its slope is $dy/dx=-k_1/K\sim {\omega }_{\infty }{x}^{-p/2}/(K\sqrt{g\alpha })$, so crests become parallel to the shoreline. Their separation in the shore-normal direction is $2\pi /|k_1|\sim 2\pi \sqrt{g\alpha }{x}^{p/2}/{\omega }_{\infty }$ and decreases to zero. These are formal ray asymptotics; for $p<2$, the slow-variation assumption eventually fails arbitrarily close to the shoreline. The sketch shows a representative shallow-water case $p=1$; it illustrates these limiting directions rather than assuming that value in the derivation.

For a two-dimensional internal gravity wave with fixed horizontal wave number $k\ne 0$, the vertical velocity amplitude satisfiesThis follows from the Linearized Boussinesq equations just as in part (a), allowing the background buoyancy frequency to vary. The WKB approximation gives locallyIt requires slowly varying stratification compared with the local vertical wavelength, in particular $|m_z|\ll m^2$, and a small-amplitude wave in the inviscid, nonrotating regime. Propagation requires $N>|\omega |$; where $N<|\omega |$, the local vertical wave number is imaginary and the solution is evanescent. Turning levels and abrupt changes need a separate connection calculation. Finite viscosity, background flow, rotation or nonlinear effects require modifications of this model.

The ray tracing equations use the local dispersion relation as a Hamiltonian:The stationary, horizontally uniform background conserves the frequency and horizontal wave number. In two dimensions the group velocity gives $dx/dz=-m/k$. For the specified buoyancy frequency and $\omega =1$, $m^2=256k^2{z}^2$. Hence for $z>0$ the two geometric internal-wave rays through the specified point areEach curve has both directed propagation senses, giving four local directed rays. Their formal intersections with $z=0$ are $x=-8$ and $x=8$.

At $z=0$, $N=\omega$ and $m=0$: this is a double-zero internal-wave turning level, since $m^2$ touches zero without changing sign. The ray becomes vertical geometrically, but its group velocity tends to zero. Taking $k>0$ without loss of generality, the downward branch hasThe formal travel time diverges logarithmically as $z\to 0$. Meanwhile $|m_z|/m^2=1/(16|k|z^2)$ grows, so the WKB approximation fails before that limit; its predicted $|m{|}^{-1/2}$ amplitude cannot be extrapolated to infinity. The spatial ray reaches a finite limiting position, but finite-wavelength behaviour near the zero must be found beyond WKB. There is no evanescent half-space supplied by this particular $N(z)$, so ordinary simple-turning-point reflection cannot be assumed solely from the local ray construction.

Use the geometrical optics conventionBecause ${\partial }_t=\epsilon {\partial }_T$ and ${\partial }_x=\epsilon {\partial }_X$, differentiating $e^{i\theta /\epsilon }$ gives an order-one physical angular frequency and order-one physical wavenumbers. Absorb a constant leading amplitude phase into $\theta$ and take $A_0$ real; for a complex amplitude the corresponding quadratic density is $|A_0{|}^2$.

Substitution into the wave equation gives the eikonal equation for a variable-speed wave at leading order:Choose the positive-frequency branch $\omega =\Omega =cK$, $K=\sqrt{k^2+{\ell }^2}$; the negative branch can be treated in exactly the same way. The imaginary first-order terms giveMultiplying by $A_0$ puts this wave-action transport for a variable-speed wave in conservation form:Its density is $\omega A_0^2$ and its flux is that density times the group velocity. Independently, equality of mixed wave phase derivatives givesThis phase compatibility in geometrical optics is essential when converting the differentiated dispersion relation into equations along rays.

For completeness let $\omega =\Omega (\mathbf{k};\mathbf{X},T)$ be a general local dispersion relation. Define ray tracing by $d\mathbf{X}/dT={\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$. Using the chain rule in ${\nabla }_{\mathbf{X}}\omega$ and the symmetry ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ givesLikewise ${\omega }_T={\Omega }_T+{\mathbf{c}}_g\cdot {\mathbf{k}}_T$ cancels its advective term because ${\mathbf{k}}_T=-\nabla \omega$. Hence the Hamiltonian ray equations for a local dispersion relation areThe derivatives on the right hold the wavevector fixed. For this wave equation they reduce to

When $c=c(Y)$, the ray invariants are $k$ and $\omega$, while $d\ell /dT=-K{c}_Y<0$. For a monochromatic component with these fixed invariants write $\theta =kX-\omega T+S(Y)$, soUsing the stated independence of $A_0$ from $X,T$, the wave-action transport for a variable-speed wave becomes $(c^2\ell A_0^2{)}_Y=0$. ThereforeThis is the stationary wave amplitude in a stratified wavespeed; $\mu$ is the constant for that component or ray family.

At the proposed simple turning point of a variable-speed wave, take $c_s=c(Y_s)=\omega /k$ and $c_s^{\prime }=c_Y(Y_s)>0$. On the chosen positive branch this entails $k>0$. DefineAs the ray approaches from below, $\ell \sim [q(Y_s-Y){]}^{1/2}$ andThe amplitude diverges and the vertical group velocity $c^2\ell /\omega$ vanishes. The WKB approximation also fails its slow-variation condition $\epsilon |{\ell }_Y|/{\ell }^2\ll 1$ when $Y_s-Y=O({\epsilon }^{2/3})$. The turning-point enhancement of wave amplitude shows that this divergence is a failure of the outer ray approximation, rather than an actual infinite solution.

To derive the inner wavefield directly, separate the conserved horizontal wave phase and frequency in the original differential equation:Dividing by $c^2$ gives ${\epsilon }^2{\Psi }_{YY}+2{\epsilon }^2(c_Y/c){\Psi }_Y+Q(Y)\Psi =0$. Set $Y-Y_s=\delta \eta$. The second derivative is of size ${\epsilon }^2/{\delta }^2$, whereas $Q\sim -q\delta \eta$, so the Airy scaling at a variable-speed wave turning point isThe first-derivative term is smaller by $O(\delta )$, and higher coefficients in the expansion of $Q$ give the same relative order. Thus the general leading inner profile is a linear combination of the two Airy functions $\mathrm{Ai}(\eta )$ and $\mathrm{Bi}(\eta )$. For decaying Airy continuation and ray reflection, if the physical continuation is bounded and decays into $Y>Y_s$, its growing $\mathrm{Bi}$ coefficient is zero. In that usual evanescent continuation the wavefield has the formWithout a condition in the evanescent region the corresponding $\mathrm{Bi}$ term remains admissible locally; the given existence of a wavefield below the turning point alone does not select its coefficient.

Finally, the outer amplitude at distance $\delta$ is $O({\delta }^{-1/4})=O({\epsilon }^{-1/6})$ for $\mu =O(1)$ and fixed nonzero $q$. The oscillatory Airy turning-point connection formula has exactly the same $(-\eta {)}^{-1/4}$ envelope on the allowed side. Therefore the inner wavefield has magnitude $O({\epsilon }^{-1/6})$. The decaying Airy continuation matches an incident and reflected pair of rays, so a single upward $\ell >0$ branch must be supplemented by its reflected branch near the turning point; no formal coefficient matching is needed to identify this magnitude.

For $u=8\mathrm{cm}$ and negative focal length $f=-8\mathrm{cm}$,The transverse magnification is $M=-v/u=+1/2$, soThus the virtual image is upright, half the object's height, and $4.0\mathrm{cm}$ from the lens on the object's side. A parallel incident light ray diverges after the lens as though it originated at the near optical focus; the central light ray continues undeviated in the ideal thin lens model. Their backward extensions meet at the image tip, $(-4,1)\mathrm{cm}$.

Take the thin lens at $x=0$, with the object on the left and light rays travelling to the right. The positive object distance is $u=12\mathrm{cm}$ and the focal length is $f=8\mathrm{cm}$. The thin lens equation givesThe transverse magnification and signed image height areThus the real image is $24\mathrm{cm}$ beyond the lens, inverted and $10\mathrm{cm}$ high. For the ray tracing, a light ray parallel to the optical axis emerges through the far optical focus; a light ray through the optical centre continues straight in the ideal thin lens model. They intersect at $(24,-10)\mathrm{cm}$, independently reading off both answers from the diagram.

The calculation and diagram use the ideal paraxial approximation implicit in the thin lens equation. The comparatively large object height does not supply enough information to calculate optical aberrations of a real lens.

A real image lies where the outgoing light rays from each object point actually intersect in geometrical optics. A screen or detector placed at that image plane receives the focused pattern. The screen reveals the real image; it does not create the intersection. A viewer can also see a real image without a screen if their eye receives the diverging light rays beyond the intersection. ^[ref].

For a converging thin lens and real object with $u>f>0$, the real image has $v>0$ and signed transverse magnification $m<0$. Inversion is a property of this particular setup, not the definition of a real image. With $f=10\mathrm{cm}$ and $h_o=0.50\mathrm{cm}$, the following are predictions:

To test both image position and transverse magnification, use an optical bench with an illuminated arrow or millimetre target, a converging optical lens of nominal focal length $10\mathrm{cm}$, a movable white screen and a ruler. Measure corresponding target and image intervals, rather than guessing the whole image size when it exceeds the screen. PASCO's Experiment 12 compares distance-based and size-based magnification; the following procedure uses several object distances, as in the Northern Virginia laboratory. ^[ref]; ^[ref].

Keep the optical lens, illumination and central target detail fixed between trials. Approach best focus from both sides and record the interval over which sharpness is difficult to distinguish; repeat focusing to estimate its reproducibility. A modest central aperture stop tests whether peripheral light rays change the result, but also reduces brightness. Spherical aberration, chromatic aberration and diffraction can broaden the optimum; a thick optical lens also makes the chosen reference plane imperfect. Treat nominal $f$ as an initial prediction, and report position and interval-reading limits instead of forcing exact agreement. ^[ref].

Moving the object toward the front optical focus from outside makes the real image larger and farther away. A failed screen search near $u=f$ can therefore mean that the real image is beyond the bench, rather than that a virtual image has formed. At exactly $u=f$, the outgoing bundle from each point is parallel and no finite screen position focuses it.

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].