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An achromatic doublet is an achromatic lens with two optical lens elements. In the contact thin lens model, their optical powers add and their primary colour changes cancel according to achromatic doublet power balance. The conventional positive achromatic doublet uses a converging crown glass element and a weaker diverging flint glass element; these names describe a common choice, not a universal ordering of every modern optical glass composition.

Achromatic doublet power balance chooses two signed optical powers to meet two goals: obtain the required focal length and cancel the primary F–C colour difference. Model two thin lens elements in contact in air, using the paraxial approximation. Let $i=c,f$ label the crown glass and flint glass elements, and let ${\Phi }_{i,d}$ denote each element's optical power at the reference d line. The Abbe number for each material isThe subscripts label standard spectral lines: blue F near $486.1\mathrm{nm}$, red C near $656.3\mathrm{nm}$, and reference d near $587.6\mathrm{nm}$. These are three specified wavelengths, not colour bands. A larger Abbe number means less relative optical dispersion between F and C. ^[ref].

For a fixed thin lens shape in air, write the lensmaker's equation as ${\Phi }_i(\lambda )=K_i[n_i(\lambda )-1]$, with signed shape factor $K_i=1/R_{i,a}-1/R_{i,b}$. Subtracting its C value from its F value gives the primary optical power differenceThe negative element has negative $\Delta {\Phi }_i$: its blue optical power is more negative, not less negative. The contact rule for optical power therefore givesThis zero is equality of the total F and C optical powers, hence equality of their focal lengths when the total optical power is nonzero. The relation follows directly from the lensmaker's equation and the Abbe number; it is exact within the fixed-shape contact thin lens model. ^[ref]. Tatum uses a yellow D-line reference; here the Abbe number convention uses the d line consistently.

For a toy design, choose $V_c=60$, $V_f=30$, ${\Phi }_{c,d}=10\mathrm{D}$ and ${\Phi }_{f,d}=-5\mathrm{D}$. ThenThe positive element contributes $+1/6\mathrm{D}$ to the F–C difference and the negative element contributes $-1/6\mathrm{D}$. This retains the same reference focal length as the single $5\mathrm{D}$, $V_d=60$ thin lens in chromatic aberration, while cancelling its primary colour split. The values are illustrative design parameters, not a catalogue optical glass pair.

Use the same synthetic colour coordinate $q=-1/2,0,+1/2$ and toy optical dispersion as in chromatic aberration. The two element optical powers and their sum becomeAt the red endpoint, the element optical powers are $9.916667\mathrm{D}$ and $-4.916667\mathrm{D}$; at the blue endpoint, they are $10.083333\mathrm{D}$ and $-5.083333\mathrm{D}$. Both sums are $5\mathrm{D}$. In this first-order toy model alone, the sum is constant for every $q$ and the focal length remains $0.2\mathrm{m}$ throughout. Real material optical dispersion is not captured by these two straight slopes, so the flat line is not a claim that a real achromatic doublet focuses all colours perfectly.

For specified nonzero reference optical power ${\Phi }_d$ and distinct Abbe numbers, solving the two balance equations givesThus ${\Phi }_d>0$ and $V_c>V_f>0$ require positive crown glass optical power and negative flint glass optical power. If the two finite Abbe numbers were equal, cancellation would force zero total optical power. If they are close, the required element optical powers become large compared with the total: cancelling primary colour can demand much stronger individual elements.

The contact assumption matters. For two thin lens elements separated by distance $s$ in air, the equivalent optical power is ${\Phi }_c+{\Phi }_f-s{\Phi }_c{\Phi }_f$, evaluated at each wavelength; its colour change need not satisfy the contact balance. Finite thickness, departures from the paraxial approximation, and the secondary spectrum also limit the simple prediction.^[ref]

Angular magnification is the ratio of the signed viewing angle with an optical instrument to the signed viewing angle in a stated reference arrangement. Measure both as apparent directions from the eye, positive toward the same side of the optical axis; then an inverted view has negative $M_{\theta }$. If only positive angular sizes are compared, the result is $|M_{\theta }|$ and carries no orientation information. ^[ref].

For the upright virtual image of a converging thin lens, let the eye be at $x=e\ge 0$ and take the unaided reference to be the same small object at a distance $D$ from the eye. Within the small-angle approximation,Specify $D$: keeping the object in its original position means $D=e+u$, whereas the customary magnifier rating uses a near-point reference, often $D=25\mathrm{cm}$. A reference distance is a convention, not a claim that every eye has that near point. ^[ref].

For the virtual image diagram, $m=4$, $v=-30\mathrm{cm}$ and $e=5\mathrm{cm}$. With $D=25\mathrm{cm}$, $M_{\theta }\simeq 20/7$, whereas keeping the object at $D=e+u=12.5\mathrm{cm}$ gives $M_{\theta }\simeq 10/7$. Thus neither viewing comparison gives the transverse magnification four. A person may also be unable to focus on the unaided object at the shorter reference distance.

At $u=f$, each outgoing bundle is parallel with slope $-h_o/f$; its apparent viewing direction has angle ${\theta }_i\simeq h_o/f$. Consequently $M_{\theta }\simeq D/f$ remains finite even though $m$ is undefined at a finite image plane. Equivalently, in the limit $u\to f^{-}$, the factors $m$ and $e-v$ both grow without bound but their ratio tends to $1/f$. For $f=10\mathrm{cm}$ and $D=25\mathrm{cm}$, the relaxed-view prediction is $M_{\theta }\simeq 2.5$. ^[ref].

Axial chromatic aberration, also called longitudinal chromatic aberration, is colour-dependent focus position along the optical axis. For incident light rays parallel to the optical axis, it appears as a changing focal length. The chromatic aberration diagram isolates this effect using an ideal thin lens at each colour.

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.

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.

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 $L\ge 4f$.

An optical camera can focus a white object sharply in one colour while leaving coloured fringes around its edges in another. Chromatic aberration is the dependence of image position or image scale on wavelength, caused by optical dispersion in the optical lens materials. Axial chromatic aberration separates the optical foci along the optical axis; lateral chromatic aberration changes the transverse magnification with wavelength. Moving the receiving plane can favour one colour without bringing all colours into focus. ^[ref]; ^[ref].

For ordinary optical glass in the visible range, the blue refractive index exceeds the red refractive index. For a fixed positive thin lens in air, the lensmaker's equation then givesHere $\lambda$ is wavelength, $n$ is refractive index, $\Phi$ is optical power, $f$ is focal length, and $R_a,R_b$ are the signed surface radius of curvature values in the lensmaker's equation. The blue light rays therefore bend more strongly and meet sooner. This ordering is a statement about ordinary visible optical dispersion, not a universal rule for every material and every spectral range. ^[ref].

For a worked example, take a single positive thin lens with reference optical power ${\Phi }_d=5\mathrm{D}$ and Abbe number $V_d=60$, hence reference focal length $f_d=0.2\mathrm{m}$. Its F–C optical power difference is $\Delta \Phi ={\Phi }_F-{\Phi }_C={\Phi }_d/V_d=5/60\mathrm{D}\approx 0.08333\mathrm{D}$, by the achromatic doublet power balance relation below. This difference alone does not locate the reference optical power within the F–C interval. To draw a transparent example, introduce a synthetic colour coordinate $q$, a dimensionless quantity, with red $q=-1/2$, reference $q=0$, and blue $q=+1/2$, and choose the toy optical dispersion modelThe diagram's red and blue endpoints represent the C and F endpoints of this toy model. Its reference point is placed symmetrically in optical power by construction; $q$ is not a measured wavelength scale, and the model is not measured optical glass data. The endpoint focal lengths areThus the axial chromatic aberration in this example separates the endpoint optical foci by about $3.334\mathrm{mm}$. A single receiving plane cannot coincide with both optical foci.

The useful design idea is to cancel changes of optical power while retaining a positive total optical power. An achromatic lens does this with materials having different Abbe numbers. It greatly reduces chromatic aberration, but correcting two chosen wavelengths still leaves the possibility of a secondary spectrum.

Two conjugate optical planes are object and image planes related by an imaging system: points in one are imaged onto corresponding points in the other in the paraxial approximation. For a converging thin lens forming a real image in air, their distances $u,v>0$ obey $1/u+1/v=1/f$. Exchanging $u$ and $v$ preserves this relation; this symmetry gives the two settings of the Bessel lens displacement method.

A thin lens placed immediately after an aperture can reproduce its free-propagated pattern from distance L on a nearer screen d if $1/f=1/d-1/L$. In the paraxial Fresnel integral, the lens changes the aperture's quadratic propagation phase from $k{x}^2/(2d)$ to $k{x}^2(1/d-1/f)/2=k{x}^2/(2L)$. The cross term $-kxy/d$ then matches free propagation at the old-screen coordinate $Y=Ly/d$. Intensities have the same shape under transverse scale d/L, with an overall brightness factor in a lossless lens of sufficient aperture. For a narrow uniformly illuminated slit, the far-field first-zero estimate on the new screen is $d\lambda /a$.

An electron lens uses an electric field or magnetic field to make trajectories from an object point converge towards an image point. It has no glass for the electrons to pass through. In a paraxial approximation, an effective focal length permits the same first-order image geometry used for a thin lens; finite beam angles, energy spread and misalignment limit this approximation.

The field of view is the region of object directions sampled by an optical instrument in one pointing. State whether it is horizontal, vertical or diagonal, and whether it is a full angle or a half-angle. An object-plane width in millimetres also describes a field of view, but requires a specified object distance. Neither that width nor the angular field of view is the aperture diameter or the diffraction-limited resolution.

For a centred flat rectangular photodetector of width $w$ in the focal plane, with positive effective focal length $f$, infinity focus and ideal rectilinear imaging without optical distortion, the full horizontal field of view isThe model maps an object direction $\theta$ to image height of magnitude $|y|=f|\tan \theta |$; set $|y|=w/2$ and use the inverse tangent. This retains exact right triangle geometry within the stipulated imaging model; a paraxial focal length alone does not guarantee that a real wide-field optical lens obeys it. With detector height $h$, replace $w$ by $h$ for the vertical field of view, and by $\sqrt{w^2+h^2}$ for the diagonal field of view. The small-angle approximation gives ${\Theta }_x\simeq w/f$ in radians. ^[ref].

For an original worked example, a $36\mathrm{mm}$ detector width gives ${\Theta }_x\approx 39.60^{\circ }$ at $f=50\mathrm{mm}$, and $20.41^{\circ }$ at $f=100\mathrm{mm}$. Keeping $D=25\mathrm{mm}$ makes these f/2 and f/4 respectively: the field of view changes because $f$ changes. By contrast, closing $D$ from $25$ to $12.5\mathrm{mm}$ at $f=100\mathrm{mm}$ changes f/4 to f/8 while retaining $20.41^{\circ }$ in this model. The clean separation is useful: entrance pupil area controls collection, focal ratio controls the paraxial cone, and $w/f$ controls angular image coverage. Edge clipping, optical distortion, field curvature and optical aberrations can restrict the practically usable field of view.

A proposed angular readout on an optical bench illuminates a small asymmetric target at the optical focus of a collimator, providing a bundle with a known direction. Rotate the optical camera about the centre of its entrance pupil, reading a graduated rotation stage, and record the target image position on its fixed photodetector. Centre the target to establish zero; turn left and right until its centre reaches each detector edge. The difference between stage readings measures the full horizontal field of view. Repeat with a smaller aperture stop: unchanged edge directions but reduced signal support the distinction between collection and angular coverage. Keep the same active detector width, infinity focus, illumination, integration time and gain; approach each reading from the same direction to check mechanical backlash. A finite target blurs the edge criterion, detector thresholds can imitate lost coverage, and a wide bundle must illuminate the whole entrance pupil at every tested angle. These checks distinguish a geometrical boundary from insufficient signal. The analytical diagram is the prediction for this protocol, not a record of its execution.

An optical microscope whose microscope objective is designed to form a real image at a specified finite distance. The intended image distance and optical aberration corrections belong together; mechanical tube length is a mounting specification, not automatically the image distance measured from a principal plane.^[ref]

Why can a small displacement produce a large image? For a converging thin lens, write $u_o=f_o+ϵ$ with $epsilon>0$. Substitution into the thin lens equation givesThe similar triangles construction supplies the minus sign. The example has $epsilon=2\mathrm{mm}$ and hence $m_o=-5$. Letting $epsilon$ decrease moves the real image farther away; it does not keep the microscope eyepiece in focus at a fixed position. Large transverse magnification therefore requires the correct conjugate geometry, not merely a short focal length.

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 focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

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.

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].

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 $F<0$.

Image formation assigns each object point a corresponding image point: outgoing light rays from that object point either converge there, giving a real image, or diverge as though they originated there, giving a virtual image. The requirement applies separately to each point. Bringing unrelated light rays together makes a bright patch, but does not preserve the object's pattern. This point-by-point correspondence explains how an optical camera records a scene. ^[ref].

Use the same convention as thin lens: place the optical lens at $x=0$, with incident light travelling to the right and a real object at $x=-u$, $u>0$. Write the image coordinate as $x=v$, so $v>0$ for a real image and $v<0$ for a virtual image. Heights $h_o,h_i$ are positive above the optical axis and negative below it. In air, within the paraxial approximation, a thin lens obeysHere $f$ is the signed focal length, positive for a converging optical lens and negative for a diverging optical lens; the expression for $v$ assumes $u\ne f$. Object distance $u$ is positive even though the object's coordinate is negative. These coordinate rules describe transmission through the optical lens; a mirror reverses the outgoing direction, so do not identify its image distance with this same rightward coordinate. ^[ref].

The beauty of geometrical optics is that two convenient light rays can locate a point, while a whole admitted bundle forms its image. For a converging thin lens, an incident light ray parallel to the optical axis exits through the rear optical focus; a light ray through the centre continues undeviated. Real optical lens elements do not focus every admitted light ray exactly: optical aberrations and diffraction limit the point correspondence. The diagrams below are original ideal-model constructions, not measured data. ^[ref].