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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]

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.

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

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

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

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

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

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

For thin contacting components, optical powers add. At wavelength $\lambda$ the total optical power is $\Phi (\lambda )={\rho }_1[n_1(\lambda )-1]+{\rho }_2[n_2(\lambda )-1]$. An achromatic lens has ${\Phi }_F={\Phi }_C$, henceFor normally dispersive optical glasses, both index differences are positive, so the two components have opposite curvature factors and opposite optical powers. This uses the thin Lensmaker's equation; finite thickness and separated principal planes require a more detailed design. If the denominator vanishes, use the undivided power-balance equation rather than this ratio. Two-wavelength correction alone does not eliminate the residual chromatic aberration at every intermediate wavelength.

At the intermediate $d$ reference wavelength, the lensmaker's equation gives $1/f_{id}={\rho }_i(n_{id}-1)$. Therefore the curvature-factor ratio from the previous part impliesand henceThe last equality uses the Abbe number definition and expresses the achromatic doublet power balance in terms of component focal lengths.

For thin lenses in contact, optical powers add. At the two reference wavelengths, the lensmaker's equation gives total powerAn achromatic lens has ${\Phi }_F={\Phi }_C$. Subtracting and solving for the curvature-factor ratio yieldsOpposite curvature-weighted powers compensate the two materials' optical dispersion.

The requested source-image relation is the thin lens equation, using positive real-object distance $u$ and signed image distance $v$:Here $f>0$ for a converging lens, and $v<0$ denotes a virtual image. This is usually distinguished from Lensmaker's equation, $1/f=(n-1)(1/R_1-1/R_2)$ for a thin lens in air with the usual signed surface radii. The latter relates focal length to the lens material and curvature, rather than to object and image positions.^[ref]

For a magnifying glass held close to the eye, compare the angle $h/u$ with the unaided near-point angle $h/D_n$, where $D_n=25\mathrm{cm}$. Maximum usable angular magnification occurs with the virtual image at the near point, so $v=-D_n$. ThusThe object lies inside the front focal length. The emerging light rays diverge, while their backward extensions meet at an upright enlarged virtual image on the object's side. With the image at infinity instead, the relaxed-eye angular magnification would be $D_n/f=6.25$, which is smaller. A finite eye-to-lens distance $s$ changes the near-point result to $1+(D_n-s)/f$; the quoted maximum uses the conventional $s\simeq 0$ limit.^[ref]

In the paraxial approximation, apply refraction at each surface of the thin lens in air. With signed radii $R_1$ and $R_2$, the first surface obeys $n/v_1-1/u=(n-1)/R_1$, and the second obeys $1/v-n/v_1=(1-n)/R_2$. The separation of the surfaces is neglected, so the intermediate image coordinate $v_1$ cancels:This is Lensmaker's equation, with Cartesian coordinate system object coordinate $u<0$ for a real object. For an equiconvex thin lens and light travelling left to right, $R_1=R$ and $R_2=-R$. Hence

Set the red image at coordinate zero, the blue image at $\delta =0.010\mathrm{m}$, and the screen at $D=1.000\mathrm{m}$. Let the correcting thin lens be at $x$, with $\delta <x<D$. Both original images are then real objects for the correcting thin lens, with positive object distances $u_r=x$, $u_b=x-\delta$, and common positive image distance $v=D-x$. The thin lens equation givesThe refractive indices and Lensmaker's equation requireSubstitute the two focal lengths and cancel $D-x$:Thus the correcting thin lens is $100\mathrm{mm}$ beyond the red image and $900\mathrm{mm}$ before the screen. The required focal lengths and surface radius areThe blue focal length is shorter because the blue refractive index is larger; these two differing object distances are mapped to the same screen, removing the specified longitudinal chromatic aberration.

A simple converging optical lens bends different wavelengths through different angles because of optical dispersion. For a fixed shape in air, the lensmaker's equation gives $1/f\propto n(\lambda )-1$: the larger short-wavelength refractive index gives a shorter focal length. Different colours therefore form sharp images at different axial positions, causing longitudinal chromatic aberration, blur and coloured fringes in a white-light image. Off the optical axis, colour-dependent transverse magnification can also give lateral chromatic aberration. An achromatic doublet combines suitable optical glasses and curvatures to reduce this wavelength dependence.

Keep the paper's $R_1,R_2>0$ as magnitudes for the two convex faces. With the standard signed-curvature convention measured to the right, the first surface has signed radius $+R_1$ and the second $-R_2$. This explains why the requested formula contains a sum of the two reciprocal magnitudes.

Let the axial lens thickness be $\ell$, and let $d_1$ be the distance from the first vertex to the intermediate focus that its refracted rays would reach without the second surface. The first refraction at a spherical surface givesAt the second face those rays are still converging toward that intermediate focus, so its object distance is negative, $u_2=-d_2$, with $d_2=d_1-\ell$. Applying the signed spherical-surface equation again givesIn the thin lens limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:The focal length is the image distance for parallel incident light rays, $u\to \infty$, hence the lensmaker's equation isThe same surrounding refractive index $n_1$ is assumed on both sides. For $n_2>n_1$ this biconvex thin lens has positive focal length. With signed radii $R_a=R_1$, $R_b=-R_2$, the equivalent familiar expression is $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.

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