= Achromatic doublet power balance
<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 is
$$
V_i=\frac{n_{i,d}-1}{n_{i,F}-n_{i,C}}.
$$
The 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. https://media.schott.com/api/public/content/b37dbd8fa7e64662b2d0ae523ae56238?download=true&v=97f67105[SCHOTT, Optical Glass 2025, §1.1 and §10: Wavelength table]{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> difference
$$
\Delta\Phi_i=\Phi_{i,F}-\Phi_{i,C}=K_i(n_{i,F}-n_{i,C})=\frac{\Phi_{i,d}}{V_i}.
$$
The negative element has negative $\Delta\Phi_i$: its blue <optical power> is more negative, not less negative. The contact rule for <optical power> therefore gives
$$
\Phi_d=\Phi_{c,d}+\Phi_{f,d}=\frac{1}{f_d},\qquad \Delta\Phi_{\rm total}=\frac{\Phi_{c,d}}{V_c}+\frac{\Phi_{f,d}}{V_f}=0.
$$
This 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. https://phys.libretexts.org/Bookshelves/Optics/Geometric_Optics_(Tatum)/02:_Lens_and_Mirror_Calculations/2.10:_Designing_an_Achromatic_Doublet[Jeremy Tatum, University of Victoria, Geometric Optics §2.10, equations 2.10.8–2.10.11, authored text hosted by LibreTexts]{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 $\Phi_{f,d}=-5\,\mathrm D$. Then
$$
\Phi_d=(10-5)\,\mathrm D=5\,\mathrm D,\qquad f_d=\frac{1}{5\,\mathrm{m^{-1}}}=0.2\,\mathrm m,\qquad \Delta\Phi_{\rm total}=\left(\frac{10}{60}-\frac{5}{30}\right)\mathrm D=0.
$$
The 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 become
$$
\Phi_c(q)=10\left(1+\frac{q}{60}\right)\mathrm D,\qquad \Phi_f(q)=-5\left(1+\frac{q}{30}\right)\mathrm D,\qquad \Phi_{\rm total}(q)=\left(5+q\left[\frac{10}{60}-\frac{5}{30}\right]\right)\mathrm D=5\,\mathrm D.
$$
At 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.
\Image[/achromatic-doublet-power-cancellation.png]
{title=<Achromatic doublet power balance>: opposite colour changes preserve $5\,\mathrm D$ in the first-order toy model}
{description=The left panel shows each signed <optical power> and their sum against synthetic colour coordinate $q$. The right panel magnifies their changes from the reference <optical powers>: equal and opposite slopes cancel. The <crown glass> and <flint glass> labels denote toy elements with $V_c=60$ and $V_f=30$, not measured <optical glass>. Only this linear toy <optical dispersion> makes the total exactly flat throughout the interval; a real <achromatic lens> can retain a <secondary spectrum>.}
For specified nonzero reference <optical power> $\Phi_d$ and distinct <Abbe numbers>, solving the two balance equations gives
$$
\Phi_{c,d}=\frac{\Phi_d V_c}{V_c-V_f},\qquad \Phi_{f,d}=-\frac{\Phi_d V_f}{V_c-V_f}.
$$
Thus $\Phi_d>0$ 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.<MIT 2.71 and 2.710 Optics, Spring 2014 by Nicholas X. Fang/Lecture Notes on Geometrical Optics>[§J: Composite lenses]{ref}
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