 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 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]