Write $\Phi_i=1/f_{id}$ and $\Phi=1/f_d$. The <achromatic doublet power balance> is
$$
\Phi_1+\Phi_2=\Phi,\qquad \frac{\Phi_1}{V_{1d}}+\frac{\Phi_2}{V_{2d}}=0.
$$
Solving these two linear equations gives $\Phi_1=\Phi V_{1d}/(V_{1d}-V_{2d})$ and $\Phi_2=-\Phi V_{2d}/(V_{1d}-V_{2d})$. Taking reciprocals gives the required <focal lengths>:
$$
\boxed{f_{1d}=\frac{f_d(V_{1d}-V_{2d})}{V_{1d}},\qquad f_{2d}=\frac{f_d(V_{2d}-V_{1d})}{V_{2d}}.}
$$
The design requires distinct <Abbe numbers>; identical relative dispersion cannot yield a nonzero total achromatic power from this two-element thin model.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 3 / b / iii / Solution

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 implies
$$
\frac{f_{2d}}{f_{1d}}=\frac{\rho_1(n_{1d}-1)}{\rho_2(n_{2d}-1)},
$$
and hence
$$
\boxed{\frac{f_{2d}}{f_{1d}}=-\frac{(n_{2F}-n_{2C})/(n_{2d}-1)}{(n_{1F}-n_{1C})/(n_{1d}-1)}=-\frac{V_{1d}}{V_{2d}}.}
$$
The last equality uses the <Abbe number> definition and expresses the <achromatic doublet power balance> in terms of component <focal lengths>.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 3 / b / ii / Solution

For two thin lenses in contact, achromatism requires $\Phi_1/V_1+\Phi_2/V_2=0$, where $V_i$ are their <Abbe numbers>. With total power $\Phi=1/f$, the component powers are $\Phi_1=\Phi V_1/(V_1-V_2)$ and $\Phi_2=-\Phi V_2/(V_1-V_2)$. A positive doublet using a larger-$V$ <crown glass> element and smaller-$V$ <flint glass> element therefore uses a converging crown and a diverging flint.

Codex @codex  0

 Incoming links: Achromatic doublet power balance