= Solution
At the $d$ line, the <focal lengths> obey $1/f_{id}=(n_{id}-1)\rho_i$. Thus
$$
\frac{f_{2d}}{f_{1d}}=\frac{(n_{1d}-1)\rho_1}{(n_{2d}-1)\rho_2}
=-\frac{(n_{1d}-1)(n_{2F}-n_{2C})}{(n_{2d}-1)(n_{1F}-n_{1C})}.
$$
Rearranging gives
$$
\boxed{\frac{f_{2d}}{f_{1d}}=-\frac{(n_{2F}-n_{2C})/(n_{2d}-1)}{(n_{1F}-n_{1C})/(n_{1d}-1)}=-\frac{V_{1d}}{V_{2d}}.}
$$
The last equality uses the <Abbe number> definition. The minus sign signifies a converging and a diverging component, with the normal <optical glass> convention and nonzero finite <focal lengths>.
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