= Solution
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 $\Phi_F=\Phi_C$, hence
$$
\rho_1(n_{1F}-n_{1C})+\rho_2(n_{2F}-n_{2C})=0,
\qquad
\boxed{\frac{\rho_1}{\rho_2}=-\frac{n_{2F}-n_{2C}}{n_{1F}-n_{1C}}.}
$$
For 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>.
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