A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of $10\ \text{cm}$. The separation between the two lenses is $2\ \text{cm}$. The focal lengths of the component lenses
Answer: (A) $18\ \text{cm},\ 20\ \text{cm}$
For two separated lenses to have minimum spherical aberration, the separation equals the difference of their focal lengths: $d = f_1 - f_2 = 2\ \text{cm}$. All four options satisfy this, so use the combined focal length:
$$\frac1F = \frac1{f_1} + \frac1{f_2} - \frac{d}{f_1f_2} = \frac{f_1 + f_2 - d}{f_1f_2}$$
For $f_1 = 20$, $f_2 = 18$:
$$\frac1F = \frac{20 + 18 - 2}{360} = \frac{36}{360} = \frac{1}{10} \quad\Rightarrow\quad F = 10\ \text{cm}$$
(The other pairs give $F = 6$, $7$ and $9\ \text{cm}$.)
Solution by Sreeraj P, M.Sc Physics