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  4. A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :

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Q. A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of $10 \,cm$. The separation between the two lenses is $2 \,cm$. The focal lengths of the component lenses are :

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Solution:

For a convergent doublet of separated lens, we have
$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}$
where $d$ is separation between two lens, $f_{1}$ and $f_{2}$ are focal lengths of component lenses,
$f$ is resultant focal length. Therefore, Eq. $(1)$ becomes
$\frac{1}{10}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{2}{f_{1} f_{2}} \Rightarrow \frac{1}{10}=\left(\frac{f_{2}+f_{1}-2}{f_{1} f_{2}}\right)$
$ \Rightarrow f f_{2}=10 f_{2}+10 f_{1}-20 $
$\Rightarrow 10 f_{1}+10 f_{2}-f f_{2}=+20$
For $f=18 \,cm$ and $f_{2}=20 \,cm$, the above equation satisfies.