A convergent doublet of separated lenses, corrected for spherical… | JEE Main 2018 (15 Apr, Shift 2)

Q63JEE Main 2018 · 15 APR · SHIFT 2Medium

Combination of Lenses

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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Answer: (d)

Condition for minimum spherical aberration: $d = f_1 - f_2$ , so the focal lengths differ by 2 cm.
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = \frac{f_1 + f_2 - d}{f_1 f_2}$
Try 18, 20: $\frac{18 + 20 - 2}{360} = \frac{36}{360} = \frac{1}{10}$ , so $F = 10\text{ cm}$ .
So the answer is (d) 18 cm, 20 cm.

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