Ray Optics

JEE Main level, single correct, medium

228 of 295

Question 228 of 295

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm10 \mathrm{~cm}. The separation between the two lenses is 2 cm2 \mathrm{~cm}. The focal lengths of the component lenses

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Answer: (A) 18 cm,20 cm18 \mathrm{~cm}, 20 \mathrm{~cm}

For minimum spherical aberration separation, d=f1−f2=2 cmd=f_1-f_2=2 \mathrm{~cm}
Resultant focal length =F=10 cm=F=10 \mathrm{~cm}
Using 1F=1f1+1f2−df1f2\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} and solving, we get f1,f218 cm\mathrm{f}_1, \mathrm{f}_2 18 \mathrm{~cm} and 20 cm20 \mathrm{~cm} respectively.

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