A convergent doublet of separated lenses,corr | vedclass

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Question

(A) For a doublet corrected for spherical aberration,the condition for separation $d$ is $d = f_1 - f_2$. Given $d = 2\,cm$,we have $f_1 - f_2 = 2\,cm

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$18\,cm, 20\,cm$

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(A) For a doublet corrected for spherical aberration,the condition for separation $d$ is $d = f_1 - f_2$. Given $d = 2\,cm$,we have $f_1 - f_2 = 2\,cm$,or $f_1 = f_2 + 2$.The formula for the resultant focal length $F$ of two lenses separated by distance $d$ is $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.Substituting the given values $F = 10\,cm$ and $d = 2\,cm$:$\frac{1}{10} = \frac{f_2 + f_1 - d}{f_1 f_2} = \frac{f_2 + (f_2 + 2) - 2}{f_1 f_2} = \frac{2f_2}{f_1 f_2} = \frac{2}{f_1}$.Thus,$f_1 = 20\,cm$.Using $f_1 - f_2 = 2\,cm$,we get $f_2 = 20 - 2 = 18\,cm$.Therefore,the focal lengths are $18\,cm$ and $20\,cm$.

$A$ convergent doublet of separated lenses,corrected for spherical aberration,has a 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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