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(B) For two thin lenses in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Given $F = 60 \, cm$,so

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$20 \, cm, -30 \, cm$

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(B) For two thin lenses in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Given $F = 60 \, cm$,so $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60} \dots (i)$.When separated by distance $d = 10 \, cm$,the equivalent focal length $F'$ is given by $\frac{1}{F'} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$. Given $F' = 30 \, cm$,so $\frac{1}{30} = \frac{1}{60} - \frac{10}{f_1 f_2}$.Rearranging gives $\frac{10}{f_1 f_2} = \frac{1}{60} - \frac{1}{30} = -\frac{1}{60}$,so $f_1 f_2 = -600 \dots (ii)$.From $(i)$,$\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{60}$,so $f_1 + f_2 = \frac{-600}{60} = -10 \dots (iii)$.Using $(f_1 - f_2)^2 = (f_1 + f_2)^2 - 4f_1 f_2 = (-10)^2 - 4(-600) = 100 + 2400 = 2500$,we get $f_1 - f_2 = 50 \dots (iv)$.Solving $(iii)$ and $(iv)$,we get $2f_1 = 40 \implies f_1 = 20 \, cm$ and $f_2 = -30 \, cm$.

Two lenses of focal lengths $f_1$ and $f_2$ are placed in contact,and the image is formed at a distance of $60 \, cm$. When the lenses are separated by a distance of $10 \, cm$,the image is formed at a distance of $30 \, cm$. What are the values of $f_1$ and $f_2$?

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