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(B) When lenses are in contact,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Since the object is at infini

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$20 \ cm, -30 \ cm$

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(B) When lenses are in contact,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Since the object is at infinity,the image is formed at the focus $F = 60 \ cm$. Thus,$\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60}$ ... $(i)$.When the lenses are separated by $d = 10 \ cm$,the equivalent focal length $F'$ is given by $\frac{1}{F'} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$. The image shifts by $30 \ cm$ towards the combination,so the new image distance is $60 \ cm - 30 \ cm = 30 \ cm$. Thus,$\frac{1}{30} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{10}{f_1 f_2}$ ... $(ii)$.Substituting $(i)$ into $(ii)$: $\frac{1}{30} = \frac{1}{60} - \frac{10}{f_1 f_2} \implies \frac{10}{f_1 f_2} = \frac{1}{60} - \frac{1}{30} = -\frac{1}{60} \implies f_1 f_2 = -600$.From $(i)$,$\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{60} \implies f_1 + f_2 = \frac{-600}{60} = -10$.Solving $x^2 - (f_1+f_2)x + f_1 f_2 = 0 \implies x^2 + 10x - 600 = 0 \implies (x+30)(x-20) = 0$. Thus,the focal lengths are $20 \ cm$ and $-30 \ cm$.

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