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(A) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For the first planoconve

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$-2$

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(A) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.For the first planoconvex lens: $\frac{1}{f_1} = (1.5 - 1)(\frac{1}{15} - \frac{1}{\infty}) = 0.5 	imes \frac{1}{15} = \frac{1}{30} \ cm^{-1}$.For the second planoconvex lens: $\frac{1}{f_2} = (1.2 - 1)(\frac{1}{\infty} - (-\frac{1}{12})) = 0.2 	imes \frac{1}{12} = \frac{1}{60} \ cm^{-1}$.The net focal length of the combination is $\frac{1}{f_{net}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} \ cm^{-1}$,so $f_{net} = 20 \ cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,with $u = -30 \ cm$ and $f = 20 \ cm$:$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.Thus,$v = 60 \ cm$.The magnification is $m = \frac{v}{u} = \frac{60}{-30} = -2$.

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