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(A) For the eyepiece:Given $v_e = -25 \ cm$ (final image at least distance of distinct vision),$f_e = 6.25 \ cm$.Using the lens formula $\frac{1}{v_e}

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$-2.5$

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(A) For the eyepiece:Given $v_e = -25 \ cm$ (final image at least distance of distinct vision),$f_e = 6.25 \ cm$.Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25} \implies -\frac{1}{u_e} = \frac{1}{6.25} + \frac{1}{25} = \frac{4+1}{25} = \frac{5}{25} = \frac{1}{5}$.So,$u_e = -5 \ cm$.The distance between the lenses is $L = 15 \ cm$,so the image distance for the objective lens is $v_o = L - |u_e| = 15 - 5 = 10 \ cm$.For the objective lens:$v_o = 10 \ cm$,$f_o = 2 \ cm$.Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{10} - \frac{1}{u_o} = \frac{1}{2} \implies -\frac{1}{u_o} = \frac{1}{2} - \frac{1}{10} = \frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}$.Therefore,$u_o = -2.5 \ cm$.The object should be placed $2.5 \ cm$ in front of the objective lens.

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