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(D) For a simple microscope,the magnification is given by $m = 1 + \frac{D}{f_0}$.Given $m = 6$ and $D = 25 \, cm$,we have $6 = 1 + \frac{25}{f_0}$,wh

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$25$

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(D) For a simple microscope,the magnification is given by $m = 1 + \frac{D}{f_0}$.Given $m = 6$ and $D = 25 \, cm$,we have $6 = 1 + \frac{25}{f_0}$,which implies $5 = \frac{25}{f_0}$,so $f_0 = 5 \, cm$.For a compound microscope,the total magnification $M$ for an image at infinity is given by $M = \frac{L \cdot D}{f_0 \cdot f_e}$,where $L$ is the tube length.Given $M = 2 	imes 6 = 12$,$L = 0.6 \, m = 60 \, cm$,$D = 25 \, cm$,and $f_0 = 5 \, cm$,we substitute these values into the formula:$12 = \frac{60 	imes 25}{5 	imes f_e}$.$12 = \frac{1500}{5 	imes f_e} = \frac{300}{f_e}$.$f_e = \frac{300}{12} = 25 \, cm$.

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