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(C) Given: Focal length of objective $f_o = 2.0 \, cm$,focal length of eyepiece $f_e = 6.25 \, cm$,distance between lenses $L = 15 \, cm$,and least di

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$20$

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(C) Given: Focal length of objective $f_o = 2.0 \, cm$,focal length of eyepiece $f_e = 6.25 \, cm$,distance between lenses $L = 15 \, cm$,and least distance of distinct vision $D = 25 \, cm$.For the eyepiece,the final image is at $v_e = -25 \, cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{6.25} = \frac{-1 - 4}{25} = \frac{-5}{25} = -\frac{1}{5} \implies u_e = -5 \, cm$.The distance of the image formed by the objective from the eyepiece is $v_o = L - |u_e| = 15 - 5 = 10 \, cm$.Using the lens formula for the objective $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{10} - \frac{1}{2} = \frac{1 - 5}{10} = -\frac{4}{10} \implies u_o = -2.5 \, cm$.The magnifying power $M$ is given by $M = \frac{v_o}{|u_o|} \left( 1 + \frac{D}{f_e} ight)$:$M = \frac{10}{2.5} \left( 1 + \frac{25}{6.25} ight) = 4 	imes (1 + 4) = 4 	imes 5 = 20$.

$A$ compound microscope consists of an objective lens of focal length $2.0 \, cm$ and an eyepiece of focal length $6.25 \, cm$ separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at the least distance of distinct vision $(25 \, cm)$?

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