$A$ person with a normal near point $(25 \;cm)$ using a compound microscope with an objective of focal length $8.0 \;mm$ and an eyepiece of focal length $2.5 \;cm$ can bring an object placed at $9.0 \;mm$ from the objective into sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

(88) Given:
Focal length of the objective lens,$f_{o} = 8.0 \;mm = 0.8 \;cm$
Focal length of the eyepiece,$f_{e} = 2.5 \;cm$
Object distance for the objective lens,$u_{o} = -9.0 \;mm = -0.9 \;cm$
Least distance of distinct vision,$d = 25 \;cm$
$1$. Finding image distance for the objective lens $(v_{o})$:
Using the lens formula $\frac{1}{v_{o}} - \frac{1}{u_{o}} = \frac{1}{f_{o}}$:
$\frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}} = \frac{1}{0.8} - \frac{1}{0.9} = \frac{0.9 - 0.8}{0.72} = \frac{0.1}{0.72} = \frac{1}{7.2}$
$v_{o} = 7.2 \;cm$
$2$. Finding object distance for the eyepiece $(u_{e})$:
The final image is formed at the near point,so $v_{e} = -25 \;cm$.
Using the lens formula $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$:
$\frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} = \frac{1}{-25} - \frac{1}{2.5} = \frac{-1 - 10}{25} = -\frac{11}{25}$
$u_{e} = -\frac{25}{11} \approx -2.27 \;cm$
$3$. Separation between the lenses $(L)$:
$L = v_{o} + |u_{e}| = 7.2 + 2.27 = 9.47 \;cm$
$4$. Magnifying power $(M)$:
$M = \frac{v_{o}}{|u_{o}|} \left(1 + \frac{d}{f_{e}}\right) = \frac{7.2}{0.9} \left(1 + \frac{25}{2.5}\right) = 8 \times (1 + 10) = 88$.