$A$ 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 $15\;cm$. How far from the objective should an object be placed in order to obtain the final image at
$(a)$ the least distance of distinct vision $(25\;cm)$,and
$(b)$ at infinity? What is the magnifying power of the microscope in each case?

(N/A) Given:
Focal length of the objective lens,$f_{1} = 2.0 \;cm$
Focal length of the eyepiece,$f_{2} = 6.25 \;cm$
Distance between the lenses,$d = 15 \;cm$
$(a)$ For the final image at the least distance of distinct vision $(d' = 25 \;cm)$:
For the eyepiece,$v_{2} = -25 \;cm$. Using the lens formula $\frac{1}{v_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$:
$\frac{1}{u_{2}} = \frac{1}{v_{2}} - \frac{1}{f_{2}} = \frac{1}{-25} - \frac{1}{6.25} = \frac{-1-4}{25} = -\frac{5}{25} = -\frac{1}{5} \implies u_{2} = -5 \;cm$.
The image distance for the objective is $v_{1} = d - |u_{2}| = 15 - 5 = 10 \;cm$.
Using the lens formula for the objective: $\frac{1}{v_{1}} - \frac{1}{u_{1}} = \frac{1}{f_{1}}$:
$\frac{1}{u_{1}} = \frac{1}{v_{1}} - \frac{1}{f_{1}} = \frac{1}{10} - \frac{1}{2} = \frac{1-5}{10} = -\frac{4}{10} \implies u_{1} = -2.5 \;cm$.
The object should be placed $2.5 \;cm$ from the objective.
Magnifying power $m = \frac{v_{1}}{|u_{1}|} (1 + \frac{d'}{f_{2}}) = \frac{10}{2.5} (1 + \frac{25}{6.25}) = 4(1+4) = 20$.
$(b)$ For the final image at infinity:
For the eyepiece,$v_{2} = \infty$,so $u_{2} = -f_{2} = -6.25 \;cm$.
The image distance for the objective is $v_{1} = d - |u_{2}| = 15 - 6.25 = 8.75 \;cm$.
Using the lens formula for the objective: $\frac{1}{u_{1}} = \frac{1}{v_{1}} - \frac{1}{f_{1}} = \frac{1}{8.75} - \frac{1}{2} = \frac{2 - 8.75}{17.5} = -\frac{6.75}{17.5} \implies u_{1} \approx -2.59 \;cm$.
The object should be placed $2.59 \;cm$ from the objective.
Magnifying power $m = \frac{v_{1}}{|u_{1}|} (\frac{d'}{|u_{2}|}) = \frac{8.75}{2.59} (\frac{25}{6.25}) \approx 3.378 \times 4 \approx 13.51$.