An angular magnification (magnifying power) of $30X$ is desired using an objective of focal length $1.25 \, cm$ and an eyepiece of focal length $5 \, cm$. How will you set up the compound microscope?

(N/A) Focal length of the objective lens,$f_{0} = 1.25 \, cm$.
Focal length of the eyepiece,$f_{e} = 5 \, cm$.
Least distance of distinct vision,$d = 25 \, cm$.
Total angular magnification,$m = 30$.
The angular magnification of the eyepiece is $m_{e} = (1 + d/f_{e}) = (1 + 25/5) = 6$.
The angular magnification of the objective lens is $m_{0} = m / m_{e} = 30 / 6 = 5$.
Using $m_{0} = v_{0} / (-u_{0})$,we get $v_{0} = -5u_{0}$.
Applying the lens formula for the objective lens: $1/f_{0} = 1/v_{0} - 1/u_{0}$.
$1/1.25 = 1/(-5u_{0}) - 1/u_{0} = -6 / (5u_{0})$.
$u_{0} = -6/5 \times 1.25 = -1.5 \, cm$.
$v_{0} = -5 \times (-1.5) = 7.5 \, cm$.
Applying the lens formula for the eyepiece: $1/v_{e} - 1/u_{e} = 1/f_{e}$.
With $v_{e} = -25 \, cm$,$1/u_{e} = 1/(-25) - 1/5 = -6/25$.
$u_{e} = -25/6 \approx -4.17 \, cm$.
The separation between the lenses is $|v_{0}| + |u_{e}| = 7.5 + 4.17 = 11.67 \, cm$.