Answer:
According to the question,
Focal length of the given objective lens is $ f_{o}$= 8 mm = 0.8 cm
Focal length of the eyepiece is $ f_{e}$= 2.5 cm
Object distance for the given objective lens is $ u_{o}$ = -9.0 mm = -0.9 cm
Least distance of distant vision here is d=25 cm
Image distance for the eyepiece is $ v_{e}$ = -d=-25 cm
Object distance for the eyepiece is denoted by $u_{e}$.
According to the lens formula, we can obtain the value of $u_{e}$ in the following manner –
$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$
$\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}$
$\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}$
$u_{e}=-\frac{25}{11}=-2.27cm$
Using lens formula, the image distance for the objective (v) lens is obtained below –
$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$
$\frac{1}{v_{o}}=\frac{1}{f_{o}}+\frac{1}{u_{o}}$
$=\frac{1}{0.8}-\frac{1}{0.9}$
$=\frac{1}{7.2}$
vo = 7.2 cm
The separation between two lenses is determined in the following manner –
= v0 + |ue| = 7.2 + 2.27
= 9.47 cm
The magnifying power of the microscope is –
Magnifying power, M = Mo × Me
$M = \frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}}) $
$M = \frac{7.2}{0.9}(1+\frac{25}{25})$
= 8 × 11 = 88