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 –

$\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}$

$\frac{1}{v_{o}}=\frac{1}{f_{o}}+\frac{1}{u_{o}}$

$=\frac{1}{0.8}-\frac{1}{0.9}$

$=\frac{1}{7.2}$

v_o = 7.2 cm

The separation between two lenses is determined in the following manner –

= v₀ + |u_e| = 7.2 + 2.27

= 9.47 cm

The magnifying power of the microscope is –

Magnifying power, M = M_o × M_e

$M = \frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}}) $

$M = \frac{7.2}{0.9}(1+\frac{25}{25})$

= 8 × 11 = 88