A person with a normal near point (25 cm)

Solution:

Focal length of the objective lens, fo = 8 mm = 0.8 cm

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objective lens, uo = −9.0 mm = −0.9 cm

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece, ve = −d = −25 cm

Object distance for the eyepiece $=u_{\mathrm{e}}$

Using the lens formula, we can obtain the value of $u_{\mathrm{e}}$ as:

\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}

$\frac{1}{u_{\mathrm{e}}}=\frac{1}{v_{\mathrm{e}}}-\frac{1}{f_{\mathrm{e}}}$

$=\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25}$

$\therefore u_{\mathrm{e}}=-\frac{25}{11}=-2.27 \mathrm{~cm}$

We can also obtain the value of the image distance for the objective lens $\left(v_{o}\right)$ using the lens formula.

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

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

$=\frac{1}{0.8}-\frac{1}{0.9}=\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}$

$\therefore v_{\mathrm{o}}=7.2 \mathrm{~cm}$

The distance between the objective lens and the eyepiece $=\left|u_{\mathrm{c}}\right|+v_{\mathrm{o}}$

$=2.27+7.2$

$=9.47 \mathrm{~cm}$

The magnifying power of the microscope is calculated as:

$\frac{v_{\mathrm{o}}}{\left|u_{\mathrm{o}}\right|}\left(1+\frac{d}{f_{\mathrm{e}}}\right)$

$=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=8(1+10)=88$

Hence, the magnifying power of the microscope is 88.