# A person with a normal near point (25 cm)

Question:

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

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, = 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.