Question

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

Answer 1

Step 1: Draw diagram of compound microscope.


Step 2: Find the object distance for the eyepiece.

Focal length of the objective lens,
$f_o=8\text{mm}=0.8\text{cm}$
Focal length of the eyepiece, $f_e=2.5\text{cm}$
Object distance for the objective lens,
$u_o=-9.0\text{mm}=-0.9\text{cm}$
Least distance of distant vision,
$D=25\text{cm}$
Image distance for the eyepiece,
$v_e=-D=-25\text{cm}$

Using lens formula for the eyepiece,

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{-25}-\frac{1}{2.5}$
$\frac{1}{u_e}=\frac{-11}{25}\text{cm}$
$u_e=-2.27\text{cm}$


Step 3: Find the image distance for the objective lens.

Using lens formula for the eyepiece,
$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$
$\frac{1}{v_o}=\frac{1}{u_o}+\frac{1}{f_o}$
$\frac{1}{v_o}=\frac{1}{-0.9}+\frac{1}{0.8}$
$\frac{1}{v_o}=\frac{-0.8+0.9}{0.72}=\frac{0.1}{0.72}$
$v_o=7.2\text{cm}$


Step 4: Find the distance between the objective lens and the eyepiece.

Distance between objective lens and eye piece is given by
$d=v_o+|u_e|$
$d=7.2+2.27$
$d=9.47\text{cm}$


Step 5: Find the Magnification power of microscope.

Magnification of microscope at the least distance of distinct vision is given by,
$m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$
$m=\frac{7.2}{0.9}(1+\frac{25}{2.5})$
$m=8\times 11$
$m=88$.