An angular magnification (magnifying power) of 30X is desired using an objective of focal length $1.25\text{cm}$ and an eyepiece of focal length $5\text{cm}$. How will you set up the compound microscope?

Step 1: Draw ray diagram for compound microscope.


As we desired 30 times angular magnification i.e. 30 times enlarged image so we have to set the distance between the objective lens and eyepiece, which is the sum of image distance for objective lens and object distance for eyepiece.

Step 2: Find magnification produced by eyepiece.
Given,
focal length of the eyepiece, $f_e=5\text{cm}$
Least distance of distinct vision,
$D=25\text{cm}$
When the final image is formed at D, the angular magnification produced by the eyepiece of a compound microscope is
$m_e=(1+\frac{D}{f_e})$
$m_e=(1+\frac{25}{5})=6$

Step 3: Find object distance from eyepiece to get desired magnification.
For eye piece, $v_e=-D=-25$.
Using the lens formula for eyepiece, we get,
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\frac{1}{5}=\frac{1}{-25}-\frac{1}{u_e}$
$\frac{1}{u_e}=-\frac{1}{5}-\frac{1}{25}=-\frac{6}{25}$
$u_e=-\frac{25}{6}=-4.17\text{cm}$

Step 4: Find relation between image and object distance for objective lens.
Magnifications by objective and eye piece are related as
Total magnification,
$m=m_o{m}_e$
$m_o=\frac{m}{m_e}=\frac{30}{6}=5$
$-\frac{v_o}{u_o}=5$
$u_o=-\frac{v_o}{5}$

Step 5: Find image distance from objective to get desired magnification.
Using the lens formula for objective, we get,
$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$\frac{1}{1.25}=\frac{1}{v_o}-\frac{1}{(-\frac{v_o}{5})}$
$\frac{1}{1.25}=-\frac{6}{v_o}$
$v_o=-1.25\times 6$$=-7.5\text{cm}$
Hence the image will be formed at $7.5\text{cm}$ from the objective piece.
Separation between the objective lens and eyepiece,
$=|v_o|+|u_e|$
$=7.5+4.17=11.67\text{cm}$
Therefore, the separation between the objective lens and the eyepiece should be $11.67\text{cm}.$