# An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm.

Question:

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

Solution:

Focal length of the objective lens, $f_{\mathrm{o}}=1.25 \mathrm{~cm}$

Focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

$m_{\mathrm{e}}=\left(1+\frac{d}{f_{\mathrm{e}}}\right)$

$=\left(1+\frac{25}{5}\right)=6$

The angular magnification of the objective lens (mo) is related to me as:

$m_{\mathrm{o}} m_{\mathrm{e}}=m$

$m_{\mathrm{o}}=\frac{m}{m_{\mathrm{e}}}$

$=\frac{30}{6}=5$

We also have the relation:

$m_{\mathrm{o}}=\frac{\text { Image distance for the objective lens }\left(v_{\mathrm{o}}\right)}{\text { Object distance for the objective lens }\left(-u_{\mathrm{o}}\right)}$

$5=\frac{v_{\mathrm{o}}}{-u_{\mathrm{o}}}$

$\therefore v_{\mathrm{o}}=-5 u_{\mathrm{o}}$    ...(1)

Applying the lens formula for the objective lens:

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

$\frac{1}{1.25}=\frac{1}{-5 u_{\mathrm{o}}}-\frac{1}{u_{\mathrm{o}}}=\frac{-6}{5 u_{\mathrm{o}}}$

$\therefore u_{\mathrm{o}}=\frac{-6}{5} \times 1.25=-1.5 \mathrm{~cm}$

And $v_{\mathrm{o}}=-5 u_{\mathrm{o}}$

$=-5 \times(-1.5)=7.5 \mathrm{~cm}$

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

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

Where,

$v_{e}=$ Image distance for the eyepiece $=-d=-25 \mathrm{~cm}$

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

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

$=\frac{-1}{25}-\frac{1}{5}=-\frac{6}{25}$

$\therefore u_{\mathrm{e}}=-4.17 \mathrm{~cm}$

Separation between the objective lens and the eyepiece $=\left|u_{\mathrm{e}}\right|+\left|v_{\mathrm{o}}\right|$

$=4.17+7.5$

$=11.67 \mathrm{~cm}$

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.