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.
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?
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.