Question

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

Answer 1

Focal length of the objective lens, $f_o=1.25cm$

Focal length of the eyepiece, $f_e=5cm$

Least distance of distinct vision, $d=25cm$

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-

$m_e=(1+d/f_e)=6$

The angular magnification of the objective lens ($m_o$) is calculated as-

$m_o{m}_e=m\Rightarrow m_o=5$

Now, $m_o=-v_o/u_o\Rightarrow v_o=-5u_o$

Using lens formula,

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

Using above two equations, $u_o=-1.5cmand{v}_o=7.5cm$

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

Now, Image distance for the eyepiece is $v_e=-d=-25cm$

Using lens formula,

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$u_e=-4.17cm$

Separation between the objective lens and the eyepiece, $|u_e|+|v_o|=11.67cm$

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