#### Question

The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.

#### Solution

For the least distance of clear vision, the total magnification is given by:

`m=-L/`

where, *L* is the separation between the eyepiece and the objective

*f*_{o} is the focal length of the objective

*f*_{e} is the focal length of the eyepiece

*D* is the least distance for clear vision

Also, the given magnification for the eyepiece:

`m_e=5=(1+D/f_e)`

`=>5=1+20/f_e`

⇒f_{e}= 5 cm

Substituting the value of *m* and *m _{e}* in equation (1), we get:

m=m_{o}.m_{e}

`=>m_o=m/m_e=20/5=4`

Now, we have:

`m_0=L/(|f_0|)`

`=>f_0=14/4=3.5 "cm"`