Question

The total magnification produced by a compound microscope is $20$. The magnification produced by eyepiece alone in $5$. The microscope is focussed on certain object. The distance between object and eyepiece is observed to be $14cm$. If the least distance of distinct vision is $20cm$.

Answer 1

Overall magnification

$m=m_o{m}_e=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)=20$

$\frac{D}{f_e}=5$

hence

$\frac{L}{f_0}=4$

But D=20;

hence focal length of eye piece $f_e=4cm$

Distance between objective and eyepiece$=f_0+L+f_e=f_0+4f_0+f_e=14cm$

Hence focal length of objective, $f_0=2cm$

Here it is assumed that the image of objective lens is formed near to focal point of eyepiece.