Question

A microscope is having objective of focal length $1cm$ and eyepiece of focal length $6cm$. If tube length is $30cm$ and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope? Take $D=25cm$.

• A. $6$
• B. $150$
• C. $25$
• D. $125$

Answer 1

The correct option is D $125$

Magnification of objective is given by:

$m_o=-\frac{L}{f_o}$

where:

$L:$ tube length of microscope

$f_o:$ focal length of objective

Magnification of eyepiece is given by:

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

where:

$D:$ Distance of least vision

$f_e:$ focal length of eyepiece

Overall magnification is:

$m=m_o{m}_e=-\frac{LD}{f_o{f}_e}$

$|m|=\frac{25\times 30}{1\times 6}=125$