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

Answer: (B)

By compound microscope
$m=\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)$
(where length of tube $L=30cm$, focal length of objective lens $f_0=1cm$, focal length of
eye-piece $f_e=6cm,D=25cm)$
$=\frac{30}{1}\left(1+\frac{25}{6}\right)=30\times \frac{(6+25)}{6}$
$=5\times 31$
$=155cm\cong 150$