Question

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

• A. 6
• B. 150
• C. 25
• D. 125

Answer 1

Answer: (B)

150

Answer 2

By compound microscope
$m=\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)$
(where length of tube $L=30 cm$, focal length of objective lens $f_{0}=1 cm$, focal length of
eye-piece $f_{e} =6\, cm,\, D=25\, cm)$
$=\frac{30}{1}\left(1+\frac{25}{6}\right)=30 \times \frac{(6+25)}{6}$
$=5 \times 31$
$=155 cm \cong 150$