Question

A microscope is having an objective of focal length $1 \, cm$ and eyepiece of focal length $6 \, cm$ . If tube length is $30 \, cm$ and the 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. $155$
• C. $25$
• D. $125$

Answer 1

Answer: (B)

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