Q.22) A microscope has an objective of focal length 2 cm , eyepiece of focal length 4 cm and the tube length of $40 cm$. If the distance of distinct vision of eye is $25 \mathrm{cm}$, the magnification in the microscope is
A) 250
B) 100
C) 125
D) 150
Solution:
Correct Option: (3) $125$
The total magnification $M$ of a compound microscope when the final image is at the least distance of distinct vision ( $\mathrm{D}=25 \mathrm{~cm}$ ) is:
$$
M=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)
$$
Given:
- Objective focal length $f_o=2 \mathrm{~cm}$
- Eyepiece focal length $f_e=4 \mathrm{~cm}$
- Tube length $L=40 \mathrm{~cm}$
- Distance of distinct vision $D=25 \mathrm{~cm}$
$$
M=\left(\frac{40}{2}\right) \cdot\left(\frac{25}{4}\right)=20 \cdot 6.25=125
$$
Hence, the answer is option (3) 125.