Q 12-09-005NEETNEET 2025Top questionEasy
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 $25$ cm, the magnification in the microscope is
Answer: (B) $125$
For a compound microscope with the final image at infinity (normal adjustment):
$$m = \frac{L}{f_o}\cdot\frac{D}{f_e} = \frac{40}{2} \times \frac{25}{4} = 20 \times 6.25 = 125$$
Solution by Sreeraj P, M.Sc Physics