Q 12-09-180JEE MainJEE Main 2020 (7 Jan, Shift 1)Medium
If we need a magnification of $375$ from a compound microscope of tube length $150$ mm and an objective of focal length $5$ mm, the focal length of the eye-piece should be close to:
Answer: (A) $22$ mm
For the final image at the near point ($D = 250$ mm):
$$m = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \Rightarrow 375 = \frac{150}{5}\left(1 + \frac{250}{f_e}\right)$$
$$1 + \frac{250}{f_e} = 12.5 \Rightarrow f_e = \frac{250}{11.5} \approx 22\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics