Q 12-09-188JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
In a compound microscope, the magnified virtual image is formed at a distance of $25\ \text{cm}$ from the eye-piece. The focal length of its objective lens is $1\ \text{cm}$. If the magnification is $100$ and the tube length of the microscope is $20\ \text{cm}$, then the focal length of the eye-piece lens (in cm) is ______.
Numerical value type. Enter your answer.
Answer: 6.25
For the final image at the near point: $M = \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right)$.
$$100 = \frac{20}{1}\left(1 + \frac{25}{f_e}\right) \Rightarrow \frac{25}{f_e} = 4 \Rightarrow f_e = 6.25\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics