NEET (UG) 2025 — Question 2

For a compound microscope in normal adjustment, the magnification is given by $M = \frac{L}{f_o} \cdot \frac{D}{f_e}$, where $L$ is the tube length, $f_o$ is the focal length of the objective, $f_e$ is the focal length of the eyepiece, and $D$ is the least distance of distinct vision. Substitute the given values: $L = 40\,\text{cm}$, $f_o = 2\,\text{cm}$, $f_e = 4\,\text{cm}$, $D = 25\,\text{cm}$. Calculate the magnification: $M = \frac{40}{2} \times \frac{25}{4} = 20 \times 6.25 = 125$. Thus, the magnification of the microscope is 125, which corresponds to option B.