The figure shows a compound microscope. For its magnifying power to be large, the focal lengths $f_o$ (objective) and $f_e$ (eyepiece) should satisfy:

**Magnifying power of a compound microscope** (image at infinity): $M = \dfrac{L \cdot D}{f_o \cdot f_e}$, where $L$ is the tube length and $D$ is the least distance of distinct vision. **Read off the requirement.** Both $f_o$ and $f_e$ sit in the *denominator*, so both should be **small** to maximise $M$. Convention places the objective slightly smaller than the eyepiece: $f_o < f_e$. **Why option C (much smaller) is wrong** — strictly, $f_o < f_e$ but not by orders of magnitude; *both* must remain small to keep $M$ high. A very small $f_o$ with a long $f_e$ would shrink the product less than two reasonably small focal lengths. **Compare with the telescope** (next figure): there, $f_o \gg f_e$ because $M_{\text{tel}} = f_o / f_e$ — focal lengths play the opposite role.