Compound Microscope

A compound microscope magnifies in two stages: the objective forms a real, enlarged intermediate image, and the eyepiece acts as a simple magnifier on it. The total magnifying power is the product $M=m_o\times m_e\approx \frac{L}{f_o}\times \frac{D}{f_e}$, where $L$ is the tube length, $f_o$ the objective focal length, $f_e$ the eyepiece focal length, and $D$ the near-point distance. Substituting $L=20$ cm, $f_o=2$ cm, $f_e=5$ cm and $D=25$ cm gives $M=\frac{20}{2}\times \frac{25}{5}=10\times 5=50$. The two factors describe physically distinct stages: $\frac{L}{f_o}$ is the linear magnification with which the objective casts a real, inverted image inside the tube, while $\frac{D}{f_e}$ is the angular magnification with which the eyepiece, acting as a simple microscope, views that intermediate image. Because the stages act in series their effects multiply rather than add, and a very short objective focal length is favoured since it both shortens the working distance and sharply raises the first factor. The value 100 is wrong because it doubles the objective contribution by misreading the tube length. The value 25 is wrong as it uses only the eyepiece factor $D/f_e$ times unity. The value 40 is wrong since it mistakenly takes $D=20$ cm. This NCERT product rule shows why a short objective focal length dominates resolution and magnification. A consistency check confirms both stages exceed unity, and their product realistically lands in the tens for laboratory microscopes.