Q.Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.

[!TLDR] Compound microscope magnifying power M = m_o x M_e = (v_o/u_o) x (D/u_e); overall length L = v_o + u_e. [!ANSW …

A compound microscope uses two convex lenses: an OBJECTIVE (focal length f_o) facing the tiny object, and an EYE LENS (focal length f_e) the observer looks through. The object is placed between f_o and 2f_o of the objective, forming a real, inverted, magnified INTERMEDIATE image at distance v_o from the objective, positioned so this intermediate image falls just WITHIN the focus of the eye lens; the eye lens then acts on this intermediate image exactly as a SIMPLE microscope, producing a final virtual, further-magnified image.

The intermediate image's linear (lateral) magnification, produced by the objective alone, is m_o = v_o/u_o. The eye lens then contributes its OWN angular magnifying power, exactly as derived for a simple microscope, M_e = D/u_e, where u_e is the intermediate image's distance from the eye lens (this treats the intermediate image as the 'object' for the eye lens's own simple-microscope action). Since the final visual angle beta at the eye (looking through the eye lens at the final image) works out, via the two successive magnifications, to the PRODUCT of these two individual effects, the overall magnifying power of the compound microscope is M = m_o x M_e = (v_o/u_o) x (D/u_e). …

Derive the compound microscope's magnifying power as the product of the objective's linear magnification (m_o=v_o/u_o) and the eye lens's own simple-microscope angular magnification (M_e=D/u_e), a …

Adding the two magnifications (m_o + M_e) instead of correctly MULTIPLYING them, or forgetting that the eye lens's contribution M_e uses u_e (the intermed …