Practice question
Question
A compound microscope has an objective of focal length \( 1.5 \, \text{cm} \) and tube length \( 18 \,
\text{cm} \). If the eyepiece focal length is \( 6 \, \text{cm} \), what is the total magnification at
infinity?
Explanation
**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Objective magnification: m_o = (L/f_o) = (18/1.5) = 12 . Eyepiece magnification: m_e = (D/f_e) = (25/6) ≈ 4.17 . Total magnification: m = m_o × m_e = 12 × 4.17 ≈ 50 . Substituting values gives 50, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u
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