Practice question
Question
A compound microscope has an objective of focal length \( 1.5 \, \text{cm} \) and eyepiece of focal
length \( 5 \, \text{cm} \) with a tube length of \( 20 \, \text{cm} \). What is the 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) = (20/1.5) ≈ 13.33 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 13.33 × 5 ≈ 66.65 ≈ 67 . Substituting values gives 67, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v
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