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Question

A compound microscope has an objective of focal length \( 2 \, \text{cm} \) and eyepiece of focal
length \( 5 \, \text{cm} \) with a tube length of \( 18 \, \text{cm} \). What is the magnification at
infinity?

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Explanation

**Simple microscope** magnification M = 1 + D/f when image at D=25 cm near point, M = D/f when image at infinity (relaxed eye), f focal length (cm), D least distance of distinct vision 25 cm. For f=4 cm, image at 25 cm, M=1+25/4=7.25, angular magnification ratio of angle subtended by image to that by object at D. Objective magnification: m_o = (L/f_o) = (18/2) = 9 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 9 × 5 = 45 . Substituting values gives 45, which matches expected image position and magnification from mirror/lens

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