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Question

A telescope has an objective of focal length \( 100 \, \text{cm} \) and an eyepiece of focal length \(
5 \, \text{cm} \). What is its magnifying power?

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Explanation

**Microscope principle** objective of short focal length forms real image between F_e and 2F_e of eyepiece, eyepiece magnifies to virtual image at D, total magnification product, for f_o=2 cm, f_e=5 cm, L=20 cm, m_o≈10, M_e≈6, M≈60, illustrating high magnification from two stages. Magnifying power: m = (f_o/f_e) . f_o = 100 cm , f_e = 5 cm . m = (100/5) = 20 . Substituting values gives 20, 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 (mirror), confirming sign conventions and refraction principles.

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