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Question

A compound microscope has an objective of focal length \( 1.25 \, \text{cm} \) and eyepiece of focal
length \( 5 \, \text{cm} \) with a tube length of \( 15 \, \text{cm} \). What is the magnification at
infinity?

Options

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Explanation

**Two lenses in contact** behave as single lens with power sum, magnification product m = m₁×m₂. For f₁=20 cm (P₁=5 D), f₂=-20 cm (P₂=-5 D) in contact, P=0, F infinite, afocal system, beam emerges parallel, principle of corrective lenses for myopia/hypermetropia. Objective magnification: m_o = (L/f_o) = (15/1.25) = 12 . Eyepiece magnification: m_e = (D/f_e) = (25/5) = 5 . Total magnification: m = m_o × m_e = 12 × 5 = 60 . Substituting values gives 60, 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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