Practice question
Question
A convex lens (\( f = 40 \, \text{cm} \)) and a concave lens (\( f = 20 \, \text{cm} \)) are in
contact. What is the effective focal length?
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. f₁ = 40 cm , f₂ = -20 cm . (1/f) = (1/f₁) + (1/f₂) = (1/40) + (1/-20) = (1 - 2/40) = (-1/40) . f = -40 cm (diverging system). Substituting values gives -40 cm, 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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