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Question

A converging beam meets a concave lens (\( f = 20 \, \text{cm} \)) \( 5 \, \text{cm} \) before the
convergence point. What is the new image distance?

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Explanation

**Spherical refracting surface** power P = (n₂-n₁)/R, lens power sum of two surfaces. Lens maker derivation combines two refractions, sign of R₂ negative for second surface convex opposite direction, yielding 1/f positive for convex lens. Object distance: u = -5 cm (virtual object), f = -20 cm . Lens formula: (1/v) - (1/-5) = (1/-20) ⇒ (1/v) + (1/5) = (1/-20) . (1/v) = (1/-20) - (1/5) = (-1 - 4/20) = (-5/20) = (-1/4) . v = -4 cm (4 cm to the left). Substituting values gives 4 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u

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