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Question

A converging beam meets a convex lens (\( f = 20 \, \text{cm} \)) \( 8 \, \text{cm} \) 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 = -8 cm (virtual object). Focal length: f = 20 cm . Lens formula: (1/v) - (1/-8) = (1/20) ⇒ (1/v) + (1/8) = (1/20) . (1/v) = (1/20) - (1/8) = (2 - 5/40) = (-3/40) . v = -(40/3) ≈ -13.33 cm (13.33 cm to the left). Substituting values gives 13.3 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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