Practice question
Question
A converging beam meets a concave lens (\( f = 15 \, \text{cm} \)) \( 6 \, \text{cm} \) before the
convergence point. What is the new image distance?
Explanation
**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Object distance: u = -6 cm (virtual object), f = -15 cm . Lens formula: (1/v) - (1/-6) = (1/-15) ⇒ (1/v) + (1/6) = (1/-15) . (1/v) = (1/-15) - (1/6) = (-2 - 5/30) = (-7/30) . v = -(30/7) ≈ -4.29 cm (4.29 cm to the left). Substituting values gives 4.3 cm, which matches expected image position and magnification from mirror/lens
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