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Question

A converging beam meets a concave lens (\( f = 20 \, \text{cm} \)) \( 8 \, \text{cm} \) before the
convergence point. What is the new image distance?

Options

Choose one · Correct answer highlighted

Explanation

**Critical angle** C satisfies sinC = n₂/n₁, n₁>n₂, n₂=1 for air, n₁=1.52 for glass gives sinC=1/1.52=0.6579, C≈41.1°, for water n=1.33 C=48.75°. Beyond C, total internal reflection occurs, all light reflected, no refracted ray, used in optical fibers and prisms. Object distance: u = -8 cm (virtual object), 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) = (-7/40) . v = -(40/7) ≈ -5.71 cm (5.71 cm to the left). Substituting values gives 5.7 cm, which matches expected image position and magnification from mirror/lens formula 1/f

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