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Question

A converging beam meets a convex lens (\( f = 10 \, \text{cm} \)) \( 5 \, \text{cm} \) 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 = -5 cm (virtual object), f = 10 cm . Lens formula: (1/v) - (1/-5) = (1/10) ⇒ (1/v) + (1/5) = (1/10) . (1/v) = (1/10) - (1/5) = (1 - 2/10) = (-1/10) . v = -10 cm (10 cm to the left). Substituting values gives 10 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

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