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Question

A converging beam meets a concave lens (\( f = 10 \, \text{cm} \)) \( 4 \, \text{cm} \) before the
convergence point. What is the new image distance?

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Explanation

**Spherical mirror reflection** follows law θ_i = θ_r, focal length f = R/2, concave R negative, convex positive. Object beyond center C (u > 2f) forms image between F and C diminished, at C same size, between C and F magnified, beyond F forms at infinity, illustrating mirror equation. Object distance: u = -4 cm (virtual object), f = -10 cm . Lens formula: (1/v) - (1/-4) = (1/-10) ⇒ (1/v) + (1/4) = (1/-10) . (1/v) = (1/-10) - (1/4) = (-2 - 5/20) = (-7/20) . v = -(20/7) ≈ -2.86 cm (2.86 cm to the left). Substituting values gives 2.86 cm, which

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