Practice question
Question
A concave lens of focal length \( 15 \, \text{cm} \) has an object placed \( 30 \, \text{cm} \) from
it. What is the image distance?
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. Focal length: f = -15 cm (concave lens). Object distance: u = -30 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-30) = (1/-15) ⇒ (1/v) + (1/30) = (1/-15) ⇒ (1/v) = (1/-15) - (1/30) = (-2 - 1/30) = (-3/30) = (-1/10) . v = -10 cm (virtual image). Substituting values gives 10 cm, which matches expected image position and
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