Practice question
Question
A concave lens of focal length \( 25 \, \text{cm} \) has an object placed \( 50 \, \text{cm} \) from
it. What is the image distance?
Explanation
**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Focal length: f = -25 cm (concave lens). Object distance: u = -50 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-50) = (1/-25) ⇒ (1/v) + (1/50) = (1/-25) ⇒ (1/v) = (1/-25) - (1/50) = (-2 - 1/50) = (-3/50) . v = -(50/3) ≈ -16.67 cm (virtual image). Substituting values gives 16.7 cm, which matches
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