Practice question
Question
What is the critical angle for a water (\( n = 1.33 \)) to air interface?
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. Critical angle: sin i_c = (n₂/n₁) . Water ( n₁ = 1.33 ), air ( n₂ = 1 ). sin i_c = (1/1.33) ≈ 0.752 . i_c = sin⁻¹(0.752) ≈ 48.75° . Substituting values gives 49°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
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