Practice question
Question
What is the critical angle for a water-air interface (\( n_{\text{water}} = 1.33 \))?
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. 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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