For total internal reflection at glass-water interface (n_g = 1.5, n_w = 1.33), critical angle:
A48.6°
Bsin⁻¹(1.5/1.33)
C30°
Dsin⁻¹(1.33/1.5) ≈ 62.5°
Answer & Solution
Correct answer: D. sin⁻¹(1.33/1.5) ≈ 62.5°
sin θ_c = n_rare/n_dense = 1.33/1.5 ≈ 0.887. θ_c ≈ 62.5°. Larger than glass-air case (≈42°) because the n-ratio is smaller.
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