Refractive index of water with respect to air is 4/3. The critical angle for a water–air interface is approximately
A{'text': 'sin⁻¹(4/3)', 'label': 'A'}
B{'text': 'sin⁻¹(3/4) ≈ 48.6°', 'label': 'B'}
C{'text': '30°', 'label': 'C'}
D{'text': '60°', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': 'sin⁻¹(3/4) ≈ 48.6°', 'label': 'B'}
1. For TIR going from water to air: sin θc = n_air / n_water = 1 / (4/3) = 3/4.
2. θc = sin⁻¹(0.75) ≈ 48.6°.
3. (B) is undefined (sin⁻¹ > 1); (C) and (D) ignore the n-ratio.
_Source: NCERT Class 12 Physics Part II, Ch 9 §9.4 + Table 9.1_
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