NECO 2002 Physics · Question 7

If the refractive index of water is 1.33, what is the critical angle for a water-air boundary?

Options

A.36.0 °
B.40.2 °
C.41.8 °
D.48.8 °
E.62.7 °

Answer: D. 48.8 °

Explanation written with AI and checked against the answer key

Analyze the question: We need the critical angle for light going from water (denser medium) into air (rarer medium). Concept: The critical angle, c, is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°. By Snell's law: n₁ sin c = n₂ sin 90° For a water-air boundary, n₁ = 1.33 and n₂ = 1 (air), so: sin c = 1/n Solution: sin c = 1/1.33 = 0.7519 c = sin⁻¹(0.7519) Check: sin 48° = 0.7431 and sin 49° = 0.7547. Interpolating, (0.7519 − 0.7431)/(0.7547 − 0.7431) ≈ 0.0088/0.0116 ≈ 0.76, so c ≈ 48.76° ≈ 48.8°. Checking against the options: The value 48.8° matches option D. Note that 41.8° (option C) is the critical angle for glass with n = 1.5 (sin⁻¹(0.667)), a common distractor. Option C is also approximately 90° − 48.2°, so it is a trap if you mix up the angle with its complement. Answer: D. 48.8° Tip: Remember the formula sin c = 1/n. A larger refractive index means a smaller critical angle. Water (1.33) gives about 49°, glass (1.5) gives about 42°, and diamond (2.42) gives about 24°. If your answer for water is below 45° you've likely made an error.

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