Light exits a diamond (n = 2.42) into air. The angle of incidence is 15.0°. What is the angle of refraction?

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Question restatement: Light exits a diamond (n = 2.42) into air. The angle of incidence is 15.0°. What is the angle of refraction?

Option analysis:
- a. 38.8°: This is the correct result. Using Snell's law, n1 sin(theta1) = n2 sin(theta2). Here, light goes from diamond (n1 = 2.42) into air (n2 ≈ 1.00) with theta1 = 15°. So sin(theta2) = (n1/n2) * sin(15°) = 2.42 * sin(15°). Since sin(15°) ≈ 0.2588, sin(theta2) ≈ 2.42 * 0.2588 ≈ 0.626, which gives theta2 ≈ arcsin(0.626) ≈ 38.8°.
- b. 51.0°: This would imply a larger refraction angle than 45°, which would require sin(theta2) ≈ sin(51°) ≈ 0.777. That would require n1/n2 ≈ 0.777 / sin(15°) ≈ 0.777 / 0.2588 ≈ 3.0, which is inconsistent with n1/n2 = 2.42. Thus this option misapplies Snell's law.
- c. 24.6°: This is smaller than the incidence angle, suggesting total internal reflection criteria or an incorrect ratio. Since the incident medium has a higher index than the exit medium, refraction should occur at an angle larger than the incidence angle for these indices, not smaller. This indicates a miscalculation of sin(theta2).
- d. 32.2°: While closer, this value would require sin(theta2) ≈ sin(32.2°) ≈ 0.532, which would correspond to sin(theta2) = (2.42) * sin(15°) ≈ 0.626. The discrepancy shows a mismatch with the exact Snell calculation, making this option incorrect.

Conclusion (without stating directly): After applying Snell's law with n1 = 2.42, n2 = 1, and theta1 = 15°, the resulting refraction angle is approximately 38.8°, matching option a. The other options diverge from the Snell-law-derived value due to incorrect use of the refractive-index ratio or misinterpretation of how the angle changes when light exits a denser medium into air.

SPH4U - Physics 12 (2025-26) - A

Light exits a diamond (n = 2.42) into air. The angle of incidence is 15.0°. What is the angle of refraction?

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