A light ray passes from water (refractive index ( n1 = 1.33 )) into diamond (refractive index ( n2 = 2.42 )). If the angle of incidence in water is ( 300 ), what is the angle of refraction in diamond?

Question

BlackTom AI · Accepted Answer

Snell's law n1 sinθ1 = n2 sinθ2 gives the relation between incidence and refraction angles across an interface. Here n1 = 1.33 (water), n2 = 2.42 (diamond), and θ1 = 30°.

Step-by-step reasoning and calculation:
1. Write Snell's law: 1.33 · sin(30°) = 2.42 · sin(θ2).
2. Evaluate sin(30°) = 0.5, so left side = 1.33 × 0.5 = 0.665.
3. Solve for sin(θ2): sin(θ2) = 0.665 / 2.42.
4. Compute the division: 0.665 ÷ 2.42 ≈ 0.2748.
5. Take the inverse sine: θ2 = arcsin(0.2748) ≈ 15.94°.
6. Rounding to one decimal gives θ2 ≈ 15.9°.

Why this result makes physical sense: since light passes from a less dense medium (water, n = 1.33) into a denser medium (diamond, n = 2.42), the ray bends toward the normal and the refracted angle is smaller than the incident 30°.

Why the other choices are incorrect:
- a. 22.8° and b. 25.4° are significantly larger than the calculated 15.9°; those would correspond to much larger sine values than 0.2748 and contradict Snell’s law for these indices.
- c. 19.2° is also larger than the correct 15.9° and does not satisfy n1 sinθ1 = n2 sinθ2 with the given indices.

Conclusion: The computed refracted angle θ2 ≈ 15.9°, so the correct answer is d. 15.90, matching the provided answer.

A light ray passes from water (refractive index ( n1 = 1.33 )) into diamond (refractive index ( n2 = 2.42 )). If the angle of incidence in water is ( 300 ), what is the angle of refraction in diamond?单项选择题

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