A light ray in glass arrives at the glass-water interface at an angle of θ = 48° with the normal. The refracted ray in water makes an angle φ = 72° with the normal, as shown in the figure. The index of refraction of water is 1.33. The angle of incidence is now changed to θ = 37°. What is the new angle of refraction φ in the water? Hint: Start by calculating the index of refraction of the glass.

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BlackTom AI · Accepted Answer

To tackle this problem, I’ll lay out the given data and use Snell’s law, n1 sin(theta) = n2 sin(phi), step by step.

First, we have the glass-to-water interface with incidence angle theta = 48° in glass and refracted angle phi = 72° in water, with water’s index n2 = 1.33. Applying Snell’s law at this interface:
 n_glass * sin(48°) = n_water * sin(72°).
Solving for n_glass gives:
 n_glass = n_water * sin(72°) / sin(48°).
Using approximate values: sin(72°) ≈ 0.9511 and sin(48°) ≈ 0.7431, so the ratio is about 0.9511/0.7431 ≈ 1.280. Therefore n_glass ≈ 1.33 * 1.280 ≈ 1.70 (more precisely around 1.70–1.71).

Second, with the angle of incidence now theta = 37° in glass, we apply Snell’s law again to find the new refracted angle phi in water:
 n_glass * sin(37°) = n_water * sin(phi).
Compute sin(37°) ≈ 0.6018. Then the left-hand side is about 1.70 * 0.6018 ≈ 1.023.
Divide by n_water = 1.33 to solve for sin(phi): sin(phi) ≈ 1.023 / 1.33 ≈ 0.770.
Now, phi ≈ arcsin(0.770). The inverse sine of 0.770 is roughly 50.3°.

Reviewing the provided answer choices: 52°, 54.1°, 55°, 40.6°, 50.4°.

Option-by-option analysis:
- 52°: If phi were 52°, sin(52°) ≈ 0.788. This would require n_glass * sin(37°) ≈ 1.33 * 0.788 ≈ 1.048, implying n_glass ≈ 1.048 / sin(37°) ≈ 1.741, which is not consistent with the n_glass value obtained from the first interface (≈1.70). So 52° is not consistent with both interfaces.
- 54.1°: sin(54.1°) ≈ 0.811. That would demand n_glass ≈ (1.33 * 0.811) / sin(37°) ≈ 1.079 / 0.6018 ≈ 1.79, which again conflicts with the n_glass inferred from the first interface.
- 55°: sin(55°) ≈ 0.819. Using the same cross-check, n_glass would be ≈ (1.33 * 0.819) / 0.6018 ≈ 1.089 / 0.6018 ≈ 1.81, not matching the initial value near 1.70.
- 40.6°: sin(40.6°) ≈ 0.651. This would imply n_glass ≈ (1.33 * 0.651) / 0.6018 ≈ 0.867 / 0.6018 ≈ 1.44, which is significantly lower than the n_glass deduced from the first interface, so this option is inconsistent.
- 50.4°: sin(50.4°) ≈ 0.770. This matches the computed sin(phi) ≈ 0.770 from the second interface calculation, yielding n_glass ≈ (1.33 * 0.770) / 0.6018 ≈ 1.024 / 0.6018 ≈ 1.70, which aligns with the glass index obtained earlier.

Thus, the angle that satisfies both interface conditions is approximately 50.4°, and this is the value that resolves consistently with Snell’s law across the two media.

PHYSICS 1201 SU2025 (10585) Final Exam- Requires Respondus LockDown Browser

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