A ray of light in air is approaching the boundary with water (n = 1.333) at an angle of 37.1 degrees. Determine the angle of refraction of the light ray.

Question

A ray of light in air is approaching the boundary with water (n = 1.333)  at an angle of 37.1 degrees. Determine the angle of refraction of the light ray.

BlackTom AI · Accepted Answer

We are given light transitioning from air into water, so apply Snell's law: n1 sin(theta1) = n2 sin(theta2).
Here n1 is the refractive index of air, approximately 1.000, and n2 is the refractive index of water, 1.333. The incidence angle theta1 is 37.1 degrees.
Compute sin(theta2) = (n1/n2) * sin(theta1) = (1 / 1.333) * sin(37.1°).
First estimate sin(37.1°): it is about 0.603 (since sin(37°) ≈ 0.6018 and small changes in the angle produce small changes in sine).
Then divide by 1.333: 0.603 / 1.333 ≈ 0.452.
Now find theta2 by taking the arcsin of 0.452. The inverse sine of 0.452 is roughly 26.8 degrees.
Thus, the refracted angle in water is approximately 26.8 degrees.
If you want to verify, you can plug back into Snell's law: 1.000 * sin(37.1°) ≈ 1.333 * sin(26.8°). Using sin(37.1°) ≈ 0.603 and sin(26.8°) ≈ 0.452, both sides are close to 0.603, confirming consistency.

A ray of light in air is approaching the boundary with water (n = 1.333) at an angle of 37.1 degrees. Determine the angle of refraction of the light ray. 简答题

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