When a light ray in water (n = 1.33) enters glass with index of refraction 1.60, what is the greatest angle of refraction that can occur?
A) 90.0°
B) 56.2°
C) 52.2°
D) 38.7°
E) 0.0°

Question

Quizplus · Accepted Answer

The greatest angle of refraction occurs when the light ray is at the critical angle in water, just before undergoing total internal reflection. Using Snell's law, $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ is the index of refraction of water (1.33), $n_2$ is the index of refraction of glass (1.60), and $\theta_1$ is the angle of incidence in water. At the critical angle, $\theta_2 = 90^\circ$, so we solve for $\theta_1$ to find the greatest angle of refraction in glass. Rearranging Snell's law gives $\sin(\theta_1) = \frac{n_2}{n_1} \sin(\theta_2)$. Substituting $n_2 = 1.60$, $n_1 = 1.33$, and $\sin(90^\circ) = 1$, we find $\sin(\theta_1) = \frac{1.60}{1.33}$, which is not physically possible since the sine of an angle cannot exceed 1. This indicates a misunderstanding in the setup; actually, we should solve for $\theta_2$ given $\theta_1$ at the critical angle for entering glass from water. The correct approach is to find the angle of refraction $\theta_2$ when light goes from water to glass, not the critical angle for total internal reflection. Using Snell's law correctly, we find that the angle of refraction is maximized when the angle of incidence is 90°, but this leads to a calculation error in the explanation. The correct calculation involves finding the maximum angle $\theta_2$ can have when light transitions from water to glass, which involves using the inverse sine function and the correct indices of refraction, leading to an angle less than 90°, which corresponds to choice B, 56.2°, after correctly applying Snell's law and mathematical calculations.

When a Light Ray in Water (N = 1