The speed of light in a material is $0.41 \mathrm { c }$ What is the critical angle of a light ray at the interface between the material and a vacuum?
A) 24°
B) 17°
C) 19°
D) 22°

Question

Quizplus · Accepted Answer

The critical angle can be found using the formula $\sin(\theta_c) = \frac{n_2}{n_1}$, where $n_2$ is the refractive index of the second medium (vacuum, $n_2 = 1$) and $n_1$ is the refractive index of the material. Since the speed of light in the material is $0.41c$, the refractive index $n_1 = \frac{c}{0.41c} = \frac{1}{0.41}$. Solving for $\theta_c$, $\sin(\theta_c) = 0.41$, which gives $\theta_c \approx 24^\circ$.

The Speed of Light in a Material Is 0.41c0.41 \mathrm { c }0.41c

The Speed of Light in a Material Is $0.41 \mathrm { c }$