An optic fiber is made of clear plastic with an index of refraction of $1.50$. For what range of angles of incidence $\theta$ with the lateral surface of the fiber will light remain inside the plastic "guide" if it is surrounded by air?
A) $23.4 ^ { \circ }

Question

An optic fiber is made of clear plastic with an index of refraction of $1.50$. For what range of angles of incidence $\theta$ with the lateral surface of the fiber will light remain inside the plastic "guide" if it is surrounded by air?
A) $23.4 ^ { \circ } < \theta < 90 ^ { \circ }$
B) $41.8 ^ { \circ } < \theta < 90 ^ { \circ }$
C) $38.3 ^ { \circ } < \theta < 90 ^ { \circ }$
D) $21.1 ^ { \circ } < \theta < 90 ^ { \circ }$
E) $40.3 ^ { \circ } < \theta < 90 ^ { \circ }$

Quizplus · Accepted Answer

For total internal reflection to occur, the angle of incidence must be greater than the critical angle. The critical angle can be calculated using Snell's law, $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$, where $n_1$ is the index of refraction of the denser medium (plastic, 1.50), $n_2$ is the index of refraction of the less dense medium (air, approximately 1.00), and $\theta_c$ is the critical angle. Solving for $\theta_c$, we get $\sin(\theta_c) = \frac{n_2}{n_1} = \frac{1.00}{1.50}$, which gives $\theta_c \approx 41.8^\circ$. Therefore, for light to remain inside the plastic guide, the angle of incidence must be greater than this critical angle, up to 90 degrees.

An Optic Fiber Is Made of Clear Plastic with an Index