An optic fiber is made of clear plastic with index of refraction of 1.50. For what angle of incidence will light remain within the plastic "guide"?
A) 40.3°
B) 21.1°
C) 41.8°
D) 23.4°
E) 38.3°

Question

An optic fiber is made of clear plastic with index of refraction of 1.50. For what angle of incidence will light remain within the plastic "guide"?
A) >40.3°
B) >21.1°
C) >41.8°
D) >23.4°
E) >38.3°

Quizplus · Accepted Answer

To find the critical angle for total internal reflection in the optic fiber, we use Snell's law at the boundary between the plastic (with a refractive index of 1.50) and air (with a refractive index of approximately 1.00). The critical angle ($\theta_c$) can be found from the equation $\sin(\theta_c) = \frac{n_2}{n_1}$, where $n_2 = 1.00$ (refractive index of air) and $n_1 = 1.50$ (refractive index of the plastic). Solving for $\theta_c$, we get $\theta_c = \sin^{-1}(\frac{1.00}{1.50})$, which calculates to approximately 41.8°. Therefore, for angles of incidence greater than 41.8°, light will remain within the plastic guide due to total internal reflection.

An Optic Fiber Is Made of Clear Plastic with Index