An optical fiber having index of refraction 1.40 is clad with a material of index of refraction 1.25. What is the critical angle in the fiber at the fiber-cladding interface?
A) 63.2°
B) 26.8°
C) 41.8°
D) 73.0°
E) 16.5°

Question

Quizplus · Accepted Answer

The critical angle can be found using Snell's law at the point of total internal reflection, where $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$, with $n_1$ being the index of refraction of the optical fiber (1.40), $n_2$ the index of refraction of the cladding (1.25), and $\theta_c$ the critical angle. Rearranging for $\theta_c$, we get $\theta_c = \sin^{-1}(\frac{n_2}{n_1}) = \sin^{-1}(\frac{1.25}{1.40})$, which calculates to approximately 63.2°.

An Optical Fiber Having Index of Refraction 1