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 for this fiber?
A) 63.2$^{\circ}$
B) 26.8$^{\circ}$
C) 41.8$^{\circ}$
D) 73.0$^{\circ}$
E) 16.5$^{\circ}$

Question

Quizplus · Accepted Answer

The critical angle can be calculated using the formula: $ \theta_c = \sin^{-1}(\frac{n_2}{n_1}) $, where $ n_1 = 1.40 $ and $ n_2 = 1.25 $. This gives $ \theta_c = \sin^{-1}(\frac{1.25}{1.40}) \approx 63.2^\circ $.

An Optical Fiber Having Index of Refraction 1 ∘^{\circ}∘ B) 268 ∘^{\circ}∘

An Optical Fiber Having Index of Refraction 1 $^{\circ}$ B) 268 $^{\circ}$