Fiber optic cable is in the form of a cylindrical core of fiber with index of refraction 1.67 surrounded by a layer of cladding of index 1.45. What is the maximum angle for light rays inside the fiber core with respect to the cable's axis for which the light will be fully transmitted thanks to total internal reflection?
A) 29.7°
B) 60.3°
C) 36.8°
D) 53.2°

Question

Quizplus · Accepted Answer

The critical angle for total internal reflection is given by $\sin^{-1}\left(\frac{n_2}{n_1}\right)$, where $n_1$ is the index of refraction of the core and $n_2$ is the index of refraction of the cladding. In this case, $n_1=1.67$ and $n_2=1.45$. Plugging into the formula gives a critical angle of $\sin^{-1}\left(\frac{1.45}{1.67}\right)\approx 29.7°$. Therefore, any angle of incidence greater than this will result in total internal reflection within the core of the cable.

Fiber Optic Cable Is in the Form of a Cylindrical