Question

An optical fiber made of glass with an index of refraction 1.50 is coated with a plastic with index of refraction 1.30. What is the critical angle of this fiber at the glass-plastic interface?

Answer 1

To solve this problem we will use Snell's law. This law is used to calculate the angle of refraction of light by crossing the separation surface between two means of propagation of light (or any electromagnetic wave) with a different index of refraction.

$n_1 sin\theta_1 = n_2 sin\theta_2$

Where,

$n_{1,2}$ = Index of refraction for each material

$\theta_1$= Angle of incidence and refraction

A ray of light propagating in a medium with index of refraction $n_1$ hitting an angle $\theta_1$ on the surface of a medium of index $n_2$ with $n_1> n_2$ can be fully reflected inside the medium of highest refractive index:

$\theta_{critical} = Sin^{-1}(\frac{n_1}{n_2})$

$n_1$ Refractive index of rarer medium

$n_2$ Refractive index of denser medium

Replacing we have,

$\theta_{critical} = Sin^{-1}(\frac{n_1}{n_2})$

$\theta_{critical} = Sin^{-1}(\frac{1.3}{1.5})$

$\theta_{critical} = 60.07\°$

the critical angle of this fiber at the glass-plastic interface is 60.07°