The glass core of an optical fiber has an index of refraction$1.60$. The index of refraction of the cladding is$1.48$. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

The boundary prevents the ray from refracting. Instead, all of the light reflects back into the prism from the boundary. This is known as total internal reflection, or TIR for short.

A critical angle is reached when${\theta }_2=90°$. Because$\sin 90°=1$, Snell’s law$n_1\sin {\theta }_c=n_2\sin 90°$gives the critical angle of incidence as

${\theta }_{\mathrm{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

$n_1=1.60$is the index of refraction of glass core.

$n_2=1.48$is the index of refraction of cladding.

Equation that gives the angle of incidence of the light ray on the cladding at which it exhibits total internal reflection is,

${\theta }_{\mathrm{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Substitute the values:

${\theta }_c={\sin }^{-1}\left(\frac{1.48}{1.60}\right)$

Simplify:

$=67.6°$

The maximum angle of the light ray with the wall of the glass core${\theta }_{\max }$is associated to the angle of incidence${\theta }_c$is:

${\theta }_{\max }=90°-{\theta }_c$

Substitute the values:

Simplify:

$=22.3°$