Question

The glass core of an optical fiber has an index of refraction 1.60.The index of refraction of the cladding is 1.45.

What is the maximum angle a light ray can make with the wall of thecore if it is to remain inside the fiber?

Answer 1

Concepts and reason

The concept required to solve the given problem is Snell’s law of refraction.

Initially calculate the minimum angle that causes total internal reflection and then the maximum angle with the wall of the core will be calculated.

Fundamentals

Snell’s law of refraction- It states that the ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given color and for the given pair of media.

The expression for the Snell’s law of refraction is,

${n_1}\sin \left( {{\theta _1}} \right) = {n_2}\sin \left( {{\theta _2}} \right)$ …… (1)

Here, n₁ is the refractive index of the first materials, n₂ is the refractive index of the second material, ${\theta _1}$ is the angle of incidence, and ${\theta _2}$ is the angle of refraction.

Substitute 1.60 for n₁, 1.45 for n₂, and $90^\circ$ for ${\theta _2}$ .

$\begin{array}{c}\\{n_1}\sin \left( {{\theta _1}} \right) = {n_2}\sin \left( {{\theta _2}} \right)\\\\1.60\sin \left( {{\theta _1}} \right) = 1.45\sin \left( {90^\circ } \right)\\\\\left( {{\theta _1}} \right) = {\sin ^{ - 1}}\left( {0.90625} \right)\\\\ = {\rm{64}}{\rm{.99}}^\circ \\\end{array}$

Calculate the value of maximum angle by subtracting the value of minimum angle from 90 degrees.

${\theta _{\max }} = 90^\circ - \left( {{\theta _1}} \right)$

Here, ${\theta _{\max }}$ is the maximum angle and ${\theta _1}$ is the value of minimum angle.

Substitute $64.99^\circ$ for $\left( {{\theta _1}} \right)$ .

$\begin{array}{c}\\{\theta _{\max }} = 90^\circ - \left( {{\theta _1}} \right)\\\\ = 90^\circ - 64.99^\circ \\\\ = 25.01^\circ \\\end{array}$

Ans:

The maximum angle that a light ray can make with the wall of the core is $25.01^\circ$ .