Question

Question: Figure depicts a simplistic optical fiber: a plastic core$\left(n_1=1.58\right)$is surrounded by a plastic sheath $\left(n_2=1.53\right)$. A light ray is incident on one end of the fiber at angle.The ray is to undergo total internal reflection at point A, where it encounters the core–sheath boundary. (Thus there is no loss of light through that boundary.) What is the maximum value of $\mathbf{\theta }$that allows total internal reflection at A?

Figure:

Short answer

The maximum value of $\theta$that allows total internal reflection at A is $23.2^{\circ }$.

Step-by-step solution

Given

1. Refractive index of the plastic core,$n_1=1.58$
2. Refractive index of the plastic sheath,$n_2=1.53$

Understanding the concept

By using the concept of the critical angle and Snell’s law, we can find the value of $\mathbf{\theta }$that allows the total internal reflection.

Formula:

Critical angle is

${\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)$

Snell’s law is :

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

Calculate the maximum value of θ that allows total internal refection at A.

The critical angle is given by

$$\begin{gathered} {\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right) \\ {\theta }_c={\sin }^{-1}\left(\frac{1.53}{1.58}\right)=75.{547}^{\circ } \end{gathered}$$

So, by geometry, the incident angle is

$\begin{array}{l}{\mathrm{\theta }}_1=90-{\mathrm{\theta }}_{\mathrm{c}}\\ {\mathrm{\theta }}_1=90-75.{547}^{\mathrm{o}}\\ =14.{45}^{\circ }\end{array}$

Now, we have to apply Snell’s law,

${\mathrm{n}}_{\mathrm{air}}\sin \left(\mathrm{\theta }\right)={\mathrm{n}}_1\sin \left({\mathrm{\theta }}_1\right)$

Since $n_{air}=1$

$\begin{array}{l}\sin \left(\mathrm{\theta }\right)=1.58\sin \left(14.{45}^{\circ }\right)\\ sim\mathrm{\theta }=0.3942\\ \mathrm{\theta }=23.2^{\circ }\end{array}$