Chapter 23, Problem 42P

To determine

The range of angle for which total internal reflection occurs at the side surface of the material (at point $A$)when a beam of light enters the end of an optic fibre and its index of refraction is greater than $1.42$.

Answer to Problem 42P

Solution:

It occurs for all angles.

Explanation of Solution

Given:

Medium present outside the material is air. Refractive index, $n>1.42$.

Formula used:

Snell’s law:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Where,

$n_1$ is the refractive index of material $1$.

$n_2$ is the refractive index of material $2$.

${\theta }_1$ is the angle of incidence.

${\theta }_2$ is the angle of refraction.

Calculation:

Consider Snell’s law, considering medium $1$ is air i.e., $n_1$=1.

Any angle of incidence $\gamma$ is given by-

$\begin{array}{l}{n}_1\sin \alpha =1.42\sin \beta \\ \Rightarrow \sin \alpha =1.42\sin \beta \\ \gamma =90°-{\sin }^{-1}\left(\frac{\sin \alpha }{1.42}\right)\\ ={\cos }^{-1}\left(\frac{\sin \alpha }{1.42}\right)\\ >{\cos }^{-1}\left(\frac{1}{1.42}\right)\\ \approx 45.23\end{array}$

To achieve total internal reflection angle of incidence $\gamma$ should be greater than critical angle ${\gamma }_c$. Thus, the critical angle ${\gamma }_c$ is calculated as-

$\begin{array}{l}1.42\sin {\gamma }_c=\sin 90°=1\\ \Rightarrow \sin {\gamma }_c=\frac{1}{1.42}\\ \Rightarrow {\gamma }_c={\sin }^{-1}\left(\frac{1}{1.42}\right)\\ \approx 44.77\end{array}$

Thus, we see that $\gamma >{\gamma }_c$ so total internal reflection occurs at the side surface of the material (at point $A$)when a beam of light enters the end of an optic fibre.

To determine

(b)

The result when fiber were immersed in the water.

Answer to Problem 42P

Solution:

It occurs for $\alpha <21.54$

Explanation of Solution

Given:

Medium present outside the material is air. Refractive index, $n>1.42$.

Formula used:

Snell’s law:

$n_1\mathrm{Sin}{\theta }_1=n_2\mathrm{Sin}{\theta }_2$

Where,

$n_1$ is the refractive index of material $1$.

$n_2$ is the refractive index of material $2$.

${\theta }_1$ is the angle of incidence.

${\theta }_2$ is the angle of refraction.

Calculation:

When fiber were immersed in the water $n_1=1.333$ Thus, applying Snell’s law we get-

$1.42\sin \beta =1.333\sin \alpha$

$\begin{array}{l}\sin \gamma =\sqrt{1-{\cos }^2\gamma }\\ =\sqrt{1-{\sin }^2\beta }\\ =\sqrt{1-{\left(\frac{1.333\sin \alpha }{1.42}\right)}^2}\\ >\left(\frac{1.33}{1.42}\right)\end{array}$

$\begin{array}{l}\Rightarrow 1-{\left(\frac{1.33}{1.42}\right)}^2>{\left(\frac{1.33}{1.42}\right)}^2{\sin }^2\alpha \\ \Rightarrow 0.1348>{\sin }^2\alpha \\ \Rightarrow \sin \alpha <\sqrt{0.1348}\\ \Rightarrow \alpha <21.54\end{array}$

When fiber were immersed in the water, to achieve total internal reflection angle of incident should be greater than $21.54°$.