Chapter 18, Problem 103A

To determine

The critical angle for light traveling from ice to air.

To explain: Whether fiber-optic cables made of ice or made of glass do a better job of keeping light inside the cable.

Answer to Problem 103A

The critical angle for light traveling from ice to air is $49.8°$ .

The fiber-optic cables made of glass do a better job of keeping light inside the cable.

Explanation of Solution

Given:

  $n_{\text{ice}}=1.31$ .

Formula used:

The expression for Snell’s law of refraction is,

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

Here, $n_1$ is the index of refraction of the first medium, ${\theta }_1$ is the angle of incidence of the first medium, $n_2$ is the index of refraction of the second medium, and ${\theta }_2$ is the angle of incidence of the second medium.

Calculation:

Consider the index of refraction of the air $n_{\text{air}}=1.00$ .

Consider the index of refraction of the glass $n_{\text{glass}}=1.54$ .

The critical angle of refraction for iceis,

  $\begin{array}{l}\sin {\theta }_{\text{c}}=\frac{n_{\text{air}}}{n_{\text{ice}}}\\ {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1.00}{1.31}\right)\\ {\theta }_{\text{c}}=49.8°\end{array}$

The critical angle of refraction for glass is,

  $\begin{array}{l}\sin {\theta }_{\text{c}}=\frac{n_{\text{air}}}{n_{\text{glass}}}\\ {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1.00}{1.54}\right)\\ {\theta }_{\text{c}}=40.5°\end{array}$

The critical angle for glass is less when compared to the critical angle for ice.

If the critical angle is larger, then few of the rays would be totally reflected internally in ice than in a glass. So the fiber-optics made of ice would not be able to transmit as much light.Hence, the fiber-optic cables made of glass will do a better job to keep light inside the cable.

Conclusion:

Thus, the critical angle for light traveling from ice to air is $49.8°$ .

Thus, the fiber-optic cables made of glass do a better job of keeping light inside the cable.