Chapter 18, Problem 56A

To determine

To sketch:The path of the ray until it leaves the glass in the given ray diagram.

Answer to Problem 56A

The path of the ray until it leaves the glass is shown in Figure 2.

Explanation of Solution

Given:

The given ray diagram is shown below.

  

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:

The index of refraction for glass is $n_{\text{g}}=1.52$ and the index of refraction for air is $n_{\text{A}}=1.00$ .

The angle of the ray enters into the glass as follows:

  $\begin{array}{l}{n}_{\text{A}}\sin {\theta }_A=n_{\text{g}}\sin {\theta }_{\text{g}}\\ {\theta }_{\text{g}}={\sin }^{-1}\left(\frac{n_{\text{A}}\sin {\theta }_A}{n_{\text{g}}}\right)\\ {\theta }_{\text{g}}={\sin }^{-1}\left(\frac{1.00\times \sin 45°}{1.52}\right)\\ {\theta }_1=28°\end{array}$

The critical angle for float glass as follows:

  $\begin{array}{l}\sin {\theta }_{\text{c}}=\frac{n_{\text{A}}}{n_{\text{g}}}\\ {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{n_{\text{A}}}{n_{\text{g}}}\right)\\ {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1.00}{1.52}\right)\\ {\theta }_{\text{c}}=41.1°\end{array}$

The angle at which it strikes the surface is,

  $\begin{array}{l}{\theta }_{\text{r}}=90°-{\theta }_{\text{g}}\\ {\theta }_{\text{r}}=90°-28°\\ {\theta }_{\text{r}}=62°\end{array}$

Hence, the total internal reflection occurs at an angle of $62°$ .

Sketch the path of the ray.

  

Figure 2

Conclusion:

Thus, the path of the ray is shown in Figure 2.