Chapter 18, Problem 97A

To determine

The distance between the ray using ray diagram.

Answer to Problem 97A

The distance between the ray that leaves and the one that would have left if the plastic were not there is $8\text{ mm}$ .

Explanation of Solution

Given:

The thickness of the plastic sheet is $25\text{ mm}$ .

The angle of the light day strikes and leaves the plastic sheet is $45°$

The index of refraction of the air medium is $n_1=1.0003$ .

The index of refraction of the plastic medium is $n_2=1.5$ .

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 $25\text{ mm}$ sheet of plastic. Provide the ray of light strikes at an angle of $45°$ and leaves the sheet at an angle of $45°$ in different positions.

Consider the angle incidence of the air medium is ${\theta }_1=45°$ .

Consider the angle incidence of the plastic medium is ${\theta }_2$ .

The value of the angle ${\theta }_2$ is,

  $\begin{array}{l}{\theta }_2={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\\ {\theta }_2={\sin }^{-1}\left(\frac{1.0003\times \sin \left(45°\right)}{1.5}\right)\\ {\theta }_2=28°\end{array}$

Break the point of ray strikes into two right triangles at angle of $28°$ and $17°$ .

Sketch the ray diagram of the sheet of plastic as shown in Figure.

  

The length of the hypotenuse of the $28°$ triangle is,

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent}}{\text{hypotenus}}\\ \text{hypotenus}=\frac{\text{25 mm}}{\cos 28°}\\ \text{hypotenus}=28.3\text{ mm}\end{array}$

The length of the hypotenuse of the $28°$ triangle is,

  $\begin{array}{l}\sin \theta =\frac{\text{opposite}}{\text{hypotenus}}\\ \text{opposite}=\left(\text{28}\text{.3 mm}\right)\sin 17°\\ \text{opposite}=8\text{ mm}\end{array}$

Conclusion:

Thus, the distance between the ray that leaves and the one that would have left if the plastic were not there is $8\text{ mm}$ .