MODEL Represent the laser beam with a single ray and use the ray model of light.

VISUALIZE FIGURE 34.16 is a pictorial representation in which the first four steps of Tactics Box 34.1 are identified. Notice that the angle of incidence is ${\theta }_1=60\text{°}$, not the $30\text{°}$ value given in the problem.

FIGURE 34.16 The ray diagram of a laser beam passing through a sheet of glass.

SOLVE

1. Snell’s law, the final step in the Tactics Box, is $n_1\text{sin}{\theta }_1=n_2\text{sin}{\theta }_2$. Using ${\theta }_1=60\text{°}$, we find that the direction of travel in the glass is

   $$\begin{array}{ccc}\hfill {\theta }_2& =\hfill & {\text{sin}}^{-1}\left({\displaystyle \frac{n_1\text{ sin }{\theta }_1}{n_2}}\right)={\text{sin}}^{-1}\left({\displaystyle \frac{\text{sin }60\text{°}}{1.5}}\right)\hfill \\ \hfill & =\hfill & {\text{sin}}^{-1}\left(0.577\right)=35.3\text{°}\hfill \end{array}$$

2. Snell’s law at the second boundary is $n_2\text{sin}{\theta }_3=n_1\text{sin}{\theta }_4$. You can see from Figure 34.16 that the interior angles are equal: ${\theta }_3={\theta }_2=35.3\text{°}$. Thus the ray emerges back into the air traveling at angle

   $$\begin{array}{ccc}\hfill {\theta }_4& =\hfill & {\text{sin}}^{-1}\left({\displaystyle \frac{n_2\text{ sin }{\theta }_3}{n_1}}\right)={\text{sin}}^{-1}(1.5\text{ sin }35.3\text{°})\hfill \\ \hfill & =\hfill & {\text{sin}}^{-1}\left(0.867\right)=60\text{°}\hfill \end{array}$$

   This is the same as ${\theta }_1$, the original angle of incidence. The glass doesn’t change the direction of the laser beam.

3. Although the exiting laser beam is parallel to the initial laser beam, it has been displaced sideways by distance $d$. FIGURE 34.17 shows the geometry for finding $d$. From trigonometry, $d=l\text{sin}\varphi$. Further, $\varphi ={\theta }_1-{\theta }_2$ and $l=t/\text{cos}{\theta }_2$, where $t$ is the thickness of the glass. Combining these gives

   $$\begin{array}{ccc}\hfill d& =\hfill & l\text{ sin}\varphi {\displaystyle \frac{t}{\text{cos}{\theta }_2}}\text{sin}\left({\theta }_1-{\theta }_2\right)\hfill \\ \hfill & =\hfill & {\displaystyle \frac{\left(1.0\text{cm}\right)\text{sin}24.7°}{\text{cos}35.3°}}=0.51\text{cm}\hfill \end{array}$$

   The glass causes the laser beam to be displaced sideways by 0.51 cm.

   FIGURE 34.17 The laser beam is deflected sideways by distance $d$.

   

   FIGURE 34.17 Full Alternative Text

REVIEW The laser beam exits the glass still traveling in the same direction as it entered. This is a general result for light traveling through a medium with parallel sides. Notice that the displacement $d$ becomes zero in the limit $t\to 0$. This will be an important observation when we get to lenses.

MODEL Represent the laser beam with a single ray and use the ray model of light.

VISUALIZE FIGURE 34.19 uses the steps of Tactics Box 34.1 to draw a ray diagram. The ray is incident perpendicular to the front face of the prism $({\theta }_{\text{incident}}=0\text{°})$, thus it is transmitted through the first boundary without deflection. At the second boundary it is especially important to draw the normal to the surface at the point of incidence and to measure angles from the normal.

FIGURE 34.19 Pictorial representation of a laser beam passing through the prism.

SOLVE From the geometry of the triangle you can find that the laser’s angle of incidence on the hypotenuse of the prism is ${\theta }_1=30\text{°}$, the same as the apex angle of the prism. The ray exits the prism at angle ${\theta }_2$ such that the deflection is $\varphi ={\theta }_2-{\theta }_1=22.6\text{°}$. Thus ${\theta }_2=52.6\text{°}$. Knowing both angles and $n_2=1.00$ for air, we can use Snell’s law to find $n_1:$