Chapter 23, Problem 52E

(a)

To determine

The direction of the refracted ray inside the plate.

Answer to Problem 52E

The direction of the refracted ray inside the plate is $19.5°$.

Explanation of Solution

Write the expression for angle of refraction.

  $\mu =\frac{\sin {\theta }_i}{\sin {\theta }_r}$        (I)

Here, $\mu$ is the refractive index of the glass plate, ${\theta }_i$ is the angle of incidence, and ${\theta }_r$ is the angle of refraction.

Rewrite the equation (I) for refraction ray.

  $\begin{array}{c}\sin {\theta }_r=\frac{\sin {\theta }_i}{\mu }\\ {\theta }_r={\sin }^{-1}\left(\frac{\sin {\theta }_i}{\mu }\right)\end{array}$        (II)

Conclusion:

Substitute $30°$ for ${\theta }_i$ and $1.5$ for $\mu$ in equation (II) to find ${\theta }_r$.

  $\begin{array}{c}{\theta }_r={\sin }^{-1}\left[\frac{\sin \left(30°\right)}{1.5}\right]\\ ={\sin }^{-1}\left(0.333\right)\\ =19.5°\end{array}$

Therefore, the direction of the refracted ray inside the plate is $19.5°$.

(b)

To determine

The direction of the ray emerging from far side of the plate.

Answer to Problem 52E

The direction of the ray emerging from far side of the plate is $77°$.

Explanation of Solution

From part (a):

Apply the Snell’s law for angle of refraction.

  $\begin{array}{c}\sin {\theta }_4=\frac{n_3}{n_4}\sin {\theta }_3\\ {\theta }_4={\sin }^{-1}\left(\frac{n_3}{n_4}\sin {\theta }_4\right)\end{array}$        (III)

Here, $n_4$ is the refractive index of air, ${\theta }_4$ is the angle of refracted ray leaves from the prism, $n_3$ is the refractive index of glass, and ${\theta }_3$ is the angle of the reflected ray.

Conclusion:

The angle between the normal line from the figure is,

  $\begin{array}{c}{\theta }_3=60°-19.5°\\ =40.5°\end{array}$

Substitute $40.5°$ for ${\theta }_3$, $1.0$ for $n_4$, and $1.5$ for $n_3$ in equation (III) to find ${\theta }_4$.

  $\begin{array}{c}{\theta }_4={\sin }^{-1}\left[\left(\frac{1.5}{1.0}\right)\sin \left(40.5°\right)\right]\\ ={\sin }^{-1}\left(0.974\right)\\ =77°\end{array}$

Therefore, the direction of the ray emerging from far side of the plate is $77°$.

(c)

To determine

The distance of the ray shifted to its original path.

Answer to Problem 52E

The distance of the ray shifted to its original path is $3.82\text{mm}$.

Explanation of Solution

The expression for the lateral displacement is,

  $x=\frac{t\sin \left(i-r\right)}{\cos r}$        (IV)

Here, $x$ is distance of the ray shifted to its original path and $t$ is the thickness of the glass plate.

Conclusion:

Substitute $0.02\text{m}$ for $t$, $30°$ for $i$, and $19.5°$ for $r$ in equation (IV) to find $x$.

  $\begin{array}{c}x=\frac{\left(0.02\text{m}\right)\sin \left(30°-19.5°\right)}{\cos \left(19.5°\right)}\\ =\frac{\left(0.02\text{m}\right)\sin \left(10.5°\right)}{\cos \left(19.5°\right)}\\ =3.82\times {10}^{-3}\text{m}\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =3.82\text{mm}\end{array}$

Therefore, the distance of the ray shifted to its original path is $3.82\text{mm}$.