Chapter 35, Problem 29P

(a)

To determine

The angle of refraction at the first surface.

Answer to Problem 29P

The angle of refraction at the first surface is $41.5°$.

Explanation of Solution

Figure-(1)

Write the expression of Snell`s Law.

    $\begin{array}{l}{n}_{\text{air}}\cdot \sin {\theta }_1=n_2\sin {\theta }_2\\ {\theta }_2={\sin }^{-1}\left(\frac{n_{\text{air}}\cdot \sin {\theta }_1}{n_2}\right)\end{array}$                                                                                               (I)

Here, $n_{\text{air}}$ is the refractive index of air, $n_2$ is the refractive index of fused quartz, ${\theta }_1$ is the angle of incidence and ${\theta }_2$  is the angle of refraction at the first surface.

Conclusion:

Substitute $1.00$ for $n_{\text{air}}$ and $1.458$ for $n_2$ in equation (I) to calculate ${\theta }_2$.

    $\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{(1.00)\sin 75.0°}{1.458}\right)\\ =41.5°\end{array}$

Therefore, the angle of refraction at the first surface is $41.5°$.

(b)

To determine

The angle of incidence at the second surface.

Answer to Problem 29P

The angle of incidence at the second surface is $18.5°$.

Explanation of Solution

From the figure-(1)

    $\alpha +{\theta }_2=90°$                                                                                                             (II)

    $\beta +{\theta }_3=90°$                                                                                                             (III)

Here ${\theta }_3$ is the angle of incidence at the second surface and ${\theta }_2$ is the angle of refraction at the first surface.

Add equation (III) and (IV).

    $\alpha +\beta +{\theta }_2+{\theta }_3=180°$                                                                                           (IV)

From the figure in $\Delta \text{ABC}$,

    $\begin{array}{c}\alpha +\beta +60°=180°\\ \alpha +\beta =120°\end{array}$                                                                                                 (V)

Conclusion:

Substitute equation (V) in equation (IV)

    ${\theta }_2+{\theta }_3=60.0°$                                                                                                      (VI)

Substitute $41.5°$ for ${\theta }_2$ in equation (VI) to calculate ${\theta }_3$.

    $\begin{array}{c}{\theta }_3=60.0°-41.5°\\ =18.5°\end{array}$

Therefore, the angle of incidence at the second surface is $18.5°$.

(c)

To determine

The angle of refraction at the second surface.

Answer to Problem 29P

The angle of refraction at the second surface is $27.5°$.

Explanation of Solution

Write the expression of Snell`s Law.

    $\begin{array}{c}{n}_2\cdot \sin {\theta }_3=n_{\text{air}}\sin {\theta }_4\\ {\theta }_4={\sin }^{-1}\left(\frac{n_2\cdot \sin {\theta }_3}{n_{\text{air}}}\right)\end{array}$                                                                               (VII)

Here, $n_{\text{air}}$ is the refractive index of air, $n_2$ is the refractive index of fused quartz, and ${\theta }_3$ is the angle of incidence at second surface and ${\theta }_4$  is the angle of refraction at the second surface.

Conclusion:

Substitute $1.00$ for $n_{\text{air}}$ and $1.458$ for $n_2$, $18.5°$ for ${\theta }_3$ in equation (VII) to calculate ${\theta }_4$.

    $\begin{array}{c}{\theta }_4={\sin }^{-1}\left(\frac{(1.458)\sin 18.5°}{1.00}\right)\\ =27.5°\end{array}$

Therefore, the angle of refraction at the second surface is $27.5°$.

(d)

To determine

The angle between the incident rays and emerging rays.

Answer to Problem 29P

The angle between the incident and emerging rays is $42.5°$.

Explanation of Solution

Write the expression of angle between incident and emerging rays.

    $\gamma ={\theta }_1-{\theta }_2-{\theta }_3+{\theta }_4$                                                                                             (VIII)

Here, $\gamma$ is the angle between incident and emerging rays, ${\theta }_1$ is the angle of incidence at the first surface, ${\theta }_2$ is the angle of refraction at the first surface, and ${\theta }_3$ is the angle of incidence at second surface and ${\theta }_4$ is the angle of refraction at the second surface.

Conclusion:

Substitute $75.0°$ for ${\theta }_1$ and $41.5°$ for ${\theta }_2$, $18.5°$ for ${\theta }_3$ and $27.5°$ for ${\theta }_4$ in equation (VIII) to calculate $\gamma$.

    $\begin{array}{c}\gamma =75.0°-41.5°-18.5°+27.5°\\ =42.5°\end{array}$

Therefore, the angle between the incident and emerging rays is $42.5°$.