Chapter 35, Problem 75AP

(a)

To determine

The angle ${\theta }_2$ of the emerging beam.

Answer to Problem 75AP

The angle ${\theta }_2$ of the emerging beam is $53.1°$.

Explanation of Solution

Apply Snell’s law for the first layer.

    $n_1\sin {\theta }_1=n_2\sin {\theta }_2^{\text{'}}$                                                                                                      (I)

Here, $n_1$ is the refractive index of the first layer, ${\theta }_1$ is the incident angle, $n_2$ is the refractive index of the second layer and ${\theta }_2^{\text{'}}$ is the emerging angle for the second layer.

Apply Snell’s law for the second layer.

    $n_2\sin {\theta }_2^{\text{'}}=n_3\sin {\theta }_3$                                                                                                    (II)

Here, $n_2$ is the refractive index of the second layer and ${\theta }_2^{\text{'}}$ is the incident angle, $n_3$ is the refractive index of the third layer and ${\theta }_3$ is the emerging angle.

Apply Snell’s law for the third layer.

    $n_3\sin {\theta }_3=n\sin {\theta }_2$                                                                                                    (III)

Here, $n_3$ is the refractive index of the third layer, ${\theta }_3$ is the incident angle, $n$ is the refractive index of air and ${\theta }_2$ is the emerging angle.

Conclusion:

Substitute $1.60$ for $n_1$, $30°$ for ${\theta }_1$ and $1.40$ for $n_2$ in equation (I) to calculate ${\theta }_2^{\text{'}}$.

    $\begin{array}{c}\left(1.60\right)\sin \left(30°\right)=\left(1.40\right)\sin {\theta }_2^{\text{'}}\\ \sin {\theta }_2^{\text{'}}=\frac{1.60}{1.40}\sin 30°\\ {\theta }_2^{\text{'}}={\sin }^{-1}\left(0.5714\right)\\ =34.8°\end{array}$

Substitute $1.40$ for $n_2$, $34.8°$ for ${\theta }_2^{\text{'}}$ and $1.20$ for $n_3$ in equation (II)  to calculate ${\theta }_3$.

    $\begin{array}{c}\left(1.40\right)\sin \left(34.8°\right)=\left(1.20\right)\sin {\theta }_3\\ \sin {\theta }_3=\frac{1.40}{1.20}\sin 34.8°\\ {\theta }_3={\sin }^{-1}\left(0.6628\right)\\ =41.74°\end{array}$

Substitute $1.20$ for $n_3$, $41.74°$ for ${\theta }_3$ and $1$ for $n$ in equation (III) to calculate ${\theta }_2$.

    $\begin{array}{c}\left(1.20\right)\sin \left(41.74°\right)=\left(1\right)\sin {\theta }_2\\ \sin {\theta }_2=\frac{1.20}{1}\sin \left(41.74°\right)\\ {\theta }_2={\sin }^{-1}\left(0.7989\right)\\ =53.1°\end{array}$

Therefore, the angle ${\theta }_2$ of the emerging beam is $53.1°$.

(b)

To determine

The value of incident angle ${\theta }_1$ to have total internal reflection at the surface between the medium with $n=1.20$ and the medium with $n=1.00$.

Answer to Problem 75AP

The value of incident angle ${\theta }_1$ to have total internal reflection at the surface between the medium with $n=1.20$ and the medium with $n=1.00$ should be greater than $38.7°$.

Explanation of Solution

Apply Snell’s law for the third layer.

    $n_3\sin {\theta }_{\text{c}}=n\sin {\theta }_2$                                                                                                    (IV)

Here, ${\theta }_{\text{c}}$ is the critical angle.

Apply Snell’s law for the second layer.

    $n_3\sin {\theta }_{\text{i}}=n_2\sin {\theta }_{\text{r}}$                                                                                                     (V)

Here, ${\theta }_{\text{r}}$ is the refracted angle.

Apply Snell’s law for the first layer.

    $n_2\sin {\theta }_{\text{i}}^{\text{'}}=n_1\sin {\theta }_{\text{r}}^{\text{'}}$                                                                                                   (VI)

Here, ${\theta }_{\text{i}}^{\text{'}}$ is the incident angle and ${\theta }_{\text{r}}^{\text{'}}$ is the refracted angle.

Conclusion:

Substitute $1$ for $n$, $1.20$ for $n_3$ and $90°$ for ${\theta }_2$ in equation (IV) to calculate ${\theta }_{\text{c}}$.

    $\begin{array}{c}\left(1.20\right)\sin {\theta }_{\text{c}}=\left(1\right)\sin 90°\\ \sin {\theta }_{\text{c}}=\frac{1}{1.20}\sin 90°\\ {\theta }_{\text{c}}={\sin }^{-1}\left(0.8333\right)\\ =56.44°\end{array}$

Substitute $1.40$ for $n_3$, $1.20$ for $n_2$ and $56.44°$ for ${\theta }_{\text{r}}$ in equation (V) to calculate ${\theta }_{\text{i}}$.

    $\begin{array}{c}\left(1.40\right)\sin {\theta }_{\text{i}}=\left(1.20\right)\sin 56.44°\\ \sin {\theta }_{\text{i}}=\frac{1.20}{1.40}\sin 56.44°\\ {\theta }_{\text{i}}={\sin }^{-1}\left(0.7142\right)\\ =45.58°\end{array}$

Substitute $1.60$ for $n_2$ and $1.40$ for $n_1$ and $45.68°$ for ${\theta }_{\text{r}}^{\text{'}}$ in equation (VI) to calculate ${\theta }_{\text{i}}^{\text{'}}$.

    $\begin{array}{c}\left(1.60\right)\sin {\theta }_{\text{i}}^{\text{'}}=\left(1.40\right)\sin 45.68°\\ \sin {\theta }_{\text{i}}^{\text{'}}=\frac{1.40}{1.60}\sin 45.68°\\ {\theta }_{\text{i}}^{\text{'}}={\sin }^{-1}\left(0.626\right)\\ =38.7°\end{array}$

Therefore, the value of incident angle ${\theta }_1$ to have total internal reflection at the surface between the medium with $n=1.20$ and the medium with $n=1.00$ should be greater than $38.7°$.