Chapter 23, Problem 77P

(a)

To determine

The angle between the refracted ray and the glass surface.

Answer to Problem 77P

The angle ${\theta }_1$ is $50.1°$.

Explanation of Solution

The angle of between the incident ray and the glass surface is $30°$ and refractive index of the glass is $1.35$.

Write the expression for Snell’s law

  $n_{\text{i}}\sin {\theta }_{\text{i}}=n_{\text{r}}\sin {\theta }_{\text{r}}$                                                                    (I)

Here, ${\theta }_{\text{i}}$ is the angle between surface normal and the incident ray, ${\theta }_{\text{r}}$ is the angle between surface normal and the refracted ray, $n_{\text{i}}$ is the refractive index of the air and $n_{\text{r}}$ is the refractive index of the glass.

Substitute $90°-30°$ for ${\theta }_{\text{i}}$, $90°-{\theta }_1$ for ${\theta }_{\text{r}}$, $1.00$ for $n_{\text{i}}$ and $1.35$ for $n_{\text{r}}$ in (I)

  $\left(1.00\right)\sin \left(90°-30°\right)=\left(1.35\right)\sin \left(90°-{\theta }_{\text{1}}\right)$                                      (II)

Rewrite

  $\left(1.00\right)\sin \left(60°\right)=\left(1.35\right)\cos {\theta }_{\text{1}}$

Solve for ${\theta }_{\text{1}}$

  $\begin{array}{c}{\theta }_{\text{1}}={\cos }^{-1}\left[\frac{\left(1.00\right)\sin \left(60°\right)}{\left(1.35\right)}\right]\\ =50.1°\end{array}$

Thus, the angle ${\theta }_1$ is $50.1°$.

(b)

To determine

The critical angle between glass and air.

Answer to Problem 77P

The critical angle between glass and air is $53°$.

Explanation of Solution

Write the expression for critical angle between glass and air using Snell’s law

  $n_{\text{r}}\sin {\theta }_{\text{c}}=n_{\text{i}}\sin 90°$                                                                    (III)

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

Rearrange for ${\theta }_{\text{c}}$

  ${\theta }_{\text{c}}={\sin }^{-1}\frac{n_{\text{i}}}{n_{\text{r}}}$                                                                   (IV)

Substitute $1.00$ for $n_{\text{i}}$ and $1.35$ for $n_{\text{r}}$ in (IV) to find ${\theta }_{\text{c}}$

  $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1.00}{1.35}\right)\\ =47.8°\end{array}$

Thus, the critical angle between glass and air is $53°$.

(c)

To determine

The path of light.

Answer to Problem 77P

The light follows path $A$ only.

Explanation of Solution

The angle of incidence at the bottom horizontal surface is same as ${\theta }_1$ and ${\theta }_1<{\theta }_{\text{c}}$.

Since the angle of incidence is greater than the critical angle, the ray undergoes total internal reflection at the surface and hence there is no transmitted ray.

Thus, the light follows path $A$ and not path $B$ given in the figure.

(d)

To determine

The angle ${\theta }_{\text{A}}$ and ${\theta }_{\text{B}}$ for the given path of light.

Answer to Problem 77P

The angle ${\theta }_{\text{A}}$ is $39.9°$.

Explanation of Solution

The light doesn’t follow path $B$ because of total internal reflection.

Write the expression for the angle ${\theta }_{\text{A}}$ using the above figure.

  ${\theta }_{\text{A}}+{\theta }_{\text{1}}=90°$

Rearrange for ${\theta }_{\text{A}}$

  ${\theta }_{\text{A}}=90°-{\theta }_{\text{1}}$                                                                     (V)

Substitute $50.1°$ for ${\theta }_{\text{1}}$ in (V) to find ${\theta }_{\text{A}}$

  $\begin{array}{c}{\theta }_{\text{A}}=90°-50.1°\\ =39.9°\end{array}$

Thus, the angle ${\theta }_{\text{A}}$ is $39.9°$.