Chapter 34, Problem 32AP

Consider a horizontal interface between air above and glass of index of refraction 1.55 below. (a) Draw a light ray incident from the air at angle of incidence 30.0°. Determine the angles of the reflected and refracted rays and show them on the diagram. (b) What If? Now suppose the light ray is incident from the glass at an angle of 30.0°. Determine the angles of the reflected and refracted rays and show all three rays on a new diagram. (c) For rays incident from the air onto the air–glass surface, determine and tabulate the angles of reflection and refraction for all the angles of incidence at 10.0° intervals from 0° to 90.0°. (d) Do the same for light rays coming up to the interface through the glass.

To determine

The angle of the reflected and refracted rays when light ray incident from the air at angle of incidence $30°$ and show them on diagram.

Answer to Problem 32AP

The angle of reflection is $30°$ and the angle of refraction is $18.81°$.

Explanation of Solution

Given info: The index of refraction of glass is $1.55$ and the angle of incidence is $30°$.

From, law of reflection, the angle of incidence is equal to the angle of reflection.

Thus, angle of reflection is $30°$.

From Snell’s law,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ (1)

Here,

${\theta }_1$ is the angle of incidence.

${\theta }_2$ is the angle of refraction.

$n_1$ is the index of refraction of air.

$n_2$ is the index of refraction of glass.

Substitute $1$ for $n_1$, and $1.55$ for $n_2$, $30°$ for ${\theta }_1$, in equation(1) to calculate ${\theta }_2$.

$\begin{array}{c}1\times \sin 30°=1.55\sin {\theta }_2\\ \sin {\theta }_2=\frac{\sin 30°}{1.55}\\ {\theta }_2={\sin }^{-1}\left[\frac{\left(\frac{1}{2}\right)}{1.55}\right]\\ =18.81°\end{array}$

The figure below shows the angle of incidence, the angle of reflection and the angle of refraction.

Figure (1)

Conclusion:

Therefore, the angle of reflection is $30°$ and the angle of refraction is $18.81°$.

To determine

The angle of the reflected and refracted rays when light ray incident from the glass at angle of incidence $30°$ and show them on diagram

Answer to Problem 32AP

The angle of reflection is $30°$ and the angle of refraction is $50.80°$.

Explanation of Solution

Given info: The index of refraction of glass is $1.55$ and angle of incidence is $30°$.

From, law of reflection, angle of incidence is equal to angle of reflection.

Thus, angle of reflection is $30°$.

From, Snell’s law,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ (2)

Here,

$n_1$ is the index of refraction of glass.

$n_2$ is the index of refraction of air.

Substitute $1.55$ for $n_1$, $1$ for $n_2$, $30°$ for ${\theta }_1$, in equation(2) to calculate ${\theta }_2$.

$\begin{array}{c}1.55\times \sin 30°=1\times \sin {\theta }_2\\ \sin {\theta }_2=1.55\times \frac{1}{2}\\ {\theta }_2={\sin }^{-1}\left(0.775\right)\\ =50.80°\end{array}$

The figure below shows the angle of incidence, the angle of reflection and the angle of refraction.

Figure (2)

Conclusion:

Therefore, the angle of reflection is $30°$ and the angle of refraction is $50.80°$.

To determine

The angles of reflection and refraction for all angles of incidence at $10.0°$ from $0°$ to $90°$ in a tabular form for rays incident from air onto the air-glass surface.

Answer to Problem 32AP

The angle of reflection is same as the angle of incidence and the angle of refraction for rays incident from air onto the air-glass surface is increasing as angle of incidence is increasing. The table for different values of angles is,

Angle of incidence	Angle of reflection	Angle of refraction
$0$	$0$	$0$
$10$	$10$	$6.43°$
$20$	$20$	$12.74°$
$30$	$30$	$18.81°$
$40$	$40$	$24.5°$
$50$	$50$	$29.6°$
$60$	$60$	$34°$
$70$	$70$	$37.3°$
$80$	$80$	$39.4°$
$90$	$90$	$40.2°$

Explanation of Solution

From, law of reflection, the angle of reflection is equal to the angle of incidence. So for all value of angles between $0°$ to $90°$ the angle of incidence and angle of reflection are equal.

The formula to calculate the angle of refraction is,

From, Snell’s law,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ (3)

Substitute $1$ for $n_1$, $1.55$ for $n_2$, $10°$ for ${\theta }_1$, in equation (3) to calculate ${\theta }_2$.

$\begin{array}{c}1\times \sin 10°=1.55\sin {\theta }_2\\ \sin {\theta }_2=\frac{\sin 10°}{1.55}\\ {\theta }_2={\sin }^{-1}\left(\frac{0.1736}{1.55}\right)\\ =6.43°\end{array}$,

Substitute $1$ for $n_1$, and $1.55$ for $n_2$, $20°$ for ${\theta }_1$ in equation (3) to calculate ${\theta }_2$.

$\begin{array}{c}1\times \sin 20°=1.55\sin {\theta }_2\\ \sin {\theta }_2=\frac{\sin 20°}{1.55}\\ {\theta }_2={\sin }^{-1}\left(\frac{0.342}{1.55}\right)\\ =12.74°\end{array}$.

The remaining values of the angle of refraction can be calculated by the same method.

The table for angle of reflection and angle of refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ for rays incident from air onto the air-glass surface is,

Angle of incidence	Angle of reflection	Angle of refraction
$0$	$0$	$0$
$10$	$10$	$6.43°$
$20$	$20$	$12.74°$
$30$	$30$	$18.81°$
$40$	$40$	$24.5°$
$50$	$50$	$29.6°$
$60$	$60$	$34°$
$70$	$70$	$37.3°$
$80$	$80$	$39.4°$
$90$	$90$	$40.2°$

Conclusion:

Therefore, the angle of reflection is same as the angle of incidence and the of angle of refraction for rays incident from air onto the air-glass surface is increasing as angle of incidence is increasing.

The table for angle of reflection and angle of refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ for rays incident from air onto the air-glass surface is,

Angle of incidence	Angle of reflection	Angle of refraction
$0$	$0$	$0$
$10$	$10$	$6.43°$
$20$	$20$	$12.74°$
$30$	$30$	$18.81°$
$40$	$40$	$24.5°$
$50$	$50$	$29.6°$
$60$	$60$	$34°$
$70$	$70$	$37.3°$
$80$	$80$	$39.4°$
$90$	$90$	$40.2°$

To determine

The angles of reflection and refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ in a tabular form for rays coming up to the interface through the glass.

Answer to Problem 32AP

The angle of reflection is same as the angle of incidence and the of angle of refraction for rays coming up to the interface through the glass will increase up to angle of incidence $40°$, after that total internal reflection will occur and there will be no refraction.

The table for angle of reflection and angle of refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ for rays coming up to the interface through the glass is,

Angle of incidence( $°$) ( $i$)	Angle of reflection( $°$)	Angle of refraction( $°$) $={\sin }^{-1}\left(1.55\times \sin i\right)$
$0$	$0$	$0$
$10$	$10$	$15.6°$
$20$	$20$	$32°$
$30$	$30$	$50.8°$
$40$	$40$	$85°$
$50$	$50$	$\text{No Refraction}$
$60$	$60$	$\text{No Refraction}$
$70$	$70$	$\text{No Refraction}$
$80$	$80$	$\text{No Refraction}$
$90$	$90$	$\text{No Refraction}$

Explanation of Solution

From, the law of reflection, angle of reflection is equal to the angle of incidence. So, for all value of angles between $0°$ to $90°$ the angle of incidence and angle of reflection are equal.

The formula to calculate the angle of refraction is,

From, Snell’s law,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ (4)

Substitute $1.55$ for $n_1$, and $1$ for $n_2$, $10°$ for ${\theta }_1$, in equation (4) to calculate ${\theta }_2$.

$\begin{array}{c}1.55\times \sin 10°=1\times \sin {\theta }_2\\ \sin {\theta }_2=1.55\times 0.1736\\ {\theta }_2={\sin }^{-1}\left(1.55\times 0.1736\right)\\ =15.6°\end{array}$.

Substitute $1.55$ for $n_1$, and $1$ for $n_2$, $20°$ for ${\theta }_1$, in equation (4) to calculate ${\theta }_2$.

$\begin{array}{c}1.55\times \sin 20°=1\times \sin {\theta }_2\\ \sin {\theta }_2=1.55\times 0.342\\ {\theta }_2={\sin }^{-1}\left(1.55\times 0.342\right)\\ =32°\end{array}$.

The remaining values of the angle of refraction can be calculated by the same method.

The formula to calculate the critical angle is,

${\theta }_{\text{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$ (5)

Here,

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

$n_1$ is the index of refraction of glass.

$n_2$ is the index of refraction of air.

Substitute $1.55$ for $n_1$, and $1$ $n_2$ in equation (5) to calculate ${\theta }_{\text{c}}$.

$\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1}{1.55}\right)\\ =40.2°\end{array}$

The angle of incidence is greater than $40.2°$ will undergo total internal reflection. Hence, no refraction will occur.

The table for angle of reflection and angle of refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ for rays coming up to the interface through the glass is,

Angle of incidence( $°$) ( $i$)	Angle of reflection( $°$)	Angle of refraction( $°$) $={\sin }^{-1}\left(1.55\times \sin i\right)$
$0$	$0$	$0$
$10$	$10$	$15.6°$
$20$	$20$	$32°$
$30$	$30$	$50.8°$
$40$	$40$	$85°$
$50$	$50$	$\text{No Refraction}$
$60$	$60$	$\text{No Refraction}$
$70$	$70$	$\text{No Refraction}$
$80$	$80$	$\text{No Refraction}$
$90$	$90$	$\text{No Refraction}$

Conclusion:

Therefore, the angle of reflection is same as the angle of incidence and the of angle of refraction for rays coming up to the interface through the glass will increase up to angle of incidence $40°$, after that total internal reflection will occur and there will be no refraction.

The table for angle of reflection and angle of refraction for all angles of incidence at $10.0°$ intervals from $0°$ to $90°$ for rays coming up to the interface through the glass is,

Angle of incidence( $°$) ( $i$)	Angle of reflection( $°$)	Angle of refraction( $°$) $={\sin }^{-1}\left(1.55\times \sin i\right)$
$0$	$0$	$0$
$10$	$10$	$15.6°$
$20$	$20$	$32°$
$30$	$30$	$50.8°$
$40$	$40$	$85°$
$50$	$50$	$\text{No Refraction}$
$60$	$60$	$\text{No Refraction}$
$70$	$70$	$\text{No Refraction}$
$80$	$80$	$\text{No Refraction}$
$90$	$90$	$\text{No Refraction}$