Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction ${\theta }_{\text{s}}$ in the sapphire crystal.

Answer to Problem 96PQ

The angle of refraction ${\theta }_{\text{s}}$, in the sapphire crystal is $15.0°$.

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

    $n_{\text{a}}\sin {\theta }_{\text{a}}=n_{\text{s}}\sin {\theta }_{\text{s}}$                                                                             (I)

Here, $n_{\text{a}}$ is the refractive index of the air medium, $n_{\text{s}}$ is the refractive index of the sapphire crystal medium, ${\theta }_{\text{a}}$ is the angle of incidence of air and ${\theta }_{\text{s}}$ is the angle of refraction of sapphire crystal.

Solve the above equation to find ${\theta }_{\text{s}}$.

    $\begin{array}{c}{n}_{\text{a}}\sin {\theta }_{\text{a}}=n_{\text{s}}\sin {\theta }_{\text{s}}\\ \sin {\theta }_{\text{s}}=\sin {\theta }_{\text{a}}\left(\frac{n_{\text{a}}}{n_{\text{s}}}\right)\\ {\theta }_{\text{s}}={\sin }^{-1}\left(\sin {\theta }_{\text{a}}\left(\frac{n_{\text{a}}}{n_{\text{s}}}\right)\right)\end{array}$                                                            (II)

Conclusion:

Substitute 1.760 for $n_{\text{s}}$, $27.0°$ for ${\theta }_{\text{a}}$, $1.00$ for $n_{\text{a}}$ in equation (II).

Here, $n_{\text{s}}=1.495$ and ${\theta }_{\text{a}}=27.0°$ are taken from given data and $n_{\text{a}}=1.00$ for the air medium.

$\begin{array}{c}{\theta }_{\text{s}}={\sin }^{-1}\left(\sin {\theta }_{\text{a}}\left(\frac{n_{\text{a}}}{n_{\text{s}}}\right)\right)\\ ={\sin }^{-1}\left(\sin 27.0°\left(\frac{1.00}{1.760}\right)\right)\\ ={\sin }^{-1}\left(0.45390\left(0.56818\right)\right)\\ =15.0°\end{array}$

Therefore, the angle of refraction ${\theta }_{\text{s}}$, in the sapphire crystal is $15.0°$.

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$.

Explanation of Solution

Consider the diagram.

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is ${\theta }_{\text{a}}=27.0°$ and the angle of refraction of the light ray inside the crystal is ${\theta }_{\text{s}}$.

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface ${\theta }_{\text{s}}$ is equal to the angle of incidence at the sapphire interface ${\theta }_{\text{a}}$.

    ${\theta }_{\text{s}}={\theta }_{\text{a}}$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$.

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$.

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

    $n_{\text{a,1}}\sin {\theta }_{\text{a,1}}=n_{\text{s,1}}\sin {\theta }_{\text{s,1}}$ (III)

Here, $n_{\text{a}}$ is the refractive index of the air medium, $n_{\text{s}}$ is the refractive index of the sapphire crystal medium, ${\theta }_{\text{a}}$ is the angle of incidence of air and ${\theta }_{\text{s}}$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $\sin {\theta }_{\text{s,1}}$.

    $\begin{array}{c}{n}_{\text{a,1}}\sin {\theta }_{\text{a,1}}=n_{\text{s,1}}\sin {\theta }_{\text{s,1}}\\ \sin {\theta }_{\text{s,1}}=\sin {\theta }_{\text{a,1}}\left(\frac{n_{\text{a,1}}}{n_{\text{s,1}}}\right)\\ {\theta }_{\text{s,1}}={\sin }^{-1}\left(\sin {\theta }_{\text{a,1}}\left(\frac{n_{\text{a,1}}}{n_{\text{s,1}}}\right)\right)\end{array}$                                                                    (IV)

Conclusion:

Substitute $1.00$ for $n_{\text{s,1}}$, ${14.94}^{°}$ for ${\theta }_{\text{a,1}}$,1.760 for $n_{\text{a,1}}$ in equation (IV).

    $\begin{array}{c}{\theta }_{\text{s,1}}={\sin }^{-1}\left(\sin {\theta }_{\text{a,1}}\left(\frac{n_{\text{a,1}}}{n_{\text{s,1}}}\right)\right)\\ ={\sin }^{-1}\left(\sin 14.94°\left(\frac{1.760}{1.00}\right)\right)\\ ={\sin }^{-1}\left(\left(0.2578\right)\left(1.760\right)\right)\\ =27.0°\end{array}$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$.