Chapter 23, Problem 84P

(a)

To determine

The refractive index of the unknown substance.

Answer to Problem 84P

The refractive index of the unknown substance is $\underset{\_}{1.42}$.

Explanation of Solution

Given that the angle of reflection is $25.0°$, angle of refraction is $37.0°$.

According to the laws of reflection the angle of incidence equals the angle of reflection.

Write the expression of Snell’s law of refraction.

    $n_1\sin {\theta }_{\text{1}}=n_2\sin {\theta }_{\text{2}}$                                                                                                     (I)

Here, $n_1$ is the refractive index of the first medium, ${\theta }_{\text{1}}$ is the angle of incidence, $n_2$ is the refractive index of the second medium, and ${\theta }_{\text{2}}$ is the angle of refraction.

According to the law of reflection, the angle of incidence is equal to the angle of reflection. Thus, in the given case, the angle of incidence is $25.0°$. In the given system, the first medium is the unknown substance and the second medium is air ($n_2=1.000$).

Solve equation (I) for $n_1$.

    $n_1=\frac{n_2\sin {\theta }_{\text{2}}}{\sin {\theta }_1}$                                                                                                           (II)

Conclusion:

Substitute $1.000$ $n_2$, $37.0°$ for ${\theta }_2$ and $25.0°$ for ${\theta }_1$ in the equation (IV) to find $n_1$.

    $\begin{array}{c}{n}_1=\frac{1.000\sin 37.0°}{\sin 25°}\\ =1.42\end{array}$

Therefore, the refractive index of the unknown substance is $\underset{\_}{1.42}$.

(b)

To determine

The speed of light in the unknown substance.

Answer to Problem 84P

The speed of the light in the unknown substance is $\underset{\_}{2.11\times {10}^8\text{m}/\text{s}}$.

Explanation of Solution

It is obtained that the refractive index of the unknown substance is $1.42$.

Write the expression for the speed of light in a medium

    $v=\frac{c}{n}$                                                                                                                      (III)

Here, $v$ is the speed of the light in the medium, $c$ is the speed of light in vacuum, $n$ is the refractive index of the medium.

Conclusion:

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for $c$, $1.42$ for $n$ in equation (III) to find the speed of light in the unknown substance.

    $\begin{array}{c}v=\frac{3.00\times {10}^8\text{m}/\text{s}}{1.42}\\ =2.11\times {10}^8\text{m}/\text{s}\end{array}$

Therefore, the speed of the light in the unknown substance is $\underset{\_}{2.11\times {10}^8\text{m}/\text{s}}$.

(c)

To determine

The minimum angle of incidence at which the light would undergo total internal reflection.

Answer to Problem 84P

The minimum angle of incidence at which the light would undergo total internal reflection is $\underset{\_}{44.6°}$.

Explanation of Solution

When a light passes from a high refractive index medium to a lower refractive index medium, and if the angle of incidence is greater than the critical angle, the total internal reflection occurs.

Write the expression for the critical angle.

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

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

Use equation (II) in equation (IV).

    $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\frac{n_2}{\left(\frac{n_2\sin {\theta }_2}{\sin {\theta }_1}\right)}\\ ={\sin }^{-1}\frac{\sin {\theta }_1}{\sin {\theta }_2}\end{array}$                                                                                              (V)

Conclusion:

Substitute $25.0°$ for ${\theta }_1$ and $37.0°$ for ${\theta }_2$ in equation (VII) to find ${\theta }_{\text{c}}$.

    $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\left(\frac{\sin 25.0°}{\sin 37.0°}\right)\\ =44.6°\end{array}$

Therefore, the minimum angle of incidence at which the light would undergo total internal reflection is $\underset{\_}{44.6°}$.