Chapter 25, Problem 4P

(a)

To determine

The angle of refraction for the sound wave.

Answer to Problem 4P

The angle of refraction for sound wave is $87.43°$.

Explanation of Solution

Given info: The wavelength of sound wave is $589\text{nm}$ and angle of incidence is $13.0°$.

The speed of sound in air at $20°\text{C}$ is $343.216\text{m}/\text{s}$ and speed of sound in water at $25°\text{C}$ is $1531\text{m}/\text{s}$.

The expression for the Snell’s law is,

${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$

Here,

${\mu }_1$ is the refractive index of sound in air.

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

${\mu }_2$ is refractive index of sound in water.

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

Rearrange the above formula to find ${\theta }_2$.

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

The formula to calculate speed of sound in water is,

$\frac{{\mu }_1}{{\mu }_2}=\frac{v_2}{v_1}$ (2)

Here,

${\mu }_1$ is the refractive index of sound in air.

${\mu }_2$ is the refractive index of sound in water.

$v_1$ is the speed of sound in air.

$v_2$ is the speed of sound in water.

Substitute $\frac{v_2}{v_1}$ for $\frac{{\mu }_1}{{\mu }_2}$ in formula (1) as,

${\theta }_2={\sin }^{-1}\left(\frac{v_2}{v_1}\sin {\theta }_1\right)$

Substitute $343.216\text{m}/\text{s}$ for $v_1$, $1531\text{m}/\text{s}$ for $v_2$, $13.0°$ for ${\theta }_1$ in the above formula as,

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{v_2}{v_1}\sin {\theta }_1\right)\\ ={\sin }^{-1}\left(\frac{1531\text{m}/\text{s}}{343.216\text{m}/\text{s}}\sin 13°\right)\\ ={\sin }^{-1}\left(4.460\times 0.224\right)\\ =87.43°\end{array}$

Conclusion:

Therefore, the angle of refraction for the sound wave is $87.43°$

(b)

To determine

The wavelength of sound in water.

Answer to Problem 4P

The wavelength of sound in water is $2627.38\text{nm}$.

Explanation of Solution

Given info: The wavelength of sound wave is $589\text{nm}$ and angle of incidence is $13.0°$.

The formula to calculate the wavelength is,

$\frac{v_1}{{\lambda }_1}=\frac{v_2}{{\lambda }_2}$

Here,

$v_1$ is the speed of sound wave in air.

$v_2$ is the speed of sound wave in water.

${\lambda }_1$ is the wavelength of sound wave in air.

${\lambda }_2$ is the wavelength of sound in water.

Rearrange the above formula to find ${\lambda }_2$ as,

$\begin{array}{c}\frac{v_1}{{\lambda }_1}=\frac{v_2}{{\lambda }_2}\\ {\lambda }_2=\frac{v_2}{v_1}{\lambda }_1\end{array}$

Substitute $343.216\text{m}/\text{s}$ for $v_1$, $1531\text{m}/\text{s}$ for $v_2$, $589\text{nm}$ for ${\lambda }_1$ in the above formula as,

$\begin{array}{c}{\lambda }_2=\frac{v_2}{v_1}{\lambda }_1\\ =\frac{1531\text{m}/\text{s}}{343.216\text{m}/\text{s}}\left(589\text{nm}\right)\\ =2627.38\text{nm}\end{array}$

Conclusion:

Therefore, the wavelength of sound in water is $2627.38\text{nm}$

(c)

To determine

The angle of refraction.

Answer to Problem 4P

The angle of refraction is $9.67°$.

Explanation of Solution

Given info: The wavelength of sodium yellow light is $589\text{nm}$ and angle of incidence is $13.0°$.

The formula to calculate the Snell’s law is,

${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$

Here,

${\mu }_1$ is the refractive index for air.

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

${\mu }_2$ is refractive index for water.

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

Rearrange the above formula to find ${\theta }_2$ as,

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

Substitute $1$ for ${\mu }_1$, $1.33$ for ${\mu }_2$, $13°$ for ${\theta }_1$ in the above formula as,

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\sin {\theta }_1\right)\\ ={\sin }^{-1}\left(\frac{1}{1.33}\sin 13°\right)\\ ={\sin }^{-1}\left(0.751\times 0.224\right)\\ =9.67°\end{array}$

Conclusion:

Therefore, the angle of refraction is $9.67°$.

(d)

To determine

The wavelength of light in water.

Answer to Problem 4P

The wavelength of light in water is $442.85\text{nm}$.

Explanation of Solution

Given info: The wavelength of sodium yellow light is $589\text{nm}$ and angle of incidence is $13.0°$.

The formula to calculate the wavelength is,

$\frac{{\mu }_2}{{\mu }_1}=\frac{{\lambda }_1}{{\lambda }_2}$

Here,

${\mu }_1$ is the refractive index of light in air.

${\mu }_2$ is the refractive index of light in water.

${\lambda }_1$ is the wavelength of light in air.

${\lambda }_2$ is the wavelength of light in water.

Rearrange the above formula to find ${\lambda }_2$ as,

$\begin{array}{c}\frac{{\mu }_2}{{\mu }_1}=\frac{{\lambda }_1}{{\lambda }_2}\\ {\lambda }_2=\frac{{\mu }_1}{{\mu }_2}{\lambda }_1\end{array}$

Substitute $1$ for ${\mu }_1$, $1.33$ for ${\mu }_2$, $589\text{nm}$ for ${\lambda }_1$ in the above formula as,

$\begin{array}{c}{\lambda }_2=\frac{{\mu }_1}{{\mu }_2}{\lambda }_1\\ =\frac{1}{1.33}\left(589\text{nm}\right)\\ =442.85\text{nm}\end{array}$

Conclusion:

Therefore, the wavelength of light in water is $442.85\text{nm}$.

(e)

To determine

The behavior of sound and light waves.

Answer to Problem 4P

The behavior of sound and light waves is that the sound waves speeds up when travelling from rarer medium to denser medium and light rays slows down.

Explanation of Solution

Given info: The wavelength of sodium yellow light is $589\text{nm}$ and angle of incidence is $13.0°$.

The medium that has low refractive index with respect to another medium is called rarer medium and the medium that has high refractive index with respect to another medium is called denser medium.

From part (b) the wavelength of sound wave in water is larger than the wavelength of sound in air and from part (d) the wavelength of light in water is less than the wavelength of light in air.

The sound waves travelling from rarer to denser medium then the refracted sound waves bend away from the normal that the reason there is an increase in wavelength of sound waves.

The ray of light travelling from rarer to denser medium then the refracted rays bends towards the normal. So that’s the reason there is a decrease in the wavelength of light in water.

Conclusion:

Therefore, The behavior of sound and light waves is that the sound waves speeds up when travelling from rarer medium to denser medium and light rays slows down.