Question

A plane sound wave in air at $$ 20^{\circ} \mathrm{C}, $$ with wavelength $$ 589 \mathrm{mm}, $$ is incident on a smooth surface of water at $$ 25^{\circ} \mathrm{C} $$ at an angle of incidence of $$ 13.0^{\circ} . $$ Determine
(a) the angle of refraction for the sound wave and
(b) the wavelength of the sound in water. A narrow beam of sodium yellow light, with wavelength $$ 589 \mathrm{nm} $$ in a vacuum, is incident from air onto a smooth water surface at an angle of incidence of $$ 13.0^{\circ}. $$ Determine
(c) the angle of refraction and
(d) the wavelength of the light in water.
(e) Compare and contrast the behavior of the sound and light waves in this problem.

Answer 1

(a) The law of refraction $$ n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} $$ can be put into the more general form

$$\begin{aligned}\dfrac{c}{v_{1}} \sin \theta_{1} &=\dfrac{c}{v_{2}} \sin \theta_{2} \\\dfrac{\sin \theta_{1}}{v_{1}} &=\dfrac{\sin \theta_{2}}{v_{2}}\end{aligned}$$

This is equivalent to Equation $$ 35.3 . $$ This form applies to all kinds of waves that move through space.

In air at $$ 20^{\circ} \mathrm{C}, $$ the speed of sound is $$ 343 \mathrm{m} / \mathrm{s} $$. From Table $$ 17.1, $$ the speed of sound in water at $$ 25.0^{\circ} \mathrm{C} $$ is $$ 1493 \mathrm{m} / \mathrm{s} $$. The angle of incidence is $$ 13.0^{\circ}: $$

$$\begin{array}{c}\dfrac{\sin 13.0^{\circ}}{343 \mathrm{m} / \mathrm{s}}=\dfrac{\sin \theta_{2}}{1493 \mathrm{m} / \mathrm{s}} \\\theta_{2}=78.3^{\circ}\end{array}$$

(b) The wave keeps constant frequency in all media:

$$\begin{array}{l}f=\dfrac{v_{1}}{\lambda_{1}}=\dfrac{v_{2}}{\lambda_{2}} \\\lambda_{2}=\dfrac{v_{2} \lambda_{1}}{v_{1}}=\dfrac{1493 \mathrm{m} / \mathrm{s}(0.589 \mathrm{m})}{343 \mathrm{m} / \mathrm{s}}=2.56 \mathrm{m}\end{array}$$

(c) Using Snell's law,

$$n_{2} \sin \theta_{2}=n_{1} \sin \theta_{1}$$

$$\begin{array}{c}1.333 \sin \theta_{2}=1.000293 \sin 13.0^{\circ} \\\theta_{2}=9.72^{\circ}\end{array}$$

(d) $$ \quad \lambda_{2}=\dfrac{\mathrm{v}_{2} \lambda_{1}}{\mathrm{v}_{1}}=\dfrac{n_{1} \lambda_{1}}{n_{2}}=\dfrac{1.000293(589 \mathrm{nm})}{1.333}=442 \mathrm{nm} $$

(e) The light wave slows down as it move to water, but the sound wave speeds up by a large factor. The light wave bends toward the normal and its wavelength shortens, but the sound wave bends away from the normal and its wavelength increases