A plane sound wave in air at 20°C, with wavelength 589 mm, is incident on a smooth surface of water at 25°C at an angle of incidence of 11.7°.

(a) Determine the angle of refraction for the sound wave.


(b) Determine the wavelength of the sound in water.

m

A narrow beam of sodium yellow light, with wavelength 589 nm in vacuum, is incident from air onto a smooth water surface at an angle of incidence of 11.7°.

(c) Determine the angle of refraction.



(d) Determine the wavelength of the light in water.

Respuesta :

Answer:

(a). The angle of refraction for the sound wave is 61.8°.

(b). The wavelength of the sound in water is 2.56 m.

(c). The angle of refraction is 8.26°.

(d). The wavelength of the light in water is 441.75 nm.

Explanation:

Given that,

Wavelength = 589 mm

Incidence angle = 11.7°

We know that,

The speed of sound in water is GREATER than the speed of sound in air by a factor of about 4.3 times.

The speed of sound wave in water

[tex]v_{w}= 1493\ m/s[/tex]

The speed of sound wave in air at 20°C

[tex]v_{a}= 343\ m/s[/tex]

(a). We need to calculate the  angle of refraction for the sound wave

Using Snell's law

[tex]\dfrac{\sin\theta_{1}}{v_{a}}=\dfrac{\sin\theta_{2}}{v_{w}}[/tex]

Put the value into the formula

[tex]\dfrac{\sin11.7}{343}=\dfrac{\sin\theta_{2}}{1493}[/tex]

[tex]\sin\theta_{2}=\dfrac{\sin11.7}{343}\times1493[/tex]

[tex]\sin\theta_{2}=0.882[/tex]

[tex]\theta_{2}=\sin^{-1}(0.882)[/tex]

[tex]\theta_{2}=61.8^{\circ}[/tex]

The angle of refraction for the sound wave is 61.8°.

(b).  We need to calculate the wavelength of the sound in water

Using formula of wavelength

[tex]\dfrac{v_{w}}{\lambda_{w}}=\dfrac{v_{a}}{\lambda_{a}}[/tex]

Put the value into the formula

[tex]\dfrac{1493}{\lambda_{w}}=\dfrac{343}{0.589}[/tex]

[tex]\lambda_{w}=\dfrac{1493\times0.589}{343}[/tex]

[tex]\lambda_{w}=2.56\ m[/tex]

The wavelength of the sound in water is 2.56 m.

(c). We need to calculate the angle of refraction

Using formula of Snell's law

[tex]n=\dfrac{\sin\theta_{i}}{\sin\theta_{r}}[/tex]

[tex]\dfrac{4}{3}=\dfrac{\sin11.7}{\sin\theta_{r}}[/tex]

[tex]\sin\theta_{r}=\dfrac{\sin11.7\times3}{4}[/tex]

[tex]\theta_{r}=\sin^{-1}(0.1437)[/tex]

[tex]\theta_{r}=8.26^{\circ}[/tex]

The angle of refraction is 8.26°.

(d). We need to calculate the wavelength of the light in water

Using formula of wavelength

[tex]\lambda_{w}=\dfrac{\lambda_{a}}{n}[/tex]

Put the value into the formula

[tex]\lambda_{w}=\dfrac{589}{\dfrac{4}{3}}[/tex]

[tex]\lambda_{w}=441.75\ nm[/tex]

The wavelength of the light in water is 441.75 nm.

Hence, This is the required solution.