How does sound reflect from an interface at an angle?

[snorkack]

How does sound reflect from interface at an angle?
The propagation of light is determined by a single factor - index of refraction.
Sound is affected by two factors - speed and acoustic impedance.

Take the case of two substances with identical acoustic impedance but different sound speeds - one with low density and high sound speed, the other with high density and low sound speed.

The refraction is defined by sound speed for simple geometric reasons. It follows that sound must refract from interface between media of matching impedances but different sound speeds - and refraction includes total internal reflection.

For angles close to total internal reflection angle, is refraction accompanied by partial reflection on a matching impedance interface?

 

[tech99]

Can you explain the last sentence more, please?

 

[snorkack]

When light meets an interface of changing refractive index, and the angle is not such as to cause a total internal reflection, there is a partial reflection. For light, the only case of no reflection is matching refractive indices, and then there is no refraction either.
For sound, how does an interface of matched acoustic impedance, but different sound speed behave? Is refraction then accompanied by splitting of ray through partial reflection?

 

[Baluncore]

For sound, how does an interface of matched acoustic impedance, but different sound speed behave? Is refraction then accompanied by splitting of ray through partial reflection?

If the acoustic impedance is matched then there can be no partial reflection.
If not parallel to the interface, the wavefront will change direction due to the change in the speed of sound.

 

[snorkack]

For the critical angle, will the whole sound ray concentrate to the interface, with no reflection?

 

[Baluncore]

I cannot see how a single ray could be on both sides of the interface at the same time.
The phase difference would result in a shear wave along the velocity discontinuity.

 

[snorkack]

I cannot see how a single ray could be on both sides of the interface at the same time.
The phase difference would result in a shear wave along the velocity discontinuity.

Liquids cannot support shear waves.

 

[Baluncore]

Liquids cannot support shear waves.

You did not specify if the material was liquid or solid. Have you now decided ?

Consider a plane P wavefront propagating through the material along the interface. The energy on the fast side will get ahead of the energy on the slow side. That accumulating phase difference must appear as a shear on the plane of the interface.

 

[snorkack]

For simplicity, consider liquids.
With some searching, I found a good example.
From:
https://www.engineeringtoolbox.com/sound-speed-liquids-d_715.html
Water, 25 C - 1493 m/s
Chloroform, also 25 C - 984 m/s
Density of water is quoted as 997 kg/m³ at 25 C, chloroform at same 25 C - 1489 kg/m³
For round arguments, we may consider an exact match (some tinkering with temperature and solutes) and round it as lower phase 1500 kg/m³, 1000 m/s, upper phase 1000 kg/m³, 1500 m/s.

Chloroform also has a lower speed of light (n=1,44 in chloroform, 1,33 in water).
But light has no impedance - only refractive index. Light ray traveling upwards in chloroform and meeting interface with water will split into two rays - refracted and reflected, even at normal incidence. As the ray approaches more obliquely, the fraction of reflected light increases and refracted light decreases till at critical angle, where the refracted light would have to travel parallel to interface, the fraction of reflected light comes to total and refracted light to zero.

For sound plane wave traveling obliquely from higher density, lower sound speed liquid towards interface with lower density, higher sound speed liquid, if impedance matching forbids partial reflection as in case of light, what precisely happens on the limit where refracted ray must be parallel to interface? All the sound energy that traveled over wide wavefront and reached the interface over a wide area must continue traveling in vanishingly thin layer of water above.

 

[Steve4Physics]

But light has no impedance - only refractive index. Light ray traveling upwards in chloroform and meeting interface with water will split into two rays - refracted and reflected, even at normal incidence.

There may be some misconceptions here. See if this helps...

Light (or more correctly the medium it travels through) does have an impedance. E.g. see https://en.wikipedia.org/wiki/Wave_impedance.

If I remember correctly, the impedance, $Z$, of most non-conducting transparent media is given by $Z≈ \frac {Z_0}{n}$, where $Z_0$ is impedance of free space ($≈377Ω$) and $n$ is the medium's refractive index.

The relative intensities of reflected/transmitted light at an interface are described by Fresnel’s equations and depend on the polarisation of the light as well as refractive indices and the angle of incidence.

I presume there are similar equivalent equations for sound. Maybe someone can advise if they know the name of these equations.

Note, equations of the form $R = \frac {n_1 – n_2}{n_1 + n_2}$ are reflection coefficients for normal incidence, so do not apply (for example) near/at the critical angle: you would need to use Fresnel’s equations

 

[snorkack]

There may be some misconceptions here. See if this helps...

Light (or more correctly the medium it travels through) does have an impedance. E.g. see https://en.wikipedia.org/wiki/Wave_impedance.

If I remember correctly, the impedance, $Z$, of most non-conducting transparent media is given by $Z≈ \frac {Z_0}{n}$, where $Z_0$ is impedance of free space ($≈377Ω$) and $n$ is the medium's refractive index.

That´s my point, though. For light, impedance is uniquely determined by the speed of light, and is therefore not a variable independent of refractive index. For sound, the impedance is a variable independent of speed of sound. Water and chloroform can match acoustic impedances despite having different speeds of sound, but they cannot match wave impedances to light purely because they have different speeds of light.

 

[Steve4Physics]

That´s my point, though. For light, impedance is uniquely determined by the speed of light, and is therefore not a variable independent of refractive index. For sound, the impedance is a variable independent of speed of sound. Water and chloroform can match acoustic impedances despite having different speeds of sound, but they cannot match wave impedances to light purely because they have different speeds of light.

The impedance of a medium carrying light is:
$$Z = \sqrt {\frac {μ₀μᵣ}{ε₀εᵣ}}$$For most materials μᵣ≈ 1 (because they are negligibly magnetic) and the approximate expression for Z becomes:$$Z ≈ \sqrt {\frac {μ₀}{ε₀εᵣ}} = \frac {Z₀}{\sqrt εᵣ} = \frac {Z₀}{n}$$But, avoiding the approximation we must write:$$Z = \frac {Z₀\sqrt μᵣ}{n}$$so, in fact, Z depends on 2 parameters, μᵣ and n.

I think this link, which addresses a similar question, might help:
https://physics.stackexchange.com/questions/375385/confusion-about-reflection-acoustic-impedance

 

[snorkack]

Refractive index depends on μ, too.

 

[Steve4Physics]

Refractive index depends on μ, too.

Yes, I think $n = \sqrt{\mu_r \epsilon_r}$, so n depends on 2 independent parameters. But typically $\mu_r≈1$.

 

[jasonRF]

the oblique incidence acoustic wave problem really isn't so different from the electromagnetic problem - the primary difference is that electromagnetic waves have polarization.

Note that for electromagnetic wave the relationship between wave impedance ($Z$) and speed ($v$) are $$Z_{em} = \mu \, v_{em} = \sqrt{\mu/\epsilon}$$ where $\mu$ is the permiability and $\epsilon$ is permittivity. For acoustic waves the relationship is $$Z_a = \rho \, v_{a} = \sqrt{K\, \rho}$$, where $K$ is the bulk modulus and $\rho$ is the density.

In both sound and electromagnetic waves, the angle of the transmitted wave is determined by a Snell's law that depends on the incidence angle and the ratio of the wave speeds. This can be derived by phase-matching the plane waves at the interface. So for oblique (non-normal) incidence, the transmitted wave will be in a different direction than the incident wave if the wave speeds are different.

Likewise, the reflection coefficients depend on impedances of the two media, as well as the incident and transmitted angles (and hence the wave speeds). You derive the reflection coefficent (and transmission coefficient) applying the boundary conditions at the interface. I believe the form of the reflection coefficients are the same for acoustic waves and TE polarization electromagnetic waves. Note that even if the wave impedances of the two media are the same, if the wave speeds are different then there will be a reflection for oblique (non-normal) incidence since the transmitted angle is not the same as the incident angle.

For some relevant notes, see
https://mitocw.ups.edu.ec/courses/e...netics-and-applications-spring-2009/readings/

For acoustic waves see Chapter 13, equations 13.2.9 (Snell) and 13.2.19 (reflection).

For electromagnetic waves see Chapter 9, equations 9.2.26 (Snell), 9.2.28 (reflection for TE polarization), and 9.2.72 (reflection for TM polarization).

Jason

 

[snorkack]

the oblique incidence acoustic wave problem really isn't so different from the electromagnetic problem - the primary difference is that electromagnetic waves have polarization.

So do sound waves.
A difference is that the longitudinal polarization is forbidden for electromagnetic waves (leaving two transverse polarizations) but always allowed for sound (leaving only the longitudinal polarization for fluids, but all three for solids).
Do interfaces polarize reflected sound in isotropic solids exactly as they do in case of electromagnetic waves?

 

[hutchphd]

Depending upon the solid, there are a host of special waves at interfaces. It ain't simple. For instance Rayleigh Waves

 

[jasonRF]

So do sound waves.
A difference is that the longitudinal polarization is forbidden for electromagnetic waves (leaving two transverse polarizations) but always allowed for sound (leaving only the longitudinal polarization for fluids, but all three for solids).

I should have specified that I was only thinking about linear, longitudinal acoustic waves in isotropic fluids that do not support shear. After posts 7 and 9, I thought that was what you wanted to focus on. But as you point out things are more complicated in a more general solid.

Do interfaces polarize reflected sound in isotropic solids exactly as they do in case of electromagnetic waves?

I don't know, as I have never read-up on these things or tried to work it out. At first glance it isn't obvious to me that an incident shear wave would only reflect and transmit shear waves; perhaps it would also reflect and/or transmit longitudinal waves (or perhaps not)? Have you done any reading on the subject?

jason