Sound diffraction through a single slit

When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves in the corresponding experiment?

• – Ben Crowell Oct 22 '14 at 20:45

The answer is the two are very close, but there are theoretical differences. It is VERY hard to construct actual scenarios where linear acoustic diffraction differs substantially from the diffraction of light.

As noted already, linear acoustic waves are scalar pressure fields, whereas electromagnetism is a vector phenomenon, i.e. there is a polarisation state light that does not apply to acoustic waves.

1. (Likeness) The Cartesian components of the electromagnetic field vectors, as well as the Cartesian components of the potential four vector all fulfill the D'Alembert wave equation (Helmholtz equation in the time harmonic case). The propagation for linear acoustic waves is also the D'Alembert wave equation;

2. (Difference) In co-ordinate systems other than Cartesian, the components of the electromagnetic field do not in general fulfill the D'Alembert wave equation.

3. (Likeness) For many kinds of problem, 2. is not a big difference, either because light either has a predominant polarisation in a constant (non spatially varying) direction (so the only significant component is a Cartesian one), or because it is a classical probabalistic mixture of pure polarisation states with such predominant, non spatially varying direction polarisation. This idea can be generalised to any pure polarisation state: there are orthonormal (complex) transformations on the co-ordinate axes which keep the D'Alembert wave equation - see for example the Riemann-Silberstein vectors, whose Cartesian components fulfill the D'Alembert wave equation. Instead of x and y components, they have left and right circular components.

4. (Difference) The intensity of the sound field is the square of the scalar field, i.e. the square of a field fulfilling the D'Alembert wave equation. The intensity of an electromagnetic field is the inner product between the Poynting vector and the surface through which the energy flux is being defined. It can also be construed as the energy density $\frac{1}{2}\epsilon\,|\vec{E}^2| + \frac{1}{2}\mu\,|\vec{H}^2|$ times the speed of light in the medium in question. Thus it is a more general quadratic function of several different scalar fields, each fulfilling the D'Alembert wave equation.

5. (Likeness) Again in many situations, there is one predominant direction of the electromagnetic field, thus one Cartesian component seems to hold sway and so we again get a square of a single scalar field fulfilling the D'Alembert wave equation as the approximate intensity. Thus for both fields the square of a scalar is the intensity- this holds exactly for acoustics, but only approximately for electromagnetism;

6. (Difference) The main difference to my mind is the boundary conditions. For acoustics, the pressure and normal velocity component are continuous across interfaces. These are simple scalars. For the EM field, the polarisations bear critically on the behaviour at interfaces. The acoustic reflexion and transmission co-efficients bear strong likenesses to the Fresnel equations for light, but there is only one reflexion and one transmission co-efficient. For light, these two co-efficients vary widely, depending on the incident polarisation, as described by the Fresnel equations.

7. (Likeness) The Snell law, Eikonal equation and raytracing apply equally in both cases, being valid approximations under exactly analogous conditions.

Leaving the vector nature of light aside: a great deal of optical diffraction theory is scalar theory. Scalar optical diffraction theory is valid especially in the paraxial (fields comprising only a narrow spread of wavevector directions) case, and when one Cartesian component of a field vector is predominant.

There is no difference between scalar optical diffraction theory, and the theory of linear acoustic diffraction

When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves in the corresponding experiment?

I think the answer is almost, but not quite. It's a little complicated.

1. First off, there's the question of whether you're talking about the far field (Fraunhofer diffraction) or the near field (Fresnel diffraction). In the Fraunhofer approximation, we can take all the waves reaching a given point to be plane waves arriving from the same direction. But in the near field, the waves being superposed can arrive from different directions, so depending on the polarization, you can get nontrivial vector addition of the fields. This causes an actual change in the shape of the diffraction pattern, not just a renormalization.

2. Huygens' principle doesn't say anything about polarization, so if you believed in Huygens' principle, then you would think that all results in diffraction would be independent of polarization. But Huygens' principle is an approximation. One example where it fails is in the case where the slit is small compared to the wavelength: Diffraction by small holes From Huygens' principle, you'd expect that 100% of the power impinging on the slit would emerge as a diffracted wave. But in reality the transmission in this limit is proportional to $(a/\lambda)^n$, where $a$ is the width of the slit and the exponent $n$ depends on the multipolarity of the wave. As explained in an answer to the linked question, we have $n=4$ for light waves, but $n=2$ for sound. This causes a difference in normalization in the two cases.

3. Finally, a physically realistic diffracting object will interact with the polarization of the wave. For example, a conducting metal slit will interact differently with a wave depending on the direction of the wave's polarization. These are basically the same phenomena you get in a polarizing filter. If the slit has a finite thickness, it can act like a waveguide, and waveguides are sensitive to polarization.

These are three independent, well known, noncontroversial reasons why the diffraction can and will be different for light than for sound. Having said all that, I think that under most circumstances the two diffraction patterns would look almost the same. For $\lambda \ll a$, I think the normalization would even be the same, in the sense that the unitless transmission factors would both be approximately 1. It would be an interesting project to try to detect some of the differences in the lab. I don't think it would be easy at all.

• If it is truly a "corresponding experiment" than you must look at the case where the edges of the slit are not conductive since there is no obvious equivalent mechanism for sound waves. But I suppose that depends on your interpretation of "equivalent experiment". – Floris Oct 22 '14 at 21:10
• @Floris: You could also worry about the dielectric constant of the material, etc. You can't just assume a material with no electromagnetic properties -- such a material would be perfectly transparent. – Ben Crowell Oct 22 '14 at 21:21
• I don't think that assuming a perfect slit (100% absorption outside the opening) requires either "no electromagnetic properties" or "conducting at the frequency of interest" (which would result in reflection not absorption). I think your answer is a possible explanation but not the only one - and specifically I think you are assigning properties to the slit that make the experiments less "equivalent". – Floris Oct 22 '14 at 21:43

When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves ...?

The answer is no. The diffraction pattern of sound and of light behind a slit is not similar and so the mathematical description is not complete. This is because the real patterns on an observation screen (or an imaginery line for sound waves) are never the same for sound and light. The fringes from the light diffraction are not moving patterns. More, the reflection of these light (from the observation screen) does not interact with the incomming waves. The diffracted water (or sound) waves don't stand still (picture), they are moving to the left and to the right if one look on them at the position there is the screen for light fringes. If one put a screen in the reflected waves interact with the waves from the slit and produce more complicated patterns.

We have to state that the mathematical model describes for water waves a instantaneous situation. The position of the spherical water waves behind a slit depends from the postion of the wave in front of the slit at some time. Opposite situation for light, the fringes from light stand still.

As pointed out in the answer from Ben Crowell, the slit can work as a waveguide. As pointed out from Floris, the edges could "electrically conducting at the frequency of interest for the electromagnetic case". In both cases the light interacts with the slits material. This do the light with every edge too, fringes appear behind every edge. As I pointed out here, "light is a lot of photons, and if you take monochromatic light, a lot of same energy photons and they interact with the electric field of the electrons of the edge. The last point one has to imagine is that the field they "built" together is quatitized. The fringes on the observers screen show how the field is quantized."

So a mathematical model for sound or water can predict instantaneous distances between maximum and minimum points of the sperical wave behind a slit. And a matematical model for light can predict the positions of maxima and minima for fringes on an observation screen. But physically that are total different phenomena.

If you can ignore the effects of sounds attenuation in air, I believe the answer is "yes". Note that when the slits are electrically conducting at the frequency of interest for the electromagnetic case, then certain effects relating to polarization come into play as explained in Ben Crowell's answer, and the sound and light cases become more and more different as the slit dimension decreases. I would argue that since there is no equivalent mechanism (conductivity along the edge of the slit) for sound, you cannot talk of an "equivalent experiment" I that case - and that the comparison must be made with a non-conductive slit.

• I don't think this is right, for the reasons given in my answer. Note that I gave three separate mechanisms for the diffraction to behave differently. – Ben Crowell Oct 22 '14 at 21:22
• I wasn't the person who flagged it as low quality, but the original answer was only one sentence. – Ben Crowell Oct 22 '14 at 21:24