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Suppose we are talking about reflections in 2D, and let's suppose that hypothetically I have something like a weistrass function with one side completely reflective

enter image description here

How would light reflect of it's surface if we shot a beam from above to the curve (at any point)?

This may seem like a trivial question, but it gets complicated as soon as you realize since the function is not differentiable, there is no clear tangent line approximation we can talk about and hence no normal at a point. So, it's no longer clear how to apply snell's law of reflection.

Qmechanic
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tryst with freedom
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    What sort of real-world situation is this supposed to model? – ACuriousMind Jan 02 '22 at 17:12
  • Make a mirror in the shape of the weistrass function, and see the reflection which occur on it @ACuriousMind – tryst with freedom Jan 02 '22 at 17:23
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    I am entirely unconvinced non-differentiable shapes usefully model any actual mirror. – ACuriousMind Jan 02 '22 at 17:25
  • Well clearly, it is no where written that all curves drawn in real life is differentiable nicely and having well defined tangent line at all point. Hence, this is a question relating to some physic scenario which can be studied. – tryst with freedom Jan 02 '22 at 17:27
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    Possible duplicate: https://physics.stackexchange.com/q/265075/50583, since if we ignore the specific unphysical idea of the Weierstraß function modelling any surface, what remains is asking what happens during reflection at a corner/non-differentiable point. – ACuriousMind Jan 02 '22 at 17:30
  • Are you interested in what ray optics has to say or wave mechanics has to say? I think this can only be answered by wave mechanics (Maxwell's theory of electromagnetism). In such a case, I suggest you go on to math stackexchange and ask what happens (according to the wave equation) when a Gaussian beam hits a fractal boundary. – Maximal Ideal Jan 03 '22 at 19:55

2 Answers2

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Your question is phrased in the language of ray optics. But ray optics is just an approximation to the propagation of light as a wave. We can obtain the equations of ray optics by writing down a wave amplitude of the form $$ \phi(\vec{r},t) = A(\vec{r}) e^{i S(\vec{r}) - \omega t}, $$ and plugging it into the wave equation. If we make various assumptions, we can obtain the standard results of ray optics (light travels in straight lines, Snell's Laws of refraction & reflection, etc.) These assumptions basically all boil down to "the wavelength of light is small compared to any other length scale in the problem." Wikipedia has all the gory details if you're interested.

The Weierstrass function fails the assumptions required to use ray optics. The fractal nature of the surface means that the "roughness" of the surface is significant on all length scales, both greater than and less than the wavelength of the incident light. We therefore can't use Snell's Law and the other results of ray optics at all. To find out what would happen, we would have to solve the wave equation with the Weierstrass function as a boundary instead. This is left as an exercise to the reader.

Michael Seifert
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  • I don't think the question is the language of ray optics. He says "beam". I'll admit that that leaves things ambiguous. Nevertheless I think your argument works for wave optics where it is common to use macroscopic Maxwell equations. The same assumption about wavelength applies. – garyp Jan 02 '22 at 18:09
  • @garyp: The question refers to Snell's Law of Reflection and the normal to a surface, which I'm most familiar with as ray optics results. – Michael Seifert Jan 02 '22 at 18:59
  • The last sentence of the question is inapplicable here, but the rest of the question is inspiring. – dominecf Jan 02 '22 at 19:07
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This is a deeply nontrivial question, mostly because it is bringing an abstract mathematical object into interaction with real world. But not a bad one. One needs to make a clear cut where the "real world" begins.

So I first assume there is a very thin wire with diameter nonzero but negligible compared to wavelength, and it is structured to near-fractal shape down to its diameter. What is the polarization of your light?

  • P-polarization (i.e. in-plane): Light hits it, and since the wire is actually very long and thin, the current will not screen the electric field, so light dit does not scatter and passes through.
  • S-polarization (i.e. out-of plane): Light does not care about ultra-thin wires perpendicular to E. No scattering either.

A different result occurs if you extrude the Weierstrass function to become a cross-section of a structured sheet - even then the macroscopic sheet conductivity is going to be very low.

  • P-polarization - plasmonic resonances may occur between the spikes and valleys. A lot of light may be absorbed at numerous resonance frequencies. Otherwise it would be partially reflected, or partially transmitted - basically this is handled by nanoplasmonics and it will depend on whether there is one nanometer, 10 nm, or 100 nm of metal and therefore how long the surface path is.
  • S-polarization - No plasmonic resonances due to out-of-plane translational homogeneity. Light will get reflected more than in the S-polarized case. But less than from a flat mirror of the same material due to the boundary not being so abrupt.

Once wavelength is shorter than the basic period of the function, you will also observe light scattering it into high diffraction orders, as predicted by ordinary grating theory. Efficiency is going to be lower than for a smooth grating due to ohmic losses.

There are many fractal-like conductive objects in the nature down to atomic level. Typically they are absorbing much light (c.f. carbon soot).

dominecf
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    I would appreciate if downvoters shortly described what is wrong with this answer. It is based on my experience with realistic electromagnetic simulations of structured metallic surfaces. And I put some effort to describe the real physical outcome OP might expect. – dominecf Jan 02 '22 at 19:05
  • Do you have the conventional meaning of s and p reversed? In your language s is "in plane". But the plane conventionally used to define s and p is the plane of incidence, which contains the incident ray and the surface normal. The convention holds that s-polarized light is perpendicular to that plane. – garyp Jan 03 '22 at 03:33
  • Also, what is the orientation of the wire with respect to the plane of incidence? Assume for a minute that the wire is in the plane of incidence, and the light is polarized in the plane of incidence (conventional p-polarization). In that case currents will be generated and the light will be reflected. Not so for s, and s-polarized light passes by. It's not clear from your first "S-Polarization" paragraph how you draw the conclusion that light will pass by. If it has something to do with the structure of the wire, let us know that, and how it differs from the case of a conventional wire – garyp Jan 03 '22 at 03:40
  • @garyp Thanks. The s- and p-convention gets a bit more convolved (i.e. ambiguous) if one computes oblique incidence. I have fixed it. – dominecf Jan 03 '22 at 19:35
  • Thanks, now I have the setup. But I don't see why light would pass through the p-polarized first case. A uniform cross section wire (such as is found in a wire grid polarizer) will reflect. In fact, in a wire grid polarizer the wires are sub-wavelength in cross section and spacing. All of the p-polarized light is reflected. A single wire will reflect the light that hits it (loosely speaking). s-polarized light passes because the wire is too thin to generate currents, as you point out. So in the case of p-polarization you must be bringing in some effect of the fractal structure. – garyp Jan 04 '22 at 03:11
  • The scattering efficiency of p-pol light from fractal-like wire will depend on many factors, and accurate overview would make a good diploma thesis. As a rule of thumb the structured wire has to be thin, very long and curved, which all together suppresses scattering (both by resistance, electron mass and self-induction). – dominecf Jan 05 '22 at 09:14