What would happen if light hit a non-differentiable curve? 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

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.
 A: 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.
A: 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).
