your analogy works for mechanical waves like sound and water that require a medium to move so the wave travels. it isn't clear if that analogy holds for EM waves, that do not require a medium
i asked a friend more knowledgeable than myself
this is his response:
What you have in mind is Rayleigh scattering. Basically Rayleigh scattering has an intensity proportional to 1/wavelength^4 and is independent of the size of the scatterer as long as the scattering particle size << wavelength and the << means really several orders of magnitude in ratio. In truth, when scatterer size ~ wavelength all bets are off because then the physical and geometric properties of the scatter really matters. For example refractive index or dielectric constant can have profound effects. That is why microwave waveguides are sometimes filled or sometimes not.
In air where the scattering is mainly single molecular dipole radiation, Rayleigh scattering is always present for visible or near visible E-M radiation. If dust, either natural or artificial, polluting the air is very microscopic (volcanic, etc.) then violet, yellow or red sunsets result. If the dust size produced approaches the visible wavelengths in diameter, for example, then the smoke can appear either black or white depending what it is made of and there is no Rayleigh scattering. This is what is common for pollution debris just coming out of a smoke stack or for a forest fire. How black or how white depends on the density of "large" particles present and their composition.
In the final analysis you had the wavelength direction of increased scattering correct, but you erred in your criteria relative to the wavelength to particle size.
end of response