# What happens to the incident electromagnetic wave in Rayleigh scattering?

In Rayleigh scattering an incident EM wave causes an induced dipole oscillation of an atom/molecule, which in turn causes radiation at the same frequency of the incident wave. But what happens to the original wave?

I have read quite a few pages on stackexchange and wiki articles on Rayleigh scattering but nowhere does it seem to be clearly stated. Is the wave expended in inducing the dipole? Is it cancelled by something? It makes 'classical' sense to me that as you can't get something for nothing it ought to be 'spent', but I don't know how.

Possibly related: What does it mean to be 'less strongly' scattered'?

The power radiated from the dipole is proportional to wavelength as $P\propto\omega^4$. Red light from the sun is said to be less strongly or less efficiently scattered in the atmosphere. Now since I know that more red light passes by undeviated in direct sunlight, I'm led to think that 'less strongly scattered' means that more of the incident red light somehow survives the scattering event. I can interpret this two ways:

1. Less of the incident red light wave is expended in inducing the oscillating dipole, compared with blue light. (Which perhaps explains why the radiated power is lower).
2. Less red light is 'incident' in the first place - a higher wavelength makes the scattering event less likely.

Neither option feels right and I think my misunderstanding here is related to the above question.