Is reflection just a re-emitted wavelength at the electronic scale? Does reflection require absorption in the first place?

Question

I'm researching on how different wavelengths choose to either reflect, get absorbed, or transmit through a particular material.

I've read that transmission is when waves do not interact with the material's electrons at all and that reflection is when the object still absorbs the wavelength but then emits back the same wavelength as the reflected wave.

Is reflection just a re-emitted wavelength at the electronic scale? Does reflection require absorption in the first place? I am unable to find supporting links explaining all this.

Answer 1

This is an enormous topic and what happens all depends on the frequency of the incoming em wave and the material (system) which the em wave is hitting.

The incoming oscillating electric field (em wave) interacts with the charges which are part of the material and force the charges to oscillate at the frequency of the incoming em wave.
The charges which have been force to oscillate radiate em waves of the same frequency as the incoming wave in all directions.
What you "see" is the sum of all the waves radiated from the oscillating charges which make up the reflecting surface, with some reflection, some transmission and some absorption.

In a metal the waves radiated by the oscillating free electrons are approximately $180^\circ$ out of the phase with the incoming waves and this results in almost zero net electric field in the metal and reflected waves. Put another way the incoming oscillating electric field induces currents in the surface of the metal and those oscillating current produce oscillating electric fields.

The free electrons in a metal have a natural frequency of oscillation called the plasma frequency which for sodium corresponds to a wavelength of $210 \,\rm nm$ ie well below the wavelength of visible light in the ultra-violet part of the em spectrum).
So metals reflect (and absorb due to their resistance) electromagnetic waves above the plasma wavelength.
If the incoming em wave has a frequency much higher than the plasma frequency of the metal (ie a wavelength $<210\,\rm nm$ for sodium) then the free electrons cannot respond to the speed of oscillation of the incoming waves and the waves pass through the metat - the metal is transparent to such wave.

Answer 2

Photons are quantum mechanical entities, and so are electrons, atoms and molecules. They obey the rules of quantum mechanics. The photon energy is $h*ν$ where $ν$ is the frequency of the classical electromagnetic light that a large number of such energy photons will built up. See this to understand how photons build up the classical electromagnetic wave

The interaction with solids is also dependent on the quantum mechanical nature of the atoms and molecules in the solid. A solid can have a lattice., a crystal for example.

Photon-lattice scattering can be elastic , and that is when reflections can happen, from the first layer of the lattice.

If the lattice has no energy levels with the energy $h*ν$ that would allow the atoms or molecules or the lattice itself to be excited and absorb the photon, the solid is transparent to the photon.

For both reflection and transparency it is important that the energy of individual photons does not change and also the phase its wavefunction has with all the other photons in the stream. If the phase would be different , as would happen with absorption and reemission , no images could be carried by the light, because reemission has random phases for each individual photon composing the reflected image. Photons that are absorbed get out of the stream.

Note , the quantum mechanical interactions are really happening with the electric fields that atoms and molecules set up, not with individual electrons.