Why are 90% of photons reflected at optical phonon frequencies?

[Goodver]

in the dispersion relation curve for phonon, optical branch is not zero at k = 0. Thus generally speaking solid might be transparent for all photon wavelengths except the wavelength which corresponds to the energy of an optical phonon at k = 0. Thus photons can be absorbed or reflected. It says that about 90 percents of photons are reflected. Why?

Also, It is not clear for me how photon can be reflected? It requires to change the direction of the k vector, to opposite direction. Or is it that incident photon converts to optical phonon, which then converts to optical photon in opposite direction statistically?

I am on a master level.
Thank you.

 

[PietKuip]

The momentum of infrared photons is negligible. The optical phonons near the center of the Brillouin zone are like large electrical dipole oscillators, which gives a very strong interaction with the electromagnetic waves.