I How energy of light is conserved when passing through medium

[Gevorg]

The velocity of light changes when it passes through a medium of a different refractive index. So let's suppose the light is traveling through a vacuum at a velocity c and then passes through a glass wall. Its velocity decreases while traveling through it but then speeds back up to c after passing through it. How does this not violate conservation of energy?

 

[Iliody]

Energy , in light in vacuum, only depends on inverse wavelenght (color of light), not in velocity. A light beam loss energy through the medium by absorption and by wavelength dilation when it goes out of the medium.

 

[bhobba]

What's going on as light goes through a medium is very complicated. It's way beyond my expertise except in a general way.

What happens broadly is light actually becomes quasi particles (phonons I think - but don't hold me to it) while traveling through a medium then get converted back when exiting. The picture you find in beginning texts, or thinking intuitively about it ie it get's absorbed by atoms that go to a higher energy state then spontaneously emit and travel through that way and hence are slowed down is evidently wrong - and there is a simple reason it must be wrong but I can't recall it.

ZapperZ has written extensively on this eg
https://www.physicsforums.com/threads/light-and-mediums.27359/

He, or someone with a similar level of knowledge of such things, is the right person to answer this question.

Thanks
Bill

 

[Demystifier]

What happens broadly is light actually becomes quasi particles (phonons I think - but don't hold me to it) while traveling through a medium then get converted back when exiting.

It's polariton, which is a mixture of photon and polarization quanta. For a given wavelength polaritons have two modes of oscillation, where typically one mode is more like photon and the other is more like phonon.

It's indeed a bit complicated, but there is a simplified answer to the OP's question. The energy of the "photon" (where the quotes denote that it is really the photon-like polariton in the medium) is
$$E=\hbar\omega$$
and the frequency $\omega$ does not change by entering or leaving the medium. The dispersion relation $\omega(k)$ depends on the medium, which means that $k$ (and hence the wavelength) depends on the medium while $\omega$ itself does not depend on the medium.