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I know it does not because it emerges out of denser medium at 300,000 KM per second, but according to $E=mc^2$ and given that speed of light decreases inside denser medium with refractive index greater than 1, does not it suggest that energy of light inside denser medium is less?

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$E=mc^2$ is not really applicable to light. It is applicable to something that has mass.

The energy of light is given by $E=h\nu$ where $\nu$ is the frequency of light and $h$ is Planck's constant, which has a value of $\approx 6.626 \times 10^{-34} J.s$

When light enters a different medium, its frequency remains the same, and of course, so does Planck's constant. Hence, obviously, its energy remains the same throughout the entire exercise.

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You might want to add that the reason E = mc^2 doesnt apply is because the "c" in the equation is not the value of how fast the thing is travelling, it's a constant. That value happens to be the speed of light in a vacuum. E = mc^2 applies to an object with mass at rest, hence not photons. – Owens Sep 5 '13 at 7:07
So how does the speed decreases then if no property of light gets altered? – srijan Sep 11 '13 at 14:14
@srijan : the speed doesn't really decrease but the light interacts with matter and this takes time. edit: well explained in the Jerry Schirmer answer – igael Dec 23 '15 at 22:15

To a first approximation, the light is still travelling at the speed of light inside of the medium. There is an apparent slowdown because, as the light ray interacts with the molecules of the medium, it diffracts, causing it to change direction randomly. This then causes its travel length to increase, which makes the speed appear to be smaller, from a macroscopic perspective.

The velocity of the light's travel does not decrease, however.

I would also state that $E=mc^{2}$ is not a valid formula for light, as that formula is only valid for stationary objects. The appropriate version of this for light would be $E^{2} = c^{2}|{\vec p}|^{2}$, and you could answer your above question by also making an appeal to conservation of momentum.

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This is not why light slows down in a medium. I dont know exactly why it happens, but my guess is that it's more to do with the group speed of light in the medium. – Owens Sep 5 '13 at 6:38
@Owens: you would certainly agree that the group speed of light in the medium is an emergent property of the optical properties in the medium, which ultimately comes down to the interactions of photons with the atoms. And you would also agree that the path length of rays is increased, as evidenced by the fact that wavelength shifts in (standard, linear) media, while frequency does not. – Jerry Schirmer Sep 5 '13 at 16:40
@Owens: it's not overall random--net momentum is conserved, and so there's a well-defined drift direction, and the litter interaction/lot of interaction thing is averaged over moles of atoms, and so everything just will average to the expectation value. It's not a perfect model, but it's a decent model, for the level of this question. Especially since any interaction with the field of the medium isn't with the field of the medium per se, but with the field created in the medium caused by its polarization by the wave itself. Glass, after all, is electrically neutral. – Jerry Schirmer Sep 6 '13 at 4:45
@Owens: it's functionally the same thing as your bulk description. It really is. – Jerry Schirmer Sep 6 '13 at 5:23
@Owens: refraction IS scattering. Almost all real materials have dispersion, too. It's just athat in real materials, all of it is washed out over $O(10^{23})$ interactions, so all we see is the average. Like I said, it's not a perfect model, but it's a model that explains refraction, and it explains why wavelengths shift, but frequencies do not. You also can visibly see the path lengths increase through effects like thin film diffraction. – Jerry Schirmer Sep 6 '13 at 6:16

To touch on and finalise Jerry Schirmer's answer: the "light" in a medium is not only "light" in the ordinary, vacuum sense of the word, it is a quantum superpositon of free photons and excited matter states. A photon travelling though a medium is repeatedly undergoes the following cycle: it is fleetingly absorbed by electrons in the medium, which re-emit a new photon in its place a fantastically short time afterwards (femtoseconds or less). The process is somewhat like fluorescence, aside from that energy, momentum and angular momentum are wholly transferred to the new photon, whereas in fluorescence, energy (as betokened by the Stokes shift), momentum and angular momentum (as betokened by direction and polarisation shifts) are all transferred to the medium. The delay arising from the absorption / re-emission is what makes the light seem to propagate slowly, but you can see that no energy is lost. A slight variation on this theme is the birefringent material, where energy and momentum are wholly returned to the re-emitted photon, but some angular momentum is exchanged and light thus exerts a torque on a birefringent medium: see the second and third sections of my answer here and indeed there is a classic experiment demonstrating light's angular momentum by R. Beth, "Mechanical Detection and Measurement of the Angular Momentum of Light", Phys. Rev. 50 1936 pp115-127. But energy can still in principle conserved: in practice some mediums have attenuations, but some are fantastically small, for example, silica in the optical telecommunications window between $1350nm$ and $1550nm$ and, for the purposes of this argument, attenuations can in principle be nought.

Witness here is that the Poynting vector in the medium is the same as its freespace value: $\vec{S} = \vec{E}\wedge \vec{H}$, whereas the energy density $U = \frac{1}{2}\vec{D}\cdot\vec{E} + \frac{1}{2} \vec{B}\cdot\vec{H}$ in the medium is now higher: this is simply analogous to the steady state behaviour of a water tank with inlet and outlet pipe: transiently the water output rate can be less than that at the input whilst the tank fills up, but at steady state the two rates must balance. Likewise for the medium: the higher energy densities represent increased energy stores in the matter of the medium owing to the excited matter state parts of the total quantum superposition (there are also reflected energies at the medium's input and output which must be accounted for in an exact descripition, but the essential gist of this paragraph does not change).

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$E=h \nu$, hence $E$ is proportional to $\nu$, and $\nu$ does not change when light moves from rare to denser medium. So, it does not effect $E$.

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The Greek letter is $\nu$ for nu, not mu. – Kyle Kanos Dec 23 '15 at 16:30

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