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I think you will have an easier time viewing the index of refraction from a speed-point of view. Consider the following: The energy of a given photon is determined by its frequency (color): $E = h \nu$ (h being the Planck constant) Assuming the photon does not lose energy when entering the material, its frequency must be conserved. However, as light is an ...

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You are mixing up two different things. The refractive index is usually defined in terms of the velocity of light: $$n = \frac{c}{v}$$ where $v$ is the velocity in the medium. However the velocity is related to the frequency and wavelength by: $$v = \lambda f$$ so: $$n = \frac{\lambda_0 f_0}{\lambda f}$$ The frequency of the light, $f$, doesn't ...

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Absolutely, though the accuracy might not be the best because clouds don't exactly have hard edges and it might be hard to differentiate between the red light from scattering and the tail end of the reddened sunlight hitting the clouds. Refraction might have an effect too, though thinking about it, I don't think this would change the calculation much. ...

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Just to wrap this up: I decided to use Tovar and Casperson' approach (Link 1,Link 2) using a 3x3 matrix, since I wasn't sure about the q-parameter in Shaomin's 4x4. It boils down to that as with the non-misaligned ray transfer matrix, $$q_o = \frac{Aq_i + B}{Cq_i + D}$$ with the 3x3 beam matrix \begin{pmatrix} A & B & 0 \\ C & D & ...

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Neutrinos are weakly interacting quantum mechanical point particles, with very small mass. Refraction is a classical mechanics phenomenon, happens to waves traveling in a medium and it is a collective synergy of many photons impinging on the field of the atoms and molecules of the medium. Individual photons are not refracted but are scattered. In synergy ...

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It turns out that one photon states of the electromagnetic field can be written in a way such that the state "propagates" fulfilling Maxwell's equations. This is an exact model as I discuss this in more detail in my answer here. So we begin with a one-photon Fock state of the quantized electromagnetic field. Let's keep our discussion to one mode, so one ...

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