# The color of a refracted ray

My textbook explains the cause of the refraction of light as the "difference between the speeds of light when passing from medium 1 to medium 2". The same textbook explains that the cause of the differing speeds of light in different mediums as "the result of of interference between the incident light and the radiated( and delayed) light of the same frequency and a shorter wavelength".

My questions are:

1. Is this explanation of the refraction of light fundamentally flawed in some way?
2. If the speed of light when passing through a medium is the result of the shortening of its wavelength, why doesn't the color of a refracted ray of light change?

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• – dmckee Jul 9 '17 at 6:12
• Thank you for your answers, I see that the visible color of the refracted light remains the same due to its unchanging frequency regardless of the medium, and that color as seen to the human eye is dependent on the photon energy stimulus it receives. I see that my questions on stack exchange have been too specific to the situation in front of me, and that there is much more I could do just on this site alone to research the subject beforehand. This has been a great learning experience, and I want to become better at asking academic questions. – user47268 Jul 9 '17 at 6:49

The color will not change as there is a relative change in speed while traveling from one medium to another, for example, when a light will pass through a denser medium it'll take more time to go through it, but the speed of electromagnetic waves will remain the same irrespective of the medium.

Colour does not change as frequency stays the same. If we could embed an eye or whatever detector, it would respond to the same photon energy $E=h\nu$ (where $h$ is the Planck constant and $\nu$ is the frequency in Hertz) stimulus as it would at the exterior, to give to colour a more mechanicist meaning.

This is because upon entering a medium, the wavenumber (space periodicity) of the wave changes, but not its frequency (time periodicity), and what we call the "color" is associated to the photon's energy $E = \hbar \omega$.

The velocity becomes $c = \omega /k = \omega_0 /k = \omega_0 /(n\times k_0) = c_0 /n$. The energy remains $E = \hbar \omega = \hbar \omega_0 = E_0$

Where the subscript $0$ denotes the values in vacuum, and its absence denotes the quantities in the material.

The full treatment looks like this, assuming the light remains plane-wave like in the material of complex refractive index $\tilde{n} = n + i\times \kappa$ :

$\vec{E} = \vec{E_0} \times \exp(i~ (k~x -\omega t)) =\vec{E_0} \times \exp(i~(n~k_0~x-\omega_0 t)) \times \exp(-\kappa k_0 x)$.

Where the last factor $\exp(-\kappa k_0 x) = \exp(-\kappa \omega_0 x /c_0)$ will account for the decay of intensity due to absorption in the medium, which is why the intensity will decay faster for ultraviolet than for infrared, which penetrates deeper into materials.

• Of course, I assumed your question applied to the light inside the material. When the light beam comes out of the material again, its velocity and wavenumber are the same in magnitude as they were upon entering anyway (only the direction has changed). – Alexis Prel Jul 8 '17 at 23:25