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Transmission of light and velocity of light

  1. Apr 10, 2014 #1
    Greetings,

    When I think of light being "transmitted" through air, I have always used the idea that the light is not interacting... it is just passing through. However, I had a student ask me, if transmitted light is not interacting with air molecules, then why is the speed of light slower in air, relative to a vacuum. Being a chemist, I had to think about this...

    Clearly, light must be interacting with air, else the velocity would be the same as in vacuo. So, two questions:

    1) What is the nature of the interactions of visible light with say, O2 or N2, as it is being propagated along and

    2) Given this, is it even correct to say that light is being "transmitted" if there are these interactions?

    Thanks!
     
  2. jcsd
  3. Apr 10, 2014 #2
    Hi !

    I think part of the wave energy is transferred to the air particles and as Compton showed the wave length of the wave is reduced after that energy is transferred, and by the formula v= λ * ƒ it's clear why the velocity is lower than in vacuum.

    I'm not sure if this is the right explanation, just my guess.
     
  4. Apr 10, 2014 #3

    UltrafastPED

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    For any transparent material the relevant process is known as "Forward Coherent Scattering":

    1. "Forward" because that is the direction in which the image is transmitted
    2. "Coherent" because otherwise the image would be degraded
    3. "Scattering" because any absorption/emission process would remove certain colors

    So what is the theory? It looks just like the Huyghens wavelet mechanism for "wave transmission":
    http://en.wikipedia.org/wiki/Huygens–Fresnel_principle

    The scattering events result in a phase delay of the wavefront; this is why light speed is "slower" in air/glass/water than in a vacuum - and it also explains why light resumes its previous velocity when it exits: the scattering ends, so there is no more phase delay.

    In addition there is no change in the light spectrum before and after; the energy of the individual photons is unchanged by the elastic scattering process. And no, it is not Compton scattering - that is a high energy process.

    Richard Feynman discusses this and many other things having to do with light in his 4-part lecture:
    "QED: The strange theory of light and matter" - available in book form, or you can see the original lectures if you have about five hours to spend on this.
    http://vega.org.uk/video/subseries/8
     
  5. Apr 10, 2014 #4
    Thanks Ultrafast - that is very helpful.
     
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