Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dispersion question

  1. Feb 2, 2013 #1
    From the elementary texts, dispersion is the phenomenon where the refractive index of a medium depends on the wavelength of electromagnetic radiation through it. From what I've read, it is the wavelength of the radiation in vacuum. Also, it is said that the refractive index increases with decreasing wavelength, and increasing frequency.

    My question is simple, by wavelength, are they referring to the 'new' wavelength as the light passes through the material or are they referring to the wavelength in vacuum? About frequency, I don't really have much question about it, at least if I believe that the frequency stays constant even after going through different mediums.

    If we go by the equation [tex]\lambda=\frac{\lambda _o}{n}[/tex] I can see that they must be referring to the 'new' wavelength, but this is still a bit baffling for me. I'd like to ask more but maybe after someone replies on this. I probably just missed something.

    Edit: I'm starting to get the hunch that I'm might be using the wrong model/mathematical expression for this.
     
    Last edited: Feb 2, 2013
  2. jcsd
  3. Feb 2, 2013 #2

    Drakkith

    User Avatar

    Staff: Mentor

    If we measure the wavelength of a wave before and after entering a medium we will find that it has changed. The amount of change lets us calculate n, the refractive index. Let's say it gives us 1.1 for n. Now, we shoot a shorter wavelength through. We measure again and find that n now equal 1.10002. The refractive index is different for different wavelengths. Not by much, but it is. The greater the difference in your refractive index between different wavelengths the greater the dispersion.

    Your equation is simply the relationship between the wavelength before entering the medium, after entering the medium, and the refractive index of the medium.
     
  4. Feb 2, 2013 #3
    Looking back, I think the texts are actually referring to the wavelength in vacuum. The expression above seems to be valid at least if we are concerned with the wavelength of the light as it passes through the medium. I'm now half convinced that we cannot conclude anything about the dependence of the refractive index in the frequency of the specific radiation from the expression above.

    I actually tried to express the the vacuum wavelength in terms of speed of light and its frequency, but I'm getting contradicting results.
     
  5. Feb 2, 2013 #4

    Drakkith

    User Avatar

    Staff: Mentor

    You can use the expression above to find the refractive index of a material for different wavelengths. That will then allow you to find the dispersion.
     
  6. Feb 2, 2013 #5
    I get the notion that it's better to refer to frequency when we are talking about specific EM wave in the spectrum since it doesn't change when it propagates through mediums.

    Thanks anyway, I think I got it, the expression is valid only for a specific frequency of a wave. Now, I guess my real question is, why does the refractive index increase with increasing frequency of the wave (and consequently with decreasing wavelength of it in vacuum)?
     
    Last edited: Feb 2, 2013
  7. Feb 2, 2013 #6

    Drakkith

    User Avatar

    Staff: Mentor

    Sure, unless you are specifically dealing with the wavelength or something where the wavelength isn't changing. Talking about light in terms of wavelengths is actually pretty useful, as it's much easier to say light of 550 nm than it is to try to write the frequency down.

    Yeah, the expression only allows for one wavelength at a time. As for why the index increases with frequency, someone else will have to answer that, as it's way above my head.
     
  8. Feb 2, 2013 #7
    No problem, I was at least relieved that I got the simple stuff first. As for the frequency dependence, I guess I'll take it for granted for now. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook