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Index of refraction

  1. Apr 11, 2005 #1
    given is a graph of n(lambda) vs lambda where n is the index of refraction
    N(1000) =1.45

    Estimate Vphase and group velocity using the above info.

    i know that [tex] n = \frac{c}{v_{\phi}} = \frac{ck}{\omega} [/tex]

    i cant simply susbtitute into that above relation because the lambda given is that lambda in the vacuum and thus the lambda would be different in the medium. Also oncei found v phase how would i go about finding group velcoity, since group velocity is a derivative, do i simply pick off two points on the graph and find the slope?

    But first i need to figure out phase velocity which i cant get and i dont know if its right

    Please help!
     
  2. jcsd
  3. Apr 11, 2005 #2
    i know that [tex] V_{\phi} = \frac{\omega}{k} [/tex]
    where omega is the angular velocity of the wave and is 2 pi / T
    and k is the wave number 2pi / lambda
    however since it is in another medium with index of refraction 1.45 would the lambda be different??

    would i have to use the following relation to figure out the lambda in that respective medium isnce the wavelength given is that wavelength in a vacuum (right?)
    [tex] \frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}} = \frac{\lambda_{2}}{n_{1}} = \frac{f_{2}}{f_{1}} [/tex]
    and since in a vacuum n1 = 1 then i can find the lambda 2 the lambda in the medium?

    then i back substitute in the above relation and figure out the frequency as well.
    im not quite sure where this leads to... i am going off on a tangent
    im given n (wvelegnth) not n(angular frequency) which isn eeded to find the phase velocity
     
  4. Apr 11, 2005 #3

    xanthym

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    Science Advisor

    SOLUTION HINTS:
    Problem presents {n = n(λ)} for a given dispersive medium.
    Continue in direction you've begun to determine "vphase" and "vgroup" in the dispersive medium:

    [tex] 1: \ \ \ \ \ v_{phase} \ \ = \ \ \frac{c}{n} \ \ = \ \ \frac{\omega}{k} [/tex]

    [tex] 2: \ \ \ \ \ \Longrightarrow \ \ \ \omega \ \ = \ \ \frac{ck}{n} [/tex]

    [tex] 3: \ \ \ \ \ \ \ \ \Longrightarrow \ \ \ \frac{d\omega}{dk} \ \ = \ \ \frac{c}{n} \ \ - \ \ \frac{ck}{n^{2}}\left(\frac{dn}{dk}\right) \ \ = \ \ \frac{c}{n}\left(1 \ - \ \frac{k}{n}\left(\frac{dn}{dk}\right) \right) [/tex]

    [tex] 4: \ \ \ \ \ \ \ \Longrightarrow \ \ \ \mathbf{v_{group}} \ \ = \ \ \mathbf{v_{phase} }\left ( 1 \ - \ \frac{k}{n} \left ( \frac{dn}{dk} \right ) \right ) \ \ = \ \ \mathbf{v_{phase} }\left ( 1 \ + \ \frac{\lambda}{n} \color{red} \left ( \frac{dn}{d\lambda} \right ) \color{black} \right ) [/tex]

    Determine relationship between "vphase" and "vgroup" in the dispersive medium from Eq #4, where derivative in red is evaluated from the given function {n = n(λ)}.



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    Last edited: Apr 11, 2005
  5. Apr 11, 2005 #4
    so for number 4 the derivative is calculated by using two points and finding the slope of the graph (it did ask for an estimate, after all) ?

    is that the right way?
     
  6. Apr 11, 2005 #5

    xanthym

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    Science Advisor

    Without seeing the graph, it's difficult to judge. However, your method would be reasonable if the graph were approx linear. Incidentally, you should find that {(dn/dλ) < 0}


    ~~
     
    Last edited: Apr 11, 2005
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