Index of refraction

1. Apr 11, 2005

stunner5000pt

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 $$n = \frac{c}{v_{\phi}} = \frac{ck}{\omega}$$

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

2. Apr 11, 2005

stunner5000pt

i know that $$V_{\phi} = \frac{\omega}{k}$$
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?)
$$\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}} = \frac{\lambda_{2}}{n_{1}} = \frac{f_{2}}{f_{1}}$$
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

3. Apr 11, 2005

xanthym

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:

$$1: \ \ \ \ \ v_{phase} \ \ = \ \ \frac{c}{n} \ \ = \ \ \frac{\omega}{k}$$

$$2: \ \ \ \ \ \Longrightarrow \ \ \ \omega \ \ = \ \ \frac{ck}{n}$$

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

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

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
4. Apr 11, 2005

stunner5000pt

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?

5. Apr 11, 2005

xanthym

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}

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Last edited: Apr 11, 2005