# Homework Help: Index of refraction Vphase

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