Dispersion Relations and Refractive INdex

In summary, the refractive index of a plasma can be proven to be n = \sqrt{1- (\frac{\omega}{\omega_{p}})^{2}} with \omega_{p} = \sqrt{\frac{Ne^{2}}{m\epsilon_{0}}}, and the attenuation length can be shown to be L = \frac{c}{\omega} \sqrt{\frac{1}{(\omega_{p}/\omega)^{2}-1}}. The refractive index is related to the dispersion relation, and the attenuation length is a measure of the decrease in intensity of EM radiation inside a material.
  • #1
tomeatworld
51
0

Homework Statement


The conductivity of a plasma is defined as [itex]\sigma = i\frac{Ne^{2}}{m\omega}[/itex] where N is the electron density.

a) Prove the refractive index is: [itex]n = \sqrt{1- (\frac{\omega}{\omega_{p}})^{2}}[/itex] with [itex]\omega_{p} = \sqrt{\frac{Ne^{2}}{m\epsilon_{0}}}[/itex]

b) Show the Attenuation length is [itex]L = \frac{c}{\omega} \sqrt{\frac{1}{(\omega_{p}/\omega)^{2}-1}}[/itex]


Homework Equations


[itex]k^{2} = \mu\epsilon\omega^{2} + i\mu\sigma\omega[/itex]


The Attempt at a Solution


I can't find equations linking the refractive index to the dispersion relation. Also, don't know anything about the attenuation length.

Can I grab a push in the right direction?
Thanks

Edit: Right, for the first part, I stumbled upon the equation [itex]n^{2} = \frac{c^{2}}{\omega^{2}} k^{2}[/itex] but I get an answer inverse to the required answer, with an extra factor of [itex]c^{2}[/itex]. Can someone just verify I've almost got there or that that equation is completely wrong. Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
tomeatworld said:

Homework Statement


The conductivity of a plasma is defined as [itex]\sigma = i\frac{Ne^{2}}{m\omega}[/itex] where N is the electron density.

a) Prove the refractive index is: [itex]n = \sqrt{1- (\frac{\omega}{\omega_{p}})^{2}}[/itex] with [itex]\omega_{p} = \sqrt{\frac{Ne^{2}}{m\epsilon_{0}}}[/itex]

b) Show the Attenuation length is [itex]L = \frac{c}{\omega} \sqrt{\frac{1}{(\omega_{p}/\omega)^{2}-1}}[/itex]

Homework Equations


[itex]k^{2} = \mu\epsilon\omega^{2} + i\mu\sigma\omega[/itex]

The Attempt at a Solution


I can't find equations linking the refractive index to the dispersion relation. Also, don't know anything about the attenuation length.

Can I grab a push in the right direction?
Thanks

Edit: Right, for the first part, I stumbled upon the equation [itex]n^{2} = \frac{c^{2}}{\omega^{2}} k^{2}[/itex] but I get an answer inverse to the required answer, with an extra factor of [itex]c^{2}[/itex]. Can someone just verify I've almost got there or that that equation is completely wrong. Thanks.
I think you're right, and what you are being asked to prove is slightly wrong. It should be
[tex]n = \sqrt{1 - \biggl(\frac{\omega_p}{\omega}\biggr)^2}[/tex]
according to e.g. http://farside.ph.utexas.edu/teaching/em/lectures/node98.html (By the way, that's a good read)

As for the attenuation length: the intensity of EM radiation inside a material decreases exponentially as a function of the distance penetrated into the material. Mathematically,
[tex]I \propto e^{-x/\lambda}[/tex]
The constant [itex]\lambda[/itex] is the attenuation length. It's related to the imaginary part of the permittivity [itex]\epsilon[/itex] (and also related to the dispersion relation). Check your notes or references and see whether that helps you find anything relevant.
 

1. What is a dispersion relation?

A dispersion relation is a mathematical relationship that describes how the frequency and wavelength of a wave are related. Specifically, it relates the frequency ω and wavenumber k of a wave in a medium. In other words, it describes how the speed of a wave changes as it travels through a medium.

2. How do dispersion relations affect the refractive index?

The refractive index is a measure of how much a medium can bend light. It is affected by the dispersion relation because the speed of light in a medium is related to its refractive index. As the frequency of light changes, so does its speed, which in turn affects the refractive index.

3. What is the difference between normal and anomalous dispersion?

Normal dispersion refers to the relationship between frequency and refractive index where higher frequencies have higher refractive indices. This is the case for most materials. Anomalous dispersion, on the other hand, is when lower frequencies have higher refractive indices. This is typically seen in materials with complex molecular structures.

4. How does dispersion relate to the colors we see?

Dispersion is responsible for the phenomenon of color dispersion, where white light is split into its component colors when passing through a prism. This is because different frequencies of light have different speeds in the prism, causing them to bend at different angles and separate.

5. Can dispersion relations be used to predict the behavior of waves in different materials?

Yes, dispersion relations can be used to predict the behavior of waves in different materials. By understanding the dispersion relation of a material, we can determine how light will behave as it passes through that material, including how much it will bend and how different frequencies will be affected.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
11
Views
1K
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
812
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top