1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What does it mean when it says for long wavelengths?

  1. Apr 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Light is incidence on a metal surface. For long wavelengths find E and B.

    What does it mean when it says for long wavelengths?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think it just means that the EM wave can be considered classically, i.e. no photoelectric effect type stuff. Just a guess though, your description is pretty sketchy.
     
  4. Apr 11, 2008 #3
    I think you're right. Thanks
     
  5. Apr 12, 2008 #4
    So when light is incident on an insulator or a semiconductor, what would happen?
     
  6. Apr 12, 2008 #5

    pam

    User Avatar

    In metals, the wave number [tex]k=k_0(1+i\Delta)^{1/2}[/tex].
    Delta is a dimensionless parameter: [tex]\Delta=\frac{4\pi\sigma}{\epsilon\omega}[/tex]
    in Gaussian units.
    "Long wavelength", means small omega, so Delta>>1.
    Then a wave in the metal is rapidly attenuated (in less than a wave length).
     
  7. Apr 12, 2008 #6
    So section 9.4.1 in griffith is irrelevent?

    Where can I find a derivation foryour k ?
     
  8. Apr 12, 2008 #7

    pam

    User Avatar

    Eq. (9.126) follows (in SI) from the equation I wrote. The algebra is a bit complicated.
     
  9. Apr 13, 2008 #8
    What's k0?

    What about kappa? from 9.126?
     
  10. Apr 14, 2008 #9
    How would E and B change if this was an insulator or a semiconductor?
     
  11. Apr 17, 2008 #10

    pam

    User Avatar

    What I called k_0 is just what G has outside the square root.
    It is the k yu would have if sigma=0.
     
  12. Apr 17, 2008 #11

    pam

    User Avatar

    G give (9.126), but doesn't do much with it.
    His [tex]\sigma/\epsilon\omega[/tex] is what I called [tex]\Delta[/tex].
    For a good conductor, [tex]\Delta[/tex] is large and you can approximate the square root.
    For a poor conductor, it is small and you make a different expansion.
    With these expansions, you can do a lot.
    Without them, you do what G does, which is draw a picture or give a problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What does it mean when it says for long wavelengths?
Loading...