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What does it mean when it says for long wavelengths?

  • Thread starter Nusc
  • Start date
754
2
1. Homework Statement

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. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
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.
 
754
2
I think you're right. Thanks
 
754
2
So when light is incident on an insulator or a semiconductor, what would happen?
 
pam
455
1
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).
 
754
2
So section 9.4.1 in griffith is irrelevent?

Where can I find a derivation foryour k ?
 
pam
455
1
Eq. (9.126) follows (in SI) from the equation I wrote. The algebra is a bit complicated.
 
754
2
What's k0?

What about kappa? from 9.126?
 
754
2
How would E and B change if this was an insulator or a semiconductor?
 
pam
455
1
What's k0?

What about kappa? from 9.126?
What I called k_0 is just what G has outside the square root.
It is the k yu would have if sigma=0.
 
pam
455
1
How would E and B change if this was an insulator or a semiconductor?
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.
 

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