What does it mean when it says for long wavelengths?

[Nusc]

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?

Homework Equations

The Attempt at a Solution

 

[Dick]

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.

 

[Nusc]

I think you're right. Thanks

 

[Nusc]

So when light is incident on an insulator or a semiconductor, what would happen?

 

[pam]

In metals, the wave number $$k=k_0(1+i\Delta)^{1/2}$$.
Delta is a dimensionless parameter: $$\Delta=\frac{4\pi\sigma}{\epsilon\omega}$$
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).

 

[Nusc]

So section 9.4.1 in griffith is irrelevent?

Where can I find a derivation foryour k ?

 

[pam]

Eq. (9.126) follows (in SI) from the equation I wrote. The algebra is a bit complicated.

 

[Nusc]

What's k0?

What about kappa? from 9.126?

 

[Nusc]

How would E and B change if this was an insulator or a semiconductor?

 

[pam]

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]

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 $$\sigma/\epsilon\omega$$ is what I called $$\Delta$$.
For a good conductor, $$\Delta$$ 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.