# What does it mean when it says for long wavelengths?

1. Apr 11, 2008

### Nusc

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. Apr 11, 2008

### 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.

3. Apr 11, 2008

### Nusc

I think you're right. Thanks

4. Apr 12, 2008

### Nusc

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

5. Apr 12, 2008

### 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).

6. Apr 12, 2008

### Nusc

So section 9.4.1 in griffith is irrelevent?

Where can I find a derivation foryour k ?

7. Apr 12, 2008

### pam

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

8. Apr 13, 2008

### Nusc

What's k0?

9. Apr 14, 2008

### Nusc

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

10. Apr 17, 2008

### pam

What I called k_0 is just what G has outside the square root.
It is the k yu would have if sigma=0.

11. Apr 17, 2008

### pam

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.