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

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

Related Advanced Physics Homework Help News on Phys.org
Dick
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.

I think you're right. Thanks

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).

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.

What's k0?

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

pam
What's k0?

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.