Snell's law with a complex refractive index

1. Mar 14, 2008

zhanghe

hello everyone

Consider extinction coefficient k, n becomes N=n-ik.
the text book says NsinA=N'sinB still holds itself.

But the sinB,for exmple, may be a complex number, i want to know
how to get B?
how to understand this situation, which is the refractive angle?
the B's real part?

2. Mar 14, 2008

pam

The electric field in a plane wave E~exp[i(kx-wt)], where k=w N/c.
If N=n-ia, then E~exp[i(nw/c)x-wt)exp[-ax].
This means n is still used by Snell.
k is usually the wave number, so I used a for Im N.

3. Mar 14, 2008

zhanghe

thank you
and you mean that the refrective angle only depends on Re N--n
and has nothing to do with Im N--a(using your signal), right?

4. Mar 19, 2008

lightarrow

Why you write exp{i[(nw/c)x-wt]}*exp(-ax) and not exp{i[(nw/c)x-wt]}*exp(axw/c) ?

Last edited: Mar 19, 2008
5. Mar 20, 2008

pam

Sorry. That was a misprint. It should be axw/c.

"and you mean that the refrective angle only depends on Re N--n"
I'm afraid I oversimplified. The boundary conditions have to be applied at the interface, and the derivation done from scratch. It gets quite complicated.

6. Mar 20, 2008

zhanghe

these days ,i thought further that maybe it's only a kind of appearance to describe the polirized light phase shift at the interface of different media.
but i wonder which is the refractive angle if k ,ie absorption exist.
for example. air/silicon,whose N is 1 and 4.4-0.8i, respectively.
to Snell, 1*sin(AOI)=(4.4-0.8i)*sin(AOR),
AOR will be a complex number, how to get the real AOR in experiments?