# Snell's law with a complex refractive index

## Main Question or Discussion Point

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?

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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.

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?

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

Last edited:
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.

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?