This is not homework, I came across this problem during my research and can't seem to figure it out.
I'm trying to find the transmission reflection and absorption coefficients of a plane wave through a uniform slab with negative refractive index at normal incidence. The usual formula's don't work because I believe the boundary conditions change when light is lefthanded. I found a book on lefthanded light and it says at normal incidence I can just take the absolute value of n, epsilon, and mu and the formulas should work out fine, but I'm getting large discrepancies between using those formulas and when I simulate the slab.
the formula's for the coefficents at each face are:
r_face=(1abs(n))/(1+abs(n))
r'_face=r_face
t_face=2/(1+abs(n))
t'_face=t_face*(k'/k)
where abs(n) is the complex index of refraction with absolute valued components, r_face is the reflection coefficient of going from vacuum into the material, r'_face is from material to vacuum, t_face is the transmission coefficient of going from vacuum into the material, etc. k'=nw/c the wave vector inside the slab, and k=w/c the wave vector in vacuum.
I then plug these into formulas for the reflection and transmission through the slab that I have derived and checked. These should hold regardless of n being negative.
these are:
r=r_face+(t_face*t'_face*r'_face*exp(i*k*d))/(1(r'_face*exp(i*k*d))^2)
t=(t_face*t'_face*exp(i*k*d))/(1(r'_face*exp(i*k*d))^2)
where d is the thickness of the slab
R=r*conj(r)
T=t*conj(t)
A=1RT
Any ideas either why these do not work, or how to fix them?
Negative Index of Refraction
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 danpilon54
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 Joined: Fri Jul 20, 2007 12:10 am UTC
Negative Index of Refraction
Last edited by danpilon54 on Wed Jun 25, 2008 4:31 am UTC, edited 1 time in total.
Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.
Re: Negative Index of Refraction
where abs(n) is the complex index of refraction with absolute valued components
Maybe you have to do sqrt(n x n*) (i.e. the magnitude of the complex number). I'm almost certain that just taking the absolute values of the real and imaginary parts is not the correct thing to do. It never is.
And what do you mean "lefthanded light"? Leftcircularlypolarized? The reflection at normal incidence shouldn't care about polarization.
Asad
 RAPTORATTACK!!!
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 Joined: Sat Sep 30, 2006 8:48 pm UTC
 Location: Aroundabouts boston.
Re: Negative Index of Refraction
no idea how to help with this, just injecting this: Isnt that how fiber optics work?
Team 246 OVERCLOCKED!
Re: Negative Index of Refraction
RAPTORATTACK!!! wrote:no idea how to help with this, just injecting this: Isnt that how fiber optics work?
Nope, fiber optics work by total internal reflection in a normal, highindex material like glass.
Asad

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 Location: Lexington, Ky
Re: Negative Index of Refraction
danpilon54 wrote:where abs(n) is the complex index of refraction with absolute valued components
abs(a+jb) = sqrt(a^2+b^2)
(incidentally, this, V=IZ and P=IV is all I've learned as an undergrad EE )
 danpilon54
 Posts: 322
 Joined: Fri Jul 20, 2007 12:10 am UTC
Re: Negative Index of Refraction
so I have it working for n>0, but if I do the magnitude and not the absolute value it doesnt work anymore. That must mean that cant be right because the book said that the absolute value would work for both positive and negative index. Also taking the magnitude makes n real, making absorption 0 which should most certainly not be the case.
Also, light is righthanded in free space and in materials with positive index of refraction because the direction of propagation is given by ExB using the right hand rule. In a material with negative index, ExB points in the oppisite direction of propagation, making the phase velocity opposite the group velocity, and if you want ExB to be in the direction of propagation youll need to use the "left hand rule".
The reason this yields different boundary conditions (if I knew exactly what they change to I could probably figure all of this out) is because you have to correctly match the phases of the incident beam with that of the reflected and transmitted beams. Since you get a sort of inversion of the electric and magnetic fields you have to account for that in the boundary conditions.
I guess to summarize:
magnitude does not work, but neither does taking the absolute value of the components
allowing n to be negative in the usual formulas yields reflectivities greater than 1, which is unphysical.
I wish we did this kind of problem in my latest e&m class (we ran out of time)
Also, light is righthanded in free space and in materials with positive index of refraction because the direction of propagation is given by ExB using the right hand rule. In a material with negative index, ExB points in the oppisite direction of propagation, making the phase velocity opposite the group velocity, and if you want ExB to be in the direction of propagation youll need to use the "left hand rule".
The reason this yields different boundary conditions (if I knew exactly what they change to I could probably figure all of this out) is because you have to correctly match the phases of the incident beam with that of the reflected and transmitted beams. Since you get a sort of inversion of the electric and magnetic fields you have to account for that in the boundary conditions.
I guess to summarize:
magnitude does not work, but neither does taking the absolute value of the components
allowing n to be negative in the usual formulas yields reflectivities greater than 1, which is unphysical.
I wish we did this kind of problem in my latest e&m class (we ran out of time)
Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.
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