Why does a conductor's refractive index affect Snell's law for refraction?

[EmilyRuck]

Hello everybody!
I read that if one of the two materials involved in Snell's law is a conductor, the refraction angle $\theta_r$ is about $\pi / 2$ and is independent of the incident angle $\theta_i$ (I think $\theta_r$ will be precisely $\pi / 2$ if the conductor is ideal). My question is: why?
$\theta_i$ and $\theta_r$ are real quantities. The only other variables are the refractive indexes. How is the refractive index in a conductor? I found that we could write $\epsilon$ from the IV Maxwell's equation such that it includes the conduction:

$\nabla \times \mathbf{H} = j \omega \mathbf{D} + \mathbf{J} = j \omega \epsilon \mathbf{E} + \sigma \mathbf{E} =$

$= j \omega \epsilon ' \mathbf{E} + (\omega \epsilon '' + \sigma) \mathbf{E} =$

$= j \omega \left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right) \mathbf{E}$

So $\left( \epsilon ' - j \epsilon '' - j \displaystyle \frac{\sigma}{\omega} \right)$ is a new complex dielectric constant; its imaginary part takes into account the conduction. We can still write

$\epsilon = \epsilon_0 \epsilon_r$

by considering

$\epsilon_r = \displaystyle \frac{\epsilon '}{\epsilon_0} - j \frac{\epsilon ''}{\epsilon_0} - j \displaystyle \frac{\sigma}{\omega \epsilon_0}$

Now the refractive index is $n_2 = \sqrt{\epsilon_r}$ and is complex. But how it must be to generate in the Snell's law a $\theta_r = \pi / 2$, despite of $\theta_i$?

$\displaystyle \frac{n_1}{n_2} = \frac{\sin \theta_r}{\sin \theta_i}$

$n_1$ is the refractive index of a dielectric. If $| \epsilon '' + \sigma / \omega | \gg \epsilon '$ (this happens in a good conductor), we will have $n_2 \gg n_1$ and

$\sin \theta_r \simeq 0$

while I expected $\sin \theta_r \simeq 1$. What's wrong?
Thank you anyway!

Emily

 

[Simon Bridge]

It is because there is no electric field inside a conductor.

 

[EmilyRuck]

Yes, but fields are not considered in Snell's law. It deals only with rays (incident and refracted) and the $\theta_r$ should be forced to the value $\pi / 2$ and this should be evident without considering the fields. How?

 

[DrDu]

It is because there is no electric field inside a conductor.

Well, that depends. You can look through very thin sheets of gold and that's probably where something like Snells law is really applied.

 

[DrDu]

Hello everybody!

$\nabla \times \mathbf{H} = j \omega \mathbf{D} + \mathbf{J} = j \omega \epsilon \mathbf{E} + \sigma \mathbf{E} =$

When one uses both epsilon and sigma the convention is to take both real.
I would guess that what is really meant is that the propagation is at normal angle to the surface as in that direction the wave is most strongly damped. This corresponds to theta_r=0.

 

[Simon Bridge]

You can look through very thin sheets of gold and that's probably where something like Snells law is really applied.

OK - no electric field inside an ideal conductor.
Do you get the displacement effect characteristic of refraction or do the rays go right through only attenuated?

fields are not considered in Snell's law. It deals only with rays (incident and refracted)

The rays are normal to the EM-wave - the EM wave is a special case of an EM field.
The rays are modelling the behavior of a field. How does an ideal conductor respond to the field?

Bear in mind: not all EM wave behavior can be accounted for in a purely ray-based model. [*]

I'll leave the math to DrDu :)
One way of thinking about it -
If the material contains mobile-ish charges, then the material becomes polarized. The polarization field opposes the electric field ... in a conductor, the "polarization field" completely cancels it. In a dielectric medium, less so. Since the charges in a conductor will always arrange to cancel an electric field inside it, there is only one way left for the wave to travel.

IRL: conductors vary from the ideal situation: there is some penetration of the electric field beyond the classical surface. If the material is thinner than the penetration depth, then it is possible, in the model being discussed, for some light to get through. iirc though, it is better described as a quantum phenomena.

However, you will recall that although there is no electric field inside an ideal conductor, you can still get a field from one side to the other.

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[*] you can do it - for refraction in a metal though - if the index of refraction and the wave vector are complex valued. What you tried right? Now you have to decide what you mean by "light ray".

The light ray is usually the normal vector to the wavefronts (in the direction of propagation etc) - the wavefronts are surfaces... you get used to drawing wavefronts as surfaces of constant amplitude and phase.

In this model, however: the surfaces of constant real phase are planes whose normals make an angle equal to the angle of refraction with the interface normal, while the surfaces of constant amplitude, are planes parallel to the interface itself.

 

[EmilyRuck]

Bear in mind: not all EM wave behavior can be accounted for in a purely ray-based model. [*]

Yes, I know and you're right. But my doubt was simply regarding the math.

iirc though, it is better described as a quantum phenomena.

I completely agree.

[*] you can do it - for refraction in a metal though - if the index of refraction and the wave vector are complex valued.

This could be the case, if I consider the wave vector as a phasor.

The polarization field opposes the electric field ... in a conductor, the "polarization field" completely cancels it. In a dielectric medium, less so. Since the charges in a conductor will always arrange to cancel an electric field inside it, there is only one way left for the wave to travel.

Ok! But what about that way? Is it the direction orthogonal to the boundary or the direction parallel to the boundary?
In the attached image I drew the angles, the boundaries and the "rays" (even if now we know they cannot describe exactly and completely the problem).
Mathematically I could say that if a material is a conductor, $|n_2| \to +\infty$, so $\sin \theta_r \simeq 0$ and consequently $\theta_r \simeq 0$.
So, the "only way left for the wave to travel" seems to be that orthogonal to the boundary.
If it is so, the sentence I cited in my first post was wrong ("the refraction angle $\theta_r$ is about $\pi / 2$"). Isn't it?
This was my question, because I was not persuaded of that sentence: this is the reason I wrote in the forum.
Thank you a lot,

Emily

 

[DrDu]

I completely agree.

I don't. Wave optics is completely satisfactory here.

Emily, maybe you can give us a precise citation where you read about the pi/2 value of the angle?

 

[EmilyRuck]

Emily, maybe you can give us a precise citation where you read about the pi/2 value of the angle?

In this paper (first page).

 

[DrDu]

I find the paper quite horrible, as they do not specify exactly the problem and the geometry they are interested in. Well, what else do you expect from electrical engineers?
Anyhow. Did you realize that they don't speak of the angle of refraction but of the angle of reflection?
They also say that the wave propagates approximately normal to the surface in the conductor which is also what you expected.

 

[EmilyRuck]

I find the paper quite horrible, as they do not specify exactly the problem and the geometry they are interested in.

That's right. I implicitly considered the geometry in the attached image just a few posts above, but the paper does not specify anything.

Did you realize that they don't speak of the angle of refraction but of the angle of reflection?

Yes, but just now :s.

They also say that the wave propagates approximately normal to the surface in the conductor which is also what you expected.

And if my computation is exact, this agrees with Snell's law too.
Thank you for all your help,

Emily