Calculation of reflection coefficient

[ShayanJ]

I've got a problem asking for the reflection coefficient of a system consisting of a ideal conductor covered with a thin layer of lossless dielectric. I just don't know what to do.How can I do it?
Thanks

 

[DrDu]

The system is neither absorbing nor transmitting. Methinks that therefore the reflectivity has to be 100%.

 

[ShayanJ]

The system is neither absorbing nor transmitting. Methinks that therefore the reflectivity has to be 100%.

How can I prove those?
I mean, I know the physical reasoning, but it looks like some calculation is demanded because in the problem the width of the dielectric layer is given as a parameter [itex] \tau [/itex] and it is stated explicitly to "calculate" the coefficient!

 

[Born2bwire]

How can I prove those?
I mean, I know the physical reasoning, but it looks like some calculation is demanded because in the problem the width of the dielectric layer is given as a parameter [itex] \tau [/itex] and it is stated explicitly to "calculate" the coefficient!

This is a three layer system and the reflection coefficient will reflect the various phase changes and guided modes that can arise. Did the problem specify a specific polarization and direction of the wave?

The general solution can be found in texts like Balanis' on EM or Chew's "Waves and Fields in Inhogeneous Media.' Balanis is the more accessible text.

Is the layer thickness on the order of the wavelength? There could be a simplification if this is a problem of optical light off of a coated metal compared to the RF case of plane waves in layered media.

 

[DrDu]

This is a three layer system and the reflection coefficient will reflect the various phase changes and guided modes that can arise.

Of course it is a three layer system, but as the lowest layer is nonabsorbing and completely reflecting and the other ones non-absorbing, there are few effects to be expected besides maybe some phase and polarisation shifts.

 

[DrDu]

How can I prove those?
I mean, I know the physical reasoning, but it looks like some calculation is demanded because in the problem the width of the dielectric layer is given as a parameter [itex] \tau [/itex] and it is stated explicitly to "calculate" the coefficient!

Usually, the starting point would be to set up the incoming and outgoing waves in the different layers and specify the boundary conditions. It is crucial that in the third layer, there is only an incoming wave, but no outcoming wave.
So you can work backward from there.
In your problem, there is not even an incoming wave in the third layer, so nontrivial effects are only to be expected if the second layer is absorbing, which you explicitly excluded.

 

[ShayanJ]

This is a three layer system and the reflection coefficient will reflect the various phase changes and guided modes that can arise. Did the problem specify a specific polarization and direction of the wave?

The general solution can be found in texts like Balanis' on EM or Chew's "Waves and Fields in Inhogeneous Media.' Balanis is the more accessible text.

Is the layer thickness on the order of the wavelength? There could be a simplification if this is a problem of optical light off of a coated metal compared to the RF case of plane waves in layered media.

In fact the problem asks for the reflection coefficient for the normal incidence which makes both polarizations the same!
Nothing is assumed about the layer's thickness.

Usually, the starting point would be to set up the incoming and outgoing waves in the different layers and specify the boundary conditions. It is crucial that in the third layer, there is only an incoming wave, but no outcoming wave.
So you can work backward from there.
In your problem, there is not even an incoming wave in the third layer, so nontrivial effects are only to be expected if the second layer is absorbing, which you explicitly excluded.

The third layer is an ideal conductor, so no wave is present in it, neither an incoming wave nor an outgoing one!
As a boundary condition, the tangential component of the electric field should be continuous at the interface. Because here the third layer is an ideal conductor, the electric field, and so its tangential component, is zero inside it and so the tangential component of the electric field should be zero at the second layer too. But for normal incidence, we only have tangential components of fields which means the field is zero in the second layer!
Also we should note that there will be infinite number of reflections until the wave is reflected completely. But this point should be considered after solving the previous one!

 

[DrDu]

The third layer is an ideal conductor, so no wave is present in it, neither an incoming wave nor an outgoing one!

That's what I said.

As a boundary condition, the tangential component of the electric field should be continuous at the interface. Because here the third layer is an ideal conductor, the electric field, and so its tangential component, is zero inside it and so the tangential component of the electric field should be zero at the second layer too. But for normal incidence, we only have tangential components of fields which means the field is zero in the second layer!
!

No, E will be zero at the boundary 2/3 but not inside all of 2. You will get a standing wave and in a standing wave, E and B will be 90 degrees out of phase. So there will be a nonvanishing oscillating B at the boundary and away from the boundary E will be non-zero due to rot E= dB/dt

 

[Born2bwire]

In fact the problem asks for the reflection coefficient for the normal incidence which makes both polarizations the same!
Nothing is assumed about the layer's thickness.



The third layer is an ideal conductor, so no wave is present in it, neither an incoming wave nor an outgoing one!
As a boundary condition, the tangential component of the electric field should be continuous at the interface. Because here the third layer is an ideal conductor, the electric field, and so its tangential component, is zero inside it and so the tangential component of the electric field should be zero at the second layer too. But for normal incidence, we only have tangential components of fields which means the field is zero in the second layer!
Also we should note that there will be infinite number of reflections until the wave is reflected completely. But this point should be considered after solving the previous one!

Ok, then that greatly simplifies the whole process. But I'll expand upon what has already been stated regarding the solution. You basically start by assuming the basic solution to the waves. We can work with only the electric field since the magnetic field can be solved directly from there. So we start with an incident wave,

[tex] \mathbf{E}_{inc}(\rho,z) = E_0 e^{-ik_1z}\hat{x} [/tex]

where the boundary between air and dielectric lies at z=0 and along the x-y plane and the PEC lies at z=-t. Since we are talking about normal incidence, there is no dependence on the transverse position and the wave vector is in the -z direction. I've also implicitly assumed a -i\omega t time dependence.

So in the air layer, you have two waves, the incident and reflected wave which will be described by a reflection coefficient. In the dielectric, you will have a transmitted wave, and a reflected wave off of the PEC boundary. Now the reflected wave is going to be the reflection of the transmitted wave. In the PEC, as been stated previously, there is no wave. Thus,

[tex] \mathbf{E}_1(z) = E_1 \left[ e^{-ik_1z}\hat{x} + \mathcal{R}_{12} e^{ik_1z}\hat{x} \right][/tex]
[tex] \mathbf{E}_2(z) = E_2 \left[ e^{-ik_2z}\hat{x} + R_{23} e^{ik_2z+2ik_2t}\hat{x} \right] [/tex]
[tex] \mathbf{E}_3(z) = 0 [/tex]

where [itex]\mathcal{R}_{ij}[/itex] is the generalized reflection coefficient from the i-th layer to the j-th layer. Generalized in that they incorporate all the multiscattering effects from the layers above or below. We know that [itex]\mathcal{R}_{23} = R_{23}[/itex], where [itex]R_{ij}[/itex]is the reflection coefficients between the two layers as if they existed by themselves. We know this because there are no layers below the PEC, thus [itex]\mathcal{R}_{23} = R_{23}[/itex]. In addition, we have a factor of 2 in the 2ik_2t term because the wave has to travel a distance of t to get to the PEC boundary, and another phase factor comes in because the reflection coefficient is the ratio of the incident and reflected waves at the boundary (the incident wave also picks up its factor of ik_2t when it travels to the PEC boundary).

So you are correct that there are going to be these infinite number of reflections and transmissions amongst the layers. But you can account for these by the fact that we will have a single reflection/transmission coefficient account for all these multiscattering effects. The electromagnetic waves still must satisfy the boundary conditions along the boundaries. So you know that, for example, at the air-dielectric interface at z=0, that the tangential electric field is continuous. At the PEC boundary, the total tangential electric field is zero. So we know that

[tex] E_2e^{ik_2t} + R_{23}E_2e^{ik_2t} = 0[/tex]

So we see that [itex]R_{23} = -1[/itex].

Let's take a look at the constraint equation for the air-dielectric boundary. At the boundary, we have a downward going wave of amplitude [itex]E_2[/itex]. This downward traveling wave is due to the transmitted wave coming in from the air into the dielectric, and we have a wave that is coming up out of the dielectric and reflected off of the air. But the reflected wave is the upward traveling wave in the dielectric which is already scaled by [itex]E_2R_{23}[/itex]. Thus, at z=0, the downward traveling wave is:

[tex] E_2 = T_{12}E_1 + R_{21}R_{23}E_2e^{-2ik_2t} [/tex]

Here you can solve for [itex]E_2[/itex]. Next, construct the constraint equation for the upward traveling wave. The upward traveling wave in air is due to the GENERALIZED reflection coefficient and this is equal to the reflection of the incident wave in air on the dielectric and the transmission of the upward traveling wave from the dielectric into the air. So,

[tex] E_1\mathcal{R}_{12} = R_{12}E_1 + T_{21}E_2R_{23}e^{-2ik_2t} [/tex]

So now you can solve for the generalized reflection coefficient and that's all that there is too it. Things get slightly trickier when we have oblique angles but the process is the same. More convoluted for the cases of anisotropy.

Hopefully there isn't too many mistakes in the above. I've got a run so I can't spend anymore time proofreading. Unfortunately, I have only seen the treatment of this in graduate textbooks. However, a transmission line is a 1D wave equation. So I'm sure you can draw parallels in how to solve for the reflection coefficient in a series of transmission lines.

EDIT: The above assumes that we only have an anisotropy in the z-direction and are isotropic otherwise.

I'm still wondering if this is really the problem you are being asked as it seems like it's rather advanced from what you have stated. I could imagine that they were asking for the reflectivity which as already been stated is a trivial 100% assuming lossless dielectrics.

 

[ShayanJ]

Thanks people, your posts were really helpful.

I'm still wondering if this is really the problem you are being asked as it seems like it's rather advanced from what you have stated. I could imagine that they were asking for the reflectivity which as already been stated is a trivial 100% assuming lossless dielectrics.

This problem is from a bunch of problems from previous years physics olympiads. I mean national olympiad or meyabe a level behind that!
There was also a strange problem among its QM problems. It was finding energy eigenstates and eigenvalues for an electron in a one dimensional coulomb potential which I figured has no solution in terms of elementary functions and so can't be solved when giving the exam. But most problems are not like these!
I think they put such problems to get sure that no one solves all of the problems so maybe they get some prestige!