Understanding conceptually how a plane wave interacts with a boundary

[fred3142]

Hi,

I'd love a to have a more graphical understanding of how a plane wave interacts with a boundary. I know the maths that describes it, Fresnel's equations etc, and how Brewster's angle is derived and stuff.

I'm rather confused with the dipole concept. From what I understand, when a plane normal to the boundary hits the boundary, charge gathers on the boundary and reflects it. If the material is a dielectric though, the material will also reflect it, but with a phase delay (I imagine this to be because the dipole needs to absorb it and then re-radiate it). My issue comes when trying to understand what happens at an angle. I believe that at an angle, it'll basically be a combination of what I've described, as well as what happens when the electric field polarisation is parallel to the boundary (the polarisation is perpendicular to the plane of incidence). I don't 'get' what goes on here, what does the boundary 'do', what causes the reflection? And at Brewster's angle, the dipoles supposedly become oriented in such a way that they do not radiate perpendicular and hence nothing is reflected... How does the incoming wave affect the orientation of the dipoles? Is there a picture or animation that can show me the dipoles rotating?

I understand that the understanding I'm seeking may seem trivial because *really* this stuff doesn't work like this, and it's simple a limited (flawed) model; however, it would greatly help my understanding and I'd appreciate some help with this! I'm not really interested in a bunch of equations like most books seem to present.

Thank you.

 

[Jano L.]

I don't 'get' what goes on here, what does the boundary 'do', what causes the reflection?

In macroscopic EM theory, the boundary just imposes boundary condition on the EM field. The reflection comes as mathematical consequence of this and the Maxwell equations.

In microscopic theory, the boundary is made of charged particles. When external wave impinges on them, it makes them oscillate preferably in the direction of the polarization, and these particles then radiate their own EM waves out in all directions, but most intensity goes to directions from which the oscillations are viewed best (perpendicular to the line of sight). When the external wave is added with the elementary waves of the particles, it has part that comes toward the boundary (external wave) and part that returns back away from it (sum of the elementary waves due to the particles in the boundary).

And at Brewster's angle, the dipoles supposedly become oriented in such a way that they do not radiate perpendicular and hence nothing is reflected... How does the incoming wave affect the orientation of the dipoles?

The direction of oscillations of the charged particles in the boundary settles in the same direction in which the electric field settles. For simple dielectrics, this depends on their susceptibility and also on the polarization of the external wave and its angle of incidence. The oscillations settle in such pattern that the boundary conditions for given geometry are satisfied, and there is only one way this can happen.

It is hard to explain this from the point of view of microscopic theory, but it should be possible.

See the great article of Victor Weisskopf on light-matter interaction, he explains similar things in more detail without complicated mathematics:


https://www.physicsforums.com/attachment.php?attachmentid=63282&d=1382707763

 

[fred3142]

Thank you. Are you aware of any books which go into this? The article seems quite good too!

 

[Jano L.]

Good introduction to these phenomena is in Feynman's lectures:

http://www.feynmanlectures.caltech.edu/I_31.html

http://www.feynmanlectures.caltech.edu/I_33.html

 

[PJC01]

Hi there,

I completely understand your desire for a more graphical understanding of how a plane wave interacts with a boundary. It can be a complex concept to grasp, so I'll do my best to explain it in simpler terms.

Imagine a pool of water with a wall at one end. When a wave travels towards the wall, it will hit the wall and reflect back. The same thing happens with a plane wave hitting a boundary. The boundary acts as a barrier that reflects the wave back in the opposite direction.

Now, let's add a dipole to this scenario. A dipole is essentially two equal and opposite charges that are separated by a small distance. When a plane wave hits the dipole, it will cause the charges to vibrate in opposite directions. This vibration creates an electric field that will also reflect the wave back. This is what happens when the plane wave hits a dielectric material, as you mentioned. The dipole absorbs and re-radiates the wave, causing a phase delay.

When the plane wave hits the boundary at an angle, things get a bit more complicated. The wave will still reflect off the boundary, but it will also interact with the dipoles in a different way. At Brewster's angle, the electric field of the wave is parallel to the boundary. This means that the dipoles will not vibrate in opposite directions, but rather in the same direction as the electric field. This creates a canceling effect, where the reflected wave is greatly reduced.

To answer your question about how the incoming wave affects the orientation of the dipoles, it is important to understand that the dipoles are already present in the material. The incoming wave simply causes them to vibrate or rotate in response to the electric field of the wave. This is what creates the reflection at different angles.

I hope this helps to clarify the concept for you. As you mentioned, it is a limited model and there are many other factors at play in the interaction between a plane wave and a boundary. However, this simplified explanation should give you a better understanding of the basics. If you have any further questions, please don't hesitate to ask.