EM wave reflected or transmitted?

[asada]

As I learn in class, when EM wave goes from medium 1 to medium 2, there are 3 possibilities that can happen

1. Totally transmitted (i.e when the angle of incident is 0 degree)
2. Partially transmitted and reflected (i.e when the angle of incident is between 0 and critical angle)
3. Total internal reflected (i.e when the angle of incident is bigger than critical angle)

In my understanding, at the atomic level, reflection means that the energy of the EM wave is absorbed by the layer of atoms between medium 1 and 2. This layer of atoms then starts "shaking", and this "shaking" emits an EM wave back to medium 1. On the other hand, transmission is part of the EM wave that cannot be absorbed by this layer and just pass right through it.

So why the ratio of transmitted and reflected differ depends on the angle of incident? I don't see the connection between the angle of incident and the behavior of how much energy is transmitted or reflected here? I mean, at the atomic level, what is the distinction between a 0 degree and 50 degrees angle of incident make?

Some people may point to the polarization of the wave as the answer. I agree that the polarization theory can explain why unpolarized can become polarized by refraction or reflection, but I don't think it could explain my question.

Here is my example of why it won't work: Let say we have a polarized light in y-direction polarization go toward the surface between medium 1 and 2? Let us assume the atoms of this surface between two mediums only capable of vibrate in a certain direction and very limited in other directions, and for our example, the free moving direction is the y-direction. If I use the polarization theory then the amount of wave being transmitted is fixed: since the surface can 100% absorb the wave, it will be 100% reflected. But in practice, this doesn't happen. A lot of light still transmitted through before total internal reflection start

 

[vanhees71]

On the macroscopic level of description, which is fully appropriate here, you get the reflection and refraction properties of a plane em. wave via Maxwell's equations and the boundary conditions at the boundary of the two media. The result are the Fresnel equations:

https://en.wikipedia.org/wiki/Fresnel_equations

which answer all these questions.

 

[nasu]

The wave is not totally transmitted at normal incidence. The reflection coefficient is minimum but not zero. There is a total transmission condition for acoustic waves but I don't know if it may happen for EM waves. Anyway, it is not happening at normal incidence.

 

[sophiecentaur]

Summary:: Why when an EM wave passes through a different medium, depend on the angle of incident, it may partially transmit or be totally reflected?

In my understanding, at the atomic level, reflection means that the energy of the EM wave is absorbed by the layer of atoms between medium 1 and 2.

That's not a good model - there is (or needs to be) no dissipation at the interface. The reason / explanation for reflection is the Boundary Conditions on either side of the interface. Consider when there are different Impedances on each side. The ratio of E to H on one side will be different from the ratio on the other side. However, for the Power Flow to be the same, the Product of E and H has to be the same. The two requirements cannot be satisfied unless there is a difference between Incident and `transmitted power flow. That difference is explained by some power being reflected.

This applies to obliqe or normal incidence.

 

[asada]

That's not a good model - there is (or needs to be) no dissipation at the interface. The reason / explanation for reflection is the Boundary Conditions on either side of the interface. Consider when there are different Impedances on each side. The ratio of E to H on one side will be different from the ratio on the other side. However, for the Power Flow to be the same, the Product of E and H has to be the same. The two requirements cannot be satisfied unless there is a difference between Incident and `transmitted power flow. That difference is explained by some power being reflected.

This applies to obliqe or normal incidence.

Sorry for asking this but what is E and H? And is the Power Flow you mentioned is a theory or such? I don't remember hearing those words before in class and not sure how to Google it

 

[sophiecentaur]

Sorry for asking this but what is E and H? And is the Power Flow you mentioned is a theory or such? I don't remember hearing those words before in class and not sure how to Google it

Sorry - unsuitable level perhaps. EM waves (ElectroMagnetic) consists of varying Electric (E) and Magnetic (H) fields. This is the same for light, radio, X rays etc. etc.The ratio of E to H is different for different media (and free space too). The Power Flow / Energy Flux, carried by the wave is proportional to E times H. Bottom line is that there is always a reflected wave when a boundary is crossed. You can always see light reflected by even the cleanest piece of glass in air and that's because there's no way all the light can get through the interface. That faint reflection follows the laws of reflection (r = i angles) same as for a mirror. Read about Fresnel Equations here.

There is no implied absorption of energy at the interface - that's dodgy Physics, I'm afraid.

 

[sophiecentaur]

There is a total transmission condition for acoustic waves

I just spotted this. This isn't true. Acoustic waves are reflected at a change of medium - say when sound in air is reflected mostly when the wave hits water. There is also reflection when a sound wave in a tube hits the open end. A wave is a wave. A change in wave impedance implies reflection, for any type of wave.
In practice, there won't be many occasions where there is a simple boundary between two gases (very similar wave speeds) but a cold air layer with a hot layer above it will produce a measurable reflection.

 

[nasu]

I just spotted this. This isn't true. Acoustic waves are reflected at a change of medium - say when sound in air is reflected mostly when the wave hits water. There is also reflection when a sound wave in a tube hits the open end. A wave is a wave. A change in wave impedance implies reflection, for any type of wave.
In practice, there won't be many occasions where there is a simple boundary between two gases (very similar wave speeds) but a cold air layer with a hot layer above it will produce a measurable reflection.

Look up "intromission angle" for ultrasound waves. It does not happen for interfaces between gases but at the liquid-solid interfaces. It does not happen for any interface, the acoustic impedances must satisfy some specififc condition for the total transmission to happen. It is rutinely observed for water to steel or water to aluminum in the ultrasound non-destructie testing.
For example, here (page 59 - book page numbers)
https://ocw.tudelft.nl/wp-content/uploads/Reader_chapter_3.pdf

A summary of the condition for total transmission (intromission):
https://images.slideplayer.com/42/11472629/slides/slide_17.jpg

 

[sophiecentaur]

Look up "intromission angle" for ultrasound waves.

Thanks for that. There's always something more to learn!
It's like the Brewster Angle for polarised EM, which I should have remembered before making my generalisation. But, apart from a missing caveat, I was justified in pointing out that the reflected ray is very often left out of simple diagrams and that is very misleading.