Interference of Light: Where Does the Energy Go?

[MrFrankie]

Suppose to have a beam of light reflecting over two parallalel surfaces of a material, so that the reflectd rays overlap. There is interference and the square of the amplitude is:
r^2 * Eo^2 * sin (2Pi n d / L)
where
r is the reflection coefficient
Eo is the max energy of the field
n is the refraction index for the material between the two surfaces
d is the space between the surfaces
L is the wavelenght

The energy should be proportional to this value.
This means that, at fixed value of angle and d, only a fraction of the incident energy is reflected.
Where is the remaining part?

 

[Dale]

I am not really clear about the geometry and the equation, but any missing energy goes to heating the reflectors.

 

[MrFrankie]

No, there is no absorption of energy in the material, the wave is completely reflected or transmitted, without anelastic interaction. Anyway the absorption is a local phenomena and cannot depend on the interaction of the wave after the reflection on the second surface.

 

[MrFrankie]

The phenomenon is the typical interaction on thin layers of oil on the water surface.

 

[MrFrankie]

This is exactly what happens. Anti-reflective coating
http://wapedia.mobi/en/File:Optical-coating-2.png

 

[Dale]

No, there is no absorption of energy in the material, the wave is completely reflected or transmitted, without anelastic interaction. Anyway the absorption is a local phenomena and cannot depend on the interaction of the wave after the reflection on the second surface.

Ah, I thought that a reflection coefficient different from 1 indicated absorption rather than transmission. Then any energy missing in the reflection is transmitted.

 

[MrFrankie]

Well... I'm not sure.
Reflection and transmission amplitude depends on the relative coefficients of the material. I think it's Stokes Relations.
You have interference only on the reflected ray, as a modulation of amplitude variable with the thickness of the film and the direction of the ray.
the transmitted ray shouldn't be affected by the change of thickness, except for the density of energy that could be lower for thick films.
The problem is that you have reflection and still the waves erase each other when thay are out of the material.

 

[Dale]

The problem is that you have reflection and still the waves erase each other when thay are out of the material.

I think you may be misunderstanding something here. The waves do not erase each other, they add in superposition. There is both constructive and destructive interference in superposition, not simply destructive interference. Also, don't forget that light carries energy in both the electric field and in the magnetic field, and in a region where you have two light waves in superposition those two are not necessarily proportional to each other any more.

 

[MrFrankie]

Please can you explain me better this part?
If the rays have the same direction and same frequency, and there is destructive interference, that is a constant Pi phase, how can they have constructive interference?

Besides, the wave given by the supeposition of two waves should still be a wave, with an amplitude modulated in the phase, therefore it follows Maxwell equations.Then electric and magnetic field should be proportional, for example with c as constant in the empty space.

 

[Dale]

Please can you explain me better this part?
If the rays have the same direction and same frequency, and there is destructive interference, that is a constant Pi phase, how can they have constructive interference?

Assuming the incident wave is normal to the reflective surface then they don't have the same direction, they have opposite direction. The incident wave is propagating towards the reflector, and the reflected wave is propagating away from the reflector. They always have a constant Pi phase offset at the reflector and therefore add destructively there but a quarter of a wavelength away from the mirror they are always in phase and add constructively! This pattern continues through space, every quarter wavelength changing from constructive to destructive interference.

Besides, the wave given by the supeposition of two waves should still be a wave, with an amplitude modulated in the phase, therefore it follows Maxwell equations.Then electric and magnetic field should be proportional, for example with c as constant in the empty space.

Let's consider an incident wave that is propagating vertically downward to a perfect reflector on a level floor. Further, let's say that the E field points North where it strikes the reflector at some instant in time. Since the wave is propagating downwards by S = ExB we know that the B field points East.

Now, let's examine the reflected wave. The E-field is 180° out of phase, so it points South, and since the reflected wave is propagating upwards we can determine that the B field points East.

So, by superposition, the E-field at the reflector is 0, and the B-field at the reflector is twice the B-field of either individual wave. In other words, where the E-field shows destructive interference the B-field shows constructive interference. And where you have such interference it is not always true that the electric and magnetic fields are proportional to each other.

 

[MrFrankie]

Sorry, maybe I've not explained what the problem is.

I'm talking of the superposition of the two reflected waves, both part of the first incident wave, both having the same (wt-kr), only different for the phase, due to the different optical path.
The intensity is not function of the space of time, it only depends on constant fixed parameters, so the difference of phase you have at the C point (where the transimitted wave passes again through the surface after the reflection), doesn't change anymore. It could only change for a different frequency of the two waves, but both are part of the same coherent beam, so this cannot occur.

Relating to the Poynting vector you're right, but when you make the superposition of two waves pointing in the same direction, you cannot change only the sign of one field, otherwise the sign of S=ExB would change and the wave would point in the opposite direction.
But the two waves have the same direction.

 

[Dale]

OK, now I understand what you are talking about, but this goes back to my first reply. It is simply not possible to have a material that performs as you suggest. Specifically, the material must heat up. Here is a link that explains and derives the conservation of energy law from first principles:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

Note that the Ohmic heating term (E.j) is an inevitable consequence of any situation where the flux of energy across the boundary is not equal to the rate of change of the energy contained in the fields within the boundary. So, for a lens coating some small fraction of the energy does not pass through the coating and must therefore either be reflected or cause heating. For a correctly designed anti-reflective coating it heats the material rather than be reflected. This heating is a direct result of Maxwell's equations and cannot be avoided.

 

[MrFrankie]

Well I agree with you about the energy continuity equation, but I don't agree with other points:
1) You usually have heating of the material for the absorption of energy in roto-vibrational levels, true for infrared spectra, but in this case you are using light, so energies of few eV. Usually in transparent materials the energy gap for the conduction band is higher, so you don't have interaction of ligth.
And anyway the thickness of the material is so small that the absorption would be very low;
2) You don't have free electrons to interact with the radiation, because the materials you use are insulators. So the current density j cannot refer to some movement of fields sources ALONG the material;
3) Even if there is heating INSIDE the material, this doesn't depend on the phase of the transmitted wave. So it must be constant. The not absorbed part is transmitted or reflected. The trasnsmitted part doesn't depend on the phase, but only by the coefficients. The reflected part is the only one affected by interference.

It's like if you have that the sum of three terms must be constant, but one changes and the other two remain constant. I know that there is conservation of energy, but I cannot see where it is.

 

[Dale]

Sorry, I don't know how I can describe it better without actually working through Maxwell's equations for you. If it bothers you enough then simply work out Maxwell's equations and verify the energy conservation.

Btw, the conduction band gap is irrelevant here. This is a purely classical effect, not a quantum effect.

 

[MrFrankie]

Thank you for the explanation :-)
I will work on the Maxwell equation.

 

[Dale]

You are welcome. Also, I forgot to mention one thing.

The current in the energy conservation equation is the total current, so it includes free current as well as the polarization current or displacement current. Since you are dealing with insulators as you noted the free current will be minimal and most of the current will be polarization current.