I recently read (sorry but I don’t have a reference) that interference is not only about destructive and constructive interference but moving energy from destructive to constructive regions according to conservation of energy. The place where this really bothers me is when light is being transmitted through glass. As light moves through the glass there is forward scattering of light where there is constructive interference between the primary and secondary waves; so light propagates forward. On the other hand, the light that back-scatters (or side scatters as well) destructively interferes with the primary waves (I am assuming that the light has already passed the surface atoms so reflection has already been accounted for). Said another way, I can set up a thin film such that the reflectance is zero (R = 0) and transmittance is one (T = 1), don’t I need energy in the waves to cause destructive interference to begin with? So how does interference account for this energy exchange?
[Physics] How does interference move energy from destructive to constructive regions
interferencevisible-light
Related Solutions
The thickness of the AR coating is chosen such that the reflections from the two interfaces cancel out (at the wavelength for which the AR coating was designed):
See Anti-reflective coating in Wikipedia.
As endolith points out in the comments, to explain how the transmission is enhanced, you have to draw a few more rays in the diagram. Here's another illustration, from the Wikipedia article for Fabry–Pérot interferometer, which shows a few higher-order reflections:
For the anti-reflective coating, you choose the thickness such that R1 and R2 cancel while T1 and T2 constructively interfere. Note that this is dependent on the wavelength, the angle of incidence, and the index of refraction of whatever is being coated. With other thicknesses, you can make a high-reflectivity coating, or a coating of whatever reflectivity you want.
First, in the article you mention, it says that biological color perception is different from physical color wavelength associations.
When you say a red object, it can reflect many wavelengths of EM waves, which altogether our eye and brain perceive as red.
But let us just talk about an object that would reflect only red wavelength EM waves, so in a physical way it would be red.
What happens with that object is, that the molecular structure of the object is so that the atoms' electrons energy levels (as per QM) have available levels so the atoms can be excited and so some wavelength photons will be absorbed. Other photons will not be absorbed (because they do not fit the available energy level's differences) and so those will be reflected.
The physical color of the object, so the EM waves that come from the object are reflected and emitted both. Some of the emitted are not even visible wavelengths.
So the physical color of the object comes from:
1.reflection of red wavelength photons
2.emission of red wavelength photons
So basically the object will absorb and re-emit some photons, and will reflect some others. All of those photons together will be the physical color of the object (the ones that are in the visible wavelength). So some of the EM waves coming from the object will not even be in the visible wavelength. But you are talking about the visible red wavelength ones so I am going to talk about those wavelengths.
The photons that are reflected are usually elastically scattered, so their wavelength is the same, their energy level is the same, only their direction is changed.
The photons that are absorbed and re-emitted are sometimes not emitted at the same wavelength (some of them). It is because when an atom gets excited, and the electron moves to a higher energy level (as per QM), the electron will come down to the lower level again. But the electron can come down in multiple steps. And in that case it is emitted multiple photons with different wavelengths then the originally absorbed photon. So if a photon gets absorbed, sometimes it is re-emitted with the same wavelength, in that case the electron came down to the lower energy level in only one step.
But sometimes the electron will come down to the lower energy level in multiple steps, and sometimes the electron will be excited so (when absorbing a photon) that it will move to a higher energy level of more then one level away, so it will go up more then one level at one step (with one absorption).
So basically one absorption can have multiple re-emissions, and in that case the absorbed wavelength, in your case red, will have nothing to do with the re-emitted wavelength.
In your case, the reason the object is red (physically again, not biologically) is because:
it is reflecting and emitting red wavelength photons.
And because it is absorbing other wavelengths, non-reds.
And it is only re-emitting red wavelength photons.
Even if it is absorbing non-red wavelengths, it will still re-emit red wavelength photons.
Best Answer
The basic surprise here is that energy is always conserved when waves superimpose. This seems like it shouldn't be true, since superposition causes amplitudes to add, whereas energy is proportional to the square of the amplitude. This nonlinear relationship makes it seem like energy should not be additive, and therefore can't be conserved.
This is a generic issue for waves, not just light waves, so let's consider it first in the case of waves on a string.
For example, if you take a light string and a heavy string and join them together end to end, waves that hit the boundary will be partially reflected and partially transmitted. One thing we can predict immediately is that energy is conserved, since the interaction between neighboring bits of the strings are governed by the ordinary laws of mechanics. Since energy is conserved at all points in space and time during the evolution of the waves, it must be conserved over all.
Another helpful example is a single uniform string, with oppositely propagating sine waves of the same wavelength encountering one another and superposing. There are moments when the waves cancel and the string is flat, but the motion recovers after such a moment -- was energy destroyed and then created again? No, because the flatness of the string only implied zero potential energy. The string had kinetic energy at the moment of flatness. If each wave separately had 1 unit of KE and 1 unit of PE, for a total of 4 units, then at the moment of flatness, we have 4 units of KE (due to doubled velocity) and 0 units of PE.
So in general, the idea is that conservation of energy has to be proved from the equations of motion, which are a differential equation holding at every point in space and time. For light, these equations of motion are Maxwell's equations, and one can indeed prove conservation of energy from them.
It's a nice exercise to verify conservation of energy for the case of oppositely propagating light waves, as in the second example with the rope above.
(I corrected T = 0 to T = 1 in this quote.)
Here we can analyze the reflected wave as a superposition of waves reflected from the front and back surfaces. Amplitudes add, not energies. Therefore these two waves, each of which individually would have had, say, 1 unit of energy, together do not have to have 1+1=2 units of energy. Their amplitudes add to 0, so together they have 0 units of energy.