Two salient reasons:
- The interference is between the reflexions from two neighboring surfaces;
- The distance between these two surfaces is small - only a few wavelengths of visible light.
Point 1 means that it is only the phase difference that is important in determining the throughgoing / reflected light for a monochromatic wave. So, all monochromatic waves, whatever their random phase, produce the same interference pattern.
But we are dealing with all wavelengths. So let's look at point 2.
Near the center of the pattern, the distance between the two interfering surfaces can be approximated by a quadratic dependence on the distance $r$ from the center of the pattern. Therefore, the throughgoing intensity as a function of radius for light of wavelength $\lambda$ is proportional to:
$$\frac{1}{2}\left|1-\exp\left(i\,\frac{\pi\,r^2}{R\,\lambda}\right)\right|^2 = \sin\left(\frac{\pi\,r^2}{2\,R\,\lambda}\right)^2$$
where $R$ is the radius of curvature of the curved surface, if the interference pattern is formed by bringing a convex lens into contact with a glass flat, for example. Therefore the nulls in the pattern happen at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$.
Now we add the effect of all wavelengths. We can only see a narrow band of wavelengths, so we are looking at the sum of the interference patterns with nulls at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$ for $\lambda$ varying between $400{\rm nm}$ and $750{\rm nm}$. This means that the first null happens at a range of radiusses that varies only over a range of about $\pm 20\%$ - the "smear"width is well less than the distance between the first and second null. So, even with the spread over visible wavelengths, the first nulls line up pretty well. The second null less well and so forth. You see a series of colored nulls - the coloring is because different wavelengths have their nulls at different positions, but the nulls are still well enough aligned to see their structure. As you move further from the center, the nulls become more tightly packed and the accuracy of the alignment for all visible wavelengths becomes coarser than the null spacing, which means that we can no longer see fringes. This is exactly what happens in Newton's rings - the fringe visibility fades swiftly with increasing distance from the center.
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
Imagine you have a monochromator that you can tune to give you whatever wavelength you like. Now send light at a wavelength of 568 nm towards the interface.
What has been calculated is that destructive interference takes place, so that less of the incident radiation is reflected. Let's take an extreme case where the reflection from the first and second interfaces give reflected light of similar amplitude (but out of phase). This can be achieved by selecting an appropriate value for the refractive index of the film. In this case NO light is reflected.
So where does the energy of the incoming electromagnetic waves go? It must all be in the transmitted wave.
Now if you send light of a different wavelength, some of that is reflected and some of it is transmitted. i.e. not all of it is transmitted as in the case of light at 568 nm.
The net effect will be a greenish tinge to the transmitted light.
Edit: The OP is puzzled that the destructive interference between the reflected waves leads to enhanced transmission through the film compared with a situation that some reflection occurs. The easiest way to think about this definitely is conservation of energy. If (i) there is no reflected component at one wavelength, but (ii) there is at another, and if the Poynting vectors of the incident EM waves are the same in each case, then the power present in the transmitted EM wave must be enhanced in case (i) simply to conserve energy - to be exact, for normal incidence, the flux of the transmitted waves must equal the flux of the incident waves in case (i).
Let's take the case of normal incidence and (i) where we have managed to arrange that the amplitude of the reflected wave at the 2nd interface (and then transmitted back through the first interface) equals the amplitude of the reflected wave at the first interface. This is achieved by setting the refractive index of the oil film to be somewhere between that of the air and glass layers. In this case the summed E-field of the net reflected EM wave is zero - perfect cancellation. But we know from Faraday's law that the component of the E-field parallel to the plane of the interface must be the same immediately on either side of the interface. So the transmitted wave E-field at the first interface must have the same amplitude as the incident wave E-field.
In case (ii) we have a similarly thick film, but the wavelength is changed so we do not get destructive interference, but the amplitudes of the waves reflected at the first interface and second interface (then transmitted back through the first interface) are still equal. In this case the net reflected wave amplitude will be non-zero and have some phase shift that results in the total E-field on the air side of the interface being less than the E-field of the incident wave alone. Thus the transmitted wave also has an amplitude that is smaller than that of the incident wave.
Now you have to go through a similar consideration inside the film, but there is the additional complexity that some of the light reflected from the 2nd interface is also internally reflected off the first interface back into the oil (and so on...)! All this has to be solved in a set of simultaneous equations. However the basic fact is that in case (i) the amplitude of the EM wave transmitted through the first interface is larger than in case (ii). This in turn leads to a larger transmitted wave through the 2nd interface because the ratio of transmitted to incident light at this second interface depends only on their respective refractive indices and hence is the same in both cases.
Hopefully I have convinced you that it is easier to think about where the energy goes...