2 windows - will I see the reflections? 
I have a question regarding photons nature. Let's say I have a single source of light - regular bulb and the observer - in the same room. 
The observer looks through a glass window (normal glass window-nothing special about it) and sees his reflection, but some of the light is passing through the window.
Now I put a second window - as presented on the attached photo. Will the second window reflect some of the light back to the observer, or it will pass 100% of the light forward?
This may be pretty basic but this question sparked a discussion and there where no definitive answer.
 A: Classical electromagnetism predicts that the observer will see four first order reflections, one from each glass-air interface. The reflectivity at each interface is 4%. Multiple reflections will also occur, for which light is reflected back and forth. They are at least third order in intensity, that is, are 4% of 4% of 4% or 64 per million. There are 14 of these, so the total intensity is 0.1 per thousand or 160 times less than the first order total. The list goes on with fifth order etc.
Quantum mechanically these numbers mean the probability that an incoming photon is reflected. The same images that are predicted classically will be built up from single photons much in the way the paintings of French painter Seurat are built up from dots. The more photons that are observed the less noisy the images will be u til on the classical limit all shot noise is averaged out. 
A: It will reflect some of the light back, depending on the thickness of the glass. The partial reflection of light from the two glass windows varies from 0%-16% (that is, on average 8% of the light is reflected back by both windows combined). Definitely read QED by Feynman for more detail. 
A: This link was given to me by one of the people who attended the original discussion about this problem, so I thought I should publicly give my 2 cents. 
While optics is not my specialty, I am quite certain that the light that goes through the glass is identical to the light that was incident on the glass, except that it is lower intensity because part of it is absorbed and part is reflected. 
The confusion I think came from the mixup with light passing through the polarisation filters and not through ordinary glass. The usual, unpolarised light is a mixture of photons (i.e. light particles) of two polarisations. Polarisation is a property of every photon. For a photon going along a direction, we assign a polarisation which some direction perpendicular to the motion of the photon. This has to do with the way electric (and magnetic) field oscillate to make the beam, and interested reader is invited to google this.  
The point is that light has associated with it this property of polarisation, so a beam of light has associated to it its direction of motion and another direction perpendicular to it which we call polarisation. 
The polariser (or polarisation filter) simply allows only beams polarised in a certain direction to pass, and this direction is the property of the filter. So if we put two polarisation filters one after another, and orient both of them so that their polarising axes are aligned, then the first polariser will transmit 50% of the light (in reality it will be slightly less because there will undoubtedly absorb some of the light which is polarised correctly as well), but the second filter will transmit 100% of the light (again slightly smaller amount in reality). 
But the setup described is not analogous to this. A piece of ordinary glass does not transmit polarised light. If it would, what would be the poralisation axis? The piece of glass is so random, that even if a very small piece had a property to polarise light (it doesn't as far as I know), another piece of glass a bit further away would polarise light in some other direction, because the molecules that make the glass would all be moved a bit in relation to the first ones. 
However that being said it is actually true that reflected light are partially polarised, but this is only for the light that is reflected at an angle. In any case it is only completely polarised for a special angle (called the Brewster angle), but this is not relevant for the transmitted light anyway, so not relevant for the question. 
That's my 2 cents. Hope it helps.
A: There are a few things to clarify, this is a more QM answer:
You are asking about visible wavelength photons, so I am going to talk about those.
When the photons in your case interact with the glass lattice atoms, there is:


*

*elastic scattering, the photons keep their energy and phase

*inelastic scattering, the photons keep part of their energy and phase

*absorption, the photons give all their energy to the atom
Now when the photons interact with the glass, all three happen, but their ratio is different.
In the case of glass, there is:


*

*reflection, that is the mirror image of yourself, that is 1., elastic scattering, the photons keep their absolute energy and phase and relative energy and phase and angle, this is the only way to build an almost perfect mirror image. It is very important to understand that even with reflection, all three happens, but most of the photons get elastically scattered, and the ratio of inelastic scattering and absorption is very little.

*refraction, that is the image that goes through the glass, that is 1. too, elastic scattering, the photons keep their absolute energy, phase, and relative energy and phase and angle, this is the only way to keep an almost perfect image through the glass. It is very important to understand that with refraction, all three happen, most of the photons get elastically scattered, and the inelastically scattered and absorbed photons' ratio is very little.

*absorption, that is 3., when the photons' energy gets transferred into the vibrational energies of the glass lattice molecules, heats up the metal.

Now it is as per QM all about probabilities, but when the photons (visible wavelength) reach the first glass, 4% will be reflected, and 96% will be refracted.
What is the difference between them? The angle only. Both reflected and refracted will be elastically scattered. On this site, there is a debate whether refraction is elastic scattering or absorption and re-emission. Now it is elastic scattering. That is the only way to keep the photons' absolute energy, phase and relative energy, phase and angle. This is the only way to keep a mirror image through the glass.
So both the reflected and refracted photons get elastically scattered, 4% of them will change angle almost backwards, and 96% forwards relative to the angle of incidence. So the only difference is the angle. Why? It is the lattice structure of glass. It has a 4-96 ratio of probabilities of photons' interference (for visible wavelength) being constructive backwards (reflection) or forwards (refraction) angle. Metal has a 99-0 (for visible wavelength) ratio of probabilities of photons' interference being constructive backwards or forwards.
Now in your case, the first glass will reflect 4% of the photons, 96% will go through. The 4% will be visible for you as tour own mirror image. The 96% will go through, and at the second glass, 0.96*96% will go through, and 0.04*96% will be reflected. So when the 0.04*96% reaches the first glass backwards, 0.04*0.04*96% will be reflected backwards as a mirror image. And 0.96*0.04*96% will go through the first glass backwards, and you will see those as your own reflection too.
A: Ok. I am not an expert nor I have enough time to work through an answer comparable to many excellent answers I find here. 
Generally your logic is correct, meaning that every time the light goes through a medium like the one you are describing a percentage of the beam will pass through and the rest of it will be reflected.
Naturally, an ideal window will let the beam 100% to pass while an ideal mirror would reflect 100%. Now make sure you understand what I mean by beam.
There are three main ways to think about the nature of light: beams(as in most of early classical geometric optics), waves(modern mostly classical stuff), photons(more quantum stuff). In the problem you are describing, a beam seems like the best way to conceptualize light.
Now, obviously all the views must be correct and consistent. In your problem if you think of it as photons there are statistical formulas(let's say that each photon has a certain probability to pass through or be reflected) that will produce the same result.
To conceptualize these ideas Feynman's QED would be a natural way to go. A more mathematical/professional approach exists in every undergraduate modern physics book.
