I figured out that when I aim a laser beam on my window at an arbitrary angle, the total internal reflection doesn't occur whatsoever. What's more, the refracted beam seems to be pretty intense. It was basically done in normal conditions, just lazer and glass window, so diffraction or something like that cannot be the answer.
Can anyone explain this?
Visible Light – Why Total Internal Reflection Doesn’t Occur When It Is Supposed To
laserreflectionrefractionvisible-light
Related Solutions
First, there is a qualitative difference between metal and glass: metal is a conductor, while glass is a dielectric. Under so called "plasma frequency" EM waves do not travel in conductors except that near the surface (see skin effect). For higher frequencies (usually far above the visible light) metal is transparent and refraction does occur. Specifically, answering to your question
What about the electron configuration of a the medium changes the net effect of the absorption and re-emissions of the photons?
Metal has a substantial amount of free electrons, glass doesn't. From the EM point of view, the difference originates from the relative effect of two terms in the Maxwell equation (roughly speaking, the corresponding dimensionless parameter depends on conductivity and wave frequency). From the QM point of view, the interaction with electron gas is substantially different from the interaction with an atom.
When you have hit the critical angle in a medium that refracts and the light completely reflects, are the photons moving is the same manner as they would be in a material that always reflects? How does this connect to the question in the previous paragraph?
Again, from the EM point of view both of these processes result in an evanescent wave and are quite similar. I am not sure what exactly happens with a specific photon (as a particle) here, as the evanescent wave is a general wave effect. Perhaps someone more knowledgeable in QED can provide a QM perspective on this.
You are getting reflections from the front (glass surface) and back (mirrored) surface, including (multiple) internal reflections:
It should be obvious from this diagram that the spots will be further apart as you move to a more glancing angle of incidence. Depending on the polarization of the laser pointer, there is an angle (the Brewster angle) where you can make the front (glass) surface reflection disappear completely. This takes some experimenting.
The exact details of the intensity as a function of angle of incidence are described by the Fresnel Equations. From that Wikipedia article, here is a diagram showing how the intensity of the (front) reflection changes with angle of incidence and polarization:
This effect is independent of wavelength (except inasmuch as the refractive index is a weak function of wavelength... So different colors of light will have a slightly different Brewster angle); the only way in which laser light is different from "ordinary" light in this case is the fact that laser light is typically linearly polarized, so that the reflection coefficient for a particular angle can be changed simply by rotating the laser pointer.
As Rainer P pointed out in a comment, if there is a coefficient of reflection $c$ at the front face, then $(1-c)$ of the intensity makes it to the back; and if the coefficient of reflection at the inside of the glass/air interface is $r$, then the successive reflected beams will have intensities that decrease geometrically:
$$c, (1-c)(1-r), (1-c)(1-r)r, (1-c)(1-r)r^2, (1-c)(1-r)r^3, ...$$
Of course the reciprocity theorem tells us that when we reverse the direction of a beam, we get the same reflectivity, so $r=c$ . This means the above can be simplified; but I left it in this form to show better what interactions the rays undergo. The above also assumes perfect reflection at the silvered (back) face: it should be easy to see how you could add that term...
Best Answer
Total internal reflection can occur when light is in an optically more dense medium, about to enter a less dense medium. It could occur going from glass to air, but not the other way around.
The angle of incidence must also be greater than the critical angle.