electromagnetism

Question

Q3.

Answer 1

Hi Ola

Good questions!

First the short answers, then we follow up with why:

Q1

If the light is polarized in the plane of incidence then the angle at which no light is reflected (Brewster's angle) is given by arctangent of the ratio of refractive indices for the two media, with the index of the medium the light starts in on the bottom. Since your starting medium is air (n≈1) and the other is glass (for a nonmagnetic material, (εr)^(1/2)=n=2) you have;

θ≈arctan(2)≈63 degrees

If instead the light is polarized normal to the plane of incidence, then there will be no angle at which no light is reflected.

Q2

If the light was of circular polarization then the same Brewsters angle applies, except that instead of an angle where no light is reflected, it is merely the angle at which the least light (half the amplitude) is reflected.

Now the fun part:

Q1

When light is incident on a dielectric medium the oscillating electric field sets the more mobile electrons in the medium into motion along the direction of polarization of the incident light, since the polarization of light is defined as the direction of its electric field oscillations. The electrons slosh back and forth along that direction, creating a emanations in the electric field called dipole radiation. A picture is worth a thousand words here so I encourage you to look up dipole radiation on wiki or google.

Here is the missing piece. Dipoles do not radiate in the direction of motion of the source charge, instead sending toroidal shaped waves emanating from their sources with maximum radiation along the plane to which the motion of the charge is normal. Thus the condition for no light being reflected is that the direction the incident light sets the charges sloshing in within the dielectric be parallel to the direction that the reflected light wants to propagate (according to the law of reflection, angle of reflection equals angle of incidence except on the opposite side of the normal vector).

This condition can be expressed as angle of incidence and refraction sum to 90 degrees. Combining this with Snell's law which relates the angle of refraction to the angle of incidence and the refractive indices of the two mediums (n1 sin[θ1]=n2 sin[θ2]), we can generate Brewsters formula used above in the short answer.

θB=arctan(n2/n1)

As an aside, Snell's law can be quite neatly derived by assuming the light ray will always take the path of least time from point a to b.

I imagine you already see that for linear polarization normal to the plane of incidence there is no component of the sloshing charge that moves in the direction any reflected light will propagate, and thus no prospect of this mechanism reducing the reflected amplitude by adjusting the angle of incidence.

I suggest you draw out the plane of incidence with the incident ray coming in at the angle we found above and the transmitted ray at an angle given by Snell's law. Also draw the reflected ray (so that the normal to the boundary between mediums bisects the angle between incident and reflected rays). Take the case of in plane polarization and draw the direction of the electric field oscillations (remembering that this is not only in the plane of incidence but also perpendicular to the direction of propagation). If you do this you will see that the electric field oscillations of the transmitted ray are parallel to the reflected ray.

Q2

Circularly polarized waves can be thought of as the sum of two perpendicular linearly polarized waves that share amplitude and frequency and are properly offset in time, with one linear polarization leading the other by a quarter rotation. If you can choose one linear polarization to point normal to the plane of incidence, the other must be perpendicular to both that and the direction of propagation of the light itself. We can see from the answer to Q1 that what would happen is that one linear component contributes nothing to the reflected light while the other contributes fully. Further, the reflected light will be linearly polarized normal to the plane of incidence while the transmitted or refracted light's amplitude will be a mixture of one part this polarization for every two parts polarized in plane.

Finally a point of disambiguation. When you say "total absorption" takes place at Brewster's angle your not wrong exactly, after all the light is going into the medium with none bouncing off. But there is a distinction to be made between absorption and transmission (refraction being a form of transmission in which the light ray is bent, because the amount of this bending depends on the light's frequency, refraction also brings with it prismatic effects). Transmission (or reflection for that matter) can be thought of as absorption followed promptly by reemission without significant change in the shape of the wave, but the word absorption is best reserved for cases in which the process of absorption changes the wave, converting it into some other form before reradiating it. This is what happens for example if you were to use a magnifying glass to shine (mostly yellow) sunlight on a piece of metal until it glows red hot. The yellow light is said to be absorbed and reemitted at a different frequency (red).

Hope this helps!

Joss