| Light as a wave, by definition, is polarized since an arbitrary light wave can always be decomposed into plane waves described
by the electrical field vector
E(r,t)=E0exp{i(kr –wt)} and the corresponding expression for the magnetic field. |
|  | The electrical field vector (or the electrical displacement vector D in material) thus is
contained in some plane (the polarization plane) that contains the direction of
propagation (direction of k-vector). Looking for example at a picture we had before but now augmented
with the relevant vector quantities, we see that in this case: - Polarization
plane=(y,x) plane.
- Propagation direction=k direction=Poynting vector S direction=x direction here.
- E-field amplitude in y-direction given by E0.
- H-field amplitude in z-direction given by H0.
|
| |
 |
Ey(t) | = Ey0 · exp{i(kx – wt)} | Hz(t) | = Hz0 · exp{i(kx – wt)} |
|
|
 | This all looks well-known and clear. Usually we don't bother to go much
deeper at this point. In a Master study course, however, you now ponder the picture and the equations above for a
while. Then a few questions should come up, for example: |
| |
- Can polarization be described by a vector? Could I just use a unit vector in E0-direction?
- What happens if E0; H0 are not parallel to D0; B0,
something to be expected in anisotropic
materials?
- How do I describe polarization in the particle picture?
- Is the electrical and magnetic field really in phase as shown above?
- How large, in V/cm, is the electrical field E0
in a typical light wave. How large (roughly) is it for one photon of a given
energy?
- Let's assume you know the electrical field E0: how large is the magnetic field H0 associated with it?
- The energy a photon
brings along is EPh=hn. How is this light
energy transported by an electromagnetic wave?
It's not very difficult to conceive and to
understand these questions. How about the answers? |
|  | The bad news is: There are no easy answers for some of those questions. |
|  | The good news is: For most of what follows you don't have
to worry about the answers. I will go into details later whenever necessary. For now you can use this link to get some ideas about the answers and, even better, you could do the
simple but illuminating exercise below. |
 | Let's start with
the last question. Light flow equals energy flow. All of us who once experienced a sun burn (or shot down one of those
evil alien space ships with a Laser gun) know that. How does light energy "flow" exactly? |
|  | Well, we (should) know how much energy density W =energy /cm3 is contained in an electric or
magnetic field in vacuum (in materials use D and B instead of E and
H): We have: |
| |
|
|
[Welect; magn] | = | [Ws m3] |
|
|
 | We do not yet know, how E and H are linked. All we need to know is
that this follows "straight" from the Maxwell equations and is given by |
|
|
E0 | = | æ ç è | mrm0
ere0 | ö ÷ ø | ½ | ·
H0 | = | Zw· H0 |
|
|
|  | We defined a new quantity
Zw=E0/H0=(m0mr/e0er)½ that is called the wave
impedance of the medium. This is an apt name since Zw does have the
dimension "Ohm" (W) as required for an
impedance. |
|  | For the vacuum
impedance (er, mr=1) we get |
| |
|
|  | This is a bit strange if you think about it. We just
connected vacuum, the nothing, with a defined property that can be expressed in a
simple number! You can buy a resistor with 377 W and wonder why it somehow reflects a
property of vacuum. |
 | Be that as it may, we now define a new vector S as follows: |
| |
|
|  | The Poynting vector
has interesting properties: |
| | - S points in the direction of the propagation of the light, i.e. in
wavevector k direction as shown in the figure above.
- |S|=S=E0 · H0 ·
cos2(kr - wt), the magnitude of the Poynting vector,
gives directly the power flux in W/cm2 of an
electromagnetic wave
- The average power delivered to an area is thus .
|
| |
<S> | = | E0H0 2 | = | E02 Zw | = E02 · | æ ç è | ere0 mrm0 | ö
÷ ø | ½ | | | | | | | | | | | <S> | µ | E02 · er½ |
µ | E02 · n | | | | | |
|
|
|  | Don't mix up the power flux in W/cm2 with the specific
energy measured in in Ws/cm3or J/cm3 that is contained in fields as shown above! It's time for an exercise to get some ideas about the numbers associated with light power and field strength! |
| |
|
 | We still have a bunch of open questions from above but before
we tackle those we have to learn a bit about polarization. |
| | |
| Polarization of Light |
| |
| It is clear that an idealized plane
wave, given essentially by the exp[i(kr – wt)] term, describes a
monochromatic and fully coherent light wave
propagating in exactly one direction that is also linearly polarized by definition. |
|  | Some Laser beams may come close to emitting a plane wave matching that description but laser light is a far
cry from "normal" light. |
|  | Most real or normal light
beams, e.g. the light "beams" emanating from light bulbs, are not
polarized as we know. "Real" light from light bulb looks - very schematically - as shown below. We
are forced to assume that the electromagnetic wave has a beginning in space and time (at the light bulb, like now) and
an end. The length of the whole thing we call "coherence length". The
color (=wavelength), and the polarization plane for one of those waves with a beginning and an end, has some fixed
value but that is different from the next wave coming along. The distribution of wavelengths we call the spectrum of the light source, and the distribution of polarization planes will be random.
Each ones of theses "wavelets" is caused by an electron changing its state from higher to lower energy. |
| |
|
|  | The picture might be interpreted as
photons coming out of the light bulb. If one of those wavelets would be seen as a
completely abstract symbol for a photon, this would be true. However. photons are
not just short pieces of an electromagnetic wave, so one must not interpret this
figure as a picture of photons. |
 | We have a number of new questions to consider (never mind that we did not answer all the old ones yet): |
| | - How do we detect if and in which direction a light beam is
polarized?
- How can we polarize an unpolarized light beam
if the need arises?
- Does is matter if and how a light beam is polarized for the applications we have in mind?
- If it matters: how do we optimize the needed polarization?
- How does the polarization of a wave relate to
photon properties?
- Are there other polarization modes besides linear polarization?
| |
We are now at the point where serious optics starts. It ain't exactly easy but we will
only scratch the surface here. |
|
 | Let's look at these questions one by one. We will encounter rather difficult
topics that we only treat in a perfunctory way here. Some topics will come up later
again. |
| | |
 | 1. How do we detect if and how a light beam is polarized? |
|  | By using a polarizer, i.e. some material that transmits only that E-field component of an incoming light beam that is parallel to some
material specific polarization direction as shown below. |
| |
|  From Rogilbert; French Wikipedia |
|
|  | Turn your polarizer and the light
intensity coming out stays either constant if the incoming beam is not polarized at all (but is always lower in
intensity) or varies between zero and maximum (=incoming) intensity in a cos2 fashion if the incoming
beam is 100 % linearly polarized, as shown above. If the beam is partially
polarized you have a mix of the two extremes. |
 | The second question (2. How can we polarize an unpolarized light beam if the need arises?) is answered too: Use a polarizer! Of course we have a new question now, the really tough one. |
|  | 7. What, exactly, gives a material polarizing
properties? And how can I manipulate or engineer those properties? That's indeed a tough question as we shall
see in one of the next sub chapters. Before we try to answer this one, we look
at a few simple things first. |
 | What we
perceive now is that a standard non-polarized light beam simply consists of many intrinsically polarized beams but
with randomly distributed polarization directions. The total polarization effect then is zero. |
| | | | |
|
| |  | This is easy to perceive with the help of
the figure on the left. |
| |  | Randomly distributed polarization vectors sum up to zero=no polarization. |
| |  | Note that we now indirectly answered one of the earlier questions from
above: Yes, you can use a polarization vector like we do in this case and
most others. But beware! Polarization in general is more tricky than you might think and a simple vector is not always
enough to describe it mathematically. (Remember the problems encountered in using a polar vector for describing
rotations?). You will need a matrix if you go about it systematically. |
| |  | The picture also
makes clear why the polarized beam emerging from a polarizer has a lower intensity than the incoming beam, and also
how much lower it will be. |
| |  | If we want to make life easier we can replace a large
number of beams with random polarization directions (red vectors) by just two beams at right angles (brown
vectors). |
| |  | It is time for a quick exercise: |
| | |
| |
|
| |
 | 3. Does is matter if and how a light
beam is polarized for the applications we have in mind? |
|  | The answer is simple: No, it does not matter for many
simple or standard applications. Your glasses,
binoculars, microscopes, and so on, work perfectly well with "normal" light. With a polarization filter your
camera will make somewhat better pictures in certain circumstances but that is usually not decisive. |
|  | However, for anything a bit more sophisticated, like
all the optical things that a Material Scientist and Engineer needs for her or his
work or develops for others, it will matter very much as we will see. Here is a
table that should give you some idea: |
| |
Application | Polarization matters
a lot | Polarization doesn't matter
much | Simple optical instruments (Cameras,
binoculars, ...) | | Forget it | Better optical instruments (Cameras, .. | you use a "Pol
filter" | | Optical communications | not yet but soon | most cases now | Optical measurements | most depend on polarization | | 3-dim movies | including
circular polarization | | Laser | in many cases | some cases | LCD displays | impossible without polarization | |
|
 | 4. If it matters: how do we optimize the needed polarization? |
|  | The answer is simple again: by constantly coming up
with new or improved materials. Note the extremely simple and important truth: |
| |
If a light beam does not interact with some material, its properties will not change. |
|
 | 5. How does the polarization of a
wave relate to photon properties? |
|  | Well - the photon has a spin of 1/2! Naively put, there is a kind of (polar)
vector associated with it. So there might be a relation between polarization and spin? |
|  | Yes - up to a point. But it's not that simple!
Whatever the case may be, my advice is: forget it! Whenever polarization comes up,
only think in the wave picture. Otherwise you simply run into all kinds of
unnecessary problems. However, it is perfectly possible to describe all the effects of polarization in the particle
picture, too. There is no contradiction. |
 | We have the last
questions to consider and that will open a large can of worms, as the saying goes. Let's start with
No. 6: Are there other polarization modes besides linear polarization? |
|  | Yes there are. They are called circular and elliptical polarization. Here we go: |
| | |
| Other Modes of Polarization |
| |
 | Some idealized plane wave travelling in z direction with the electrical
field vector pointing in x-direction writes as
Ex(z,t)=E0, xexp{i(kz –wt) for the the electric field part. This idealized light wave is a solution of
the Maxwell equations for some
(idealized) conditions. |
 | Ey(z,t)=E0, yexp{i(kz
–wt + p/2); i.e. the same wave but with
the electrical field pointing in y-direction and a phase difference of p/2 (or any other you care to insert)
is also a solution, of course. |
 | If those waves travel inside
a material, all we have to do is to replace the electrical field E by D, the
dielectric displacement (and H by B, of course), and to take care of the
change in wavelength or the magnitude of k, respectively. |
|  | If the material is isotropic and linear, we have D=e0erE. That means that not only
the two waves from above are still a solution of the Maxwell equations for the case considered, so is any superposition of those (and possibly other) solutions. So whatever we are about to
discuss works just as well in "simple" materials. Most optical materials are amorphous (called
"glass"), and meet the requirement isotropic and linear in a fair approximation so we don't have to worry yet. |
|  | However, many crystalline materials are neither isotropic nor linear - and then we will be in trouble. Those materials are of course very interesting to a
Materials Scientist but the theory of what happens is difficult and we will deal with that later. |
 | Now let's add up the two
E (or D) solutions from above and see what we get. |
| |
Ex(z,t) +
Ey(z,t) |
= | E0,
xexp{i(kz –wt) | + | E0,
yexp{i(kz –wt+ p/2) |
|
|
 | The real part of this complex equation
gives the electrical field vector, we have |
| |
E | = | E0, x · cos(kz –wt) +
E0, y · sin(kz –wt) |
|
|
|  | The full electrical field
E thus is obtained by adding two vectors at right angles with magnitudes that change sinusoidally
between (±E0). The total effect is a vector with length
E0 that rotates with a circle frequency w around the z-axis. |
|  | What you get is called circular polarization for obvious reasons; it looks like shown below. Note that all of this
appears to be far more complicated than it actually is, this is made clear by the animation below which shows the
basic simplicity of what is going on. |
| |
|  (From wikipedia) |
|
 | You realize, if you think about it,
that we just opened the promised rather large can of worms. We started with simple,
intuitively clear linear polarization and now progressed to circular polarization by superimposing two special linearly polarized waves. Of course, we can produce two basic kinds of circular
polarization: left-handed and right-handed circular polarized light this way. Equally of course, we can now superimpose
all kinds of waves with all kinds of phases. Where will it
end? |
|  | Actually, it's not
as bad as it might be. Superimposing whatever you like just gives you elliptically polarized light (plus, perhaps, a background of unpolarized light). |
|  | The extremes of an ellipse are straight lines
(linearly polarized light, just superimpose the two waves without a phase
difference) and the circle (phase difference p/2; equal amplitudes). |
 | We will stop at this point. Far more
important than going into intricate details of all this polarization stuff is to consider: - Question 2 from above, not yet addressed: How do we make all these
kinds of polarization? (Answer: with special materials, of course).
- What is it good for? Who needs circular or elliptically
polarized light? (Answer: you - if you watch, e.g., 3D-movies)
|
|  | We will look very briefly at some answers in the next
chapter, where we deal with light and materials. |
| |
|
© H. Föll (Advanced Materials B, part 1 - script)