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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. |
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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.
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| Ey(t) |
= Ey0
· exp{i(kx – wt)}
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| Hz(t) |
= Hz0
· exp{i(kx – wt)}
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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: |
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- 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? |
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The bad news
is: There are no easy answers for some of those questions. |
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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. |
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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? |
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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: |
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| [Welect;
magn] |
= |
[Ws
m–3] |
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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 |
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| E0 |
= |
æ
ç
è |
mrm0
ere0 |
ö
÷
ø |
½ |
·
H0 |
= |
Zw ·
H0 |
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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. |
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For the vacuum
impedance (er,
mr = 1) we get |
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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. |
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Be that as it may, we now define a new vector S as follows: |
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The Poynting vector
has interesting properties: |
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- 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 .
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| <S> |
= |
E0H0
2 |
= |
E02
Zw |
= E02 ·
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æ
ç
è |
ere0
mrm0 |
ö
÷
ø |
½ |
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| <S> |
µ |
E02 · er½ |
µ |
E02 · n |
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Don't mix up the
power flux in
W/cm2 with the specific energy measured in in
Ws/cm3 or 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! |
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We still have a bunch of
open questions from above but before we tackle
those we have to learn a bit about polarization. |
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Polarization of
Light |
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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. |
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Some Laser beams may come close to
emitting a plane wave matching that description but laser light is a far cry
from "normal" light. |
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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. |
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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. |
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We have a number of new questions to consider (never mind that we did
not answer all the old ones yet): |
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- 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?
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We are now at the point where serious optics starts.
It ain't exactly easy but we will only scratch the surface here. |
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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. |
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1. How do we detect if and how a light beam is
polarized? |
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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. |
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From Rogilbert; French Wikipedia |
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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. |
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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. |
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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. |
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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. |
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This is easy to perceive with the help of the
figure on the left. |
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Randomly distributed polarization vectors sum up to zero = no
polarization. |
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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. |
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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. |
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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). |
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It is time for a quick exercise: |
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3. Does is matter if and how a
light beam is polarized for the applications we have in mind? |
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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. |
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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: |
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| Application |
Polarization
matters a lot |
Polarization
doesn't matter much |
Simple optical instruments
(Cameras, binoculars, ...) |
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Forget it |
Better optical instruments
(Cameras, .. |
you use a "Pol filter" |
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| Optical communications |
not yet but soon |
most cases now |
| Optical measurements |
most depend on polarization |
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| 3-dim movies |
including circular polarization |
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| Laser |
in many cases |
some cases |
| LCD displays |
impossible without polarization |
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4. If it matters: how do we
optimize the needed polarization? |
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The answer is simple again: by constantly coming
up with new or improved materials. Note the
extremely simple and important truth: |
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If a light beam does not interact with some material,
its properties will not change. |
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5. How does the polarization of a
wave relate to photon properties? |
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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? |
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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. |
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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? |
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Yes there are. They are called circular and elliptical polarization. Here we go: |
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Other Modes of Polarization |
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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. |
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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. |
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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. |
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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. |
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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. |
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Now let's add up the two
E (or D) solutions from above and see
what we get. |
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| Ex(z,t) +
Ey(z,t) |
= |
E0,
xexp{i(kz –wt) |
+ |
E0,
yexp{i(kz –wt +
p/2) |
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The real part of this complex
equation gives the electrical field vector, we have |
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| E |
= |
E0, x
· cos(kz –wt) +
E0, y · sin(kz –wt) |
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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. |
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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. |
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(From wikipedia) |
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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? |
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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). |
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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). |
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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)
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We will look very briefly at some answers in the
next chapter, where we deal with light and
materials. |
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© H. Föll (Advanced Materials B, part 1 - script)
