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Unformatted text preview: 33 Polarization 33—1 The electric vector of light In this chapter we shall consider those phenomena which depend on the fact
that the electric ﬁeld that describes the light is a vector. In previous chapters we
have not been concerned with the direction of oscillation of the electric ﬁeld, except
to note that the electric vector lies in a plane perpendicular to the direction of
propagation. The particular direction in this plane has not concerned us. We
now consider those phenomena whose central feature is the particular direction
of oscillation of the electric ﬁeld. In ideally monochromatic light, the electric ﬁeld must oscillate at a deﬁnite
frequency, but since the xcomponent and the ycomponent can oscillate independ
ently at a deﬁnite frequency, we must ﬁrst consider the resultant effect produced
by superposing two independent oscillations at right angles to each other. What
kind of electric ﬁeld is made up of an xcomponent and a ycomponent which
oscillate at the same frequency? If one adds to an xvibration a certain amount of
yvibration at the same phase, the result is a vibration in a new direction in the
xyplane. Figure 33~1 illustrates the superposition of different amplitudes for the
xvibration and the yvibration. But the resultants shown in Fig. 33—1 are not the
only possibilities; in all of these cases we have assumed that the xvibration and the yvibration are in phase, but it does not have to be that way. It could be that
the xvibration and the yvibration are out of phase. 1
+2
ly=1 Eyno ax=o taxi: zx=1 nx=1 33—1 The electric vector of light
33—2 Polarization of scattered light
33—3 Birefringence 33—4 Polarizers
335 Optical activity
33—6 The intensity of reﬂected light 33—7 Anomalous refraction Y Y 1 y y
= = E = 1 E = 1
By 1 EV 1 y Y =1 21:1 Fig. 33—1. Superposition of xvibrations and yvibrafions in phase. When the xvibration and the yvibration are not in phase, the electric ﬁeld
vector moves around in an ellipse, and we can illustrate this in a familiar way. If
we hang a ball from a support by a long string, so that it can swing freely in a
horizontal plane, it will execute sinusoidal oscillations. If we imagine horizontal
x and ycoordinates with their origin at the rest position of the ball, the ball can
swing in either the x or ydirection with the same pendulum frequency. By
selecting the proper initial displacement and initial velocity, we can set the ball in
oscillation along either the xaxis or the yaxis, or along any straight line in the
xyplane. These motions of the ball are analogous to the oscillations of the electric
ﬁeld vector illustrated in Fig. 33—1. In each instance, since the xvibrations and
the yvibrations reach their maxima and minima at the same time, the x and 32—05
cillations are in phase. But we know that the most general motion of the ball
is motion in an ellipse, which corresponds to oscillations in which the x and
ydirections are not in the same phase. The superposition of x and yvibrations
which are not in phase is illustrated in Fig. 33—2 for a variety of angles between the
phase of the xvibration and that of the yvibration. The general result is that the
electric vector moves around an ellipse. The motion in a straight line is a particular 33—1 /
x = cos cut; 1
By = cos cat; 1
f
xx = cos cut; 1
By = cos(an:+"/¢l; e 1n/o case corresponding to a phase difference of zero (or an integral multiple of 1r);
motion in a circle corresponds to equal amplitudes with a phase diﬁerence of 90°
(or any odd integral multiple of 7r/2). In Fig. 33—2 we have labeled the electric ﬁeld vectors in the x and ydirections
with complex numbers, which are a convenient representation in which to express
the phase diﬁerence. Do not confuse the real and imaginary components of the
complex electric vector in this notation with the x and y—coordinates of the ﬁeld.
The x and ycoordinates plotted in Fig. 33—1 and Fig. 33—2 are actual electric
ﬁelds that we can measure. The real and imaginary components of a complex
electric ﬁeld vector are only a mathematical convenience and have no physical signiﬁcance.
b c d e
cos «it; 1 cos wt; 1 cos cut; 1 cos at; 1
1
cos(ast+"/4); e "A sin wt; 1 cosmic0’74); ein/l cos art; 1
g h i
cosart; 1 cosat; 1 cosmt;1
sin wt; 1 cos(at+3n/4); e13“/4 cos at; 1 Fig. 33—2. Superposition of xvibrations and yvibrations with equal amplitudes but various relative phases. The components Ez and E, are expressed in both real and complex notations. Now for some terminology. Light is linearly polarized (sometimes called
plane polarized) when the electric field oscillates on a straight line; Fig. 33—1
illustrates linear polarization. When the end of the electric ﬁeld vector travels in
an ellipse, the light is elliptically polarized. When the end of the electric ﬁeld vector
travels around a circle, we have circular polarization. If the end of the electric
vector, when we look at it as the light comes straight toward us, goes around
in a counterclockwise direction, we call it righthand circular polarization.
Figure 33—2(g) illustrates righthand circular polarization, and Fig. 33—2(c) shows
lefthand circular polarization. In both cases the light is coming out of the paper.
Our convention for labeling lefthand and righthand circular polarization is
consistent with that which is used today for all the other particles in physics which
exhibit polarization (e.g., electrons). However, in some books on optics the
opposite conventions are used, so one must be careful. We have considered linearly, circularly, and elliptically polarized light, which
covers everything except for the case of unpolarized light. Now how can the light
be unpolarized when we know that it must vibrate in one or another of these
ellipses? If the light is not absolutely monochromatic, or if the x and yphases
are not kept perfectly together, so that the electric vector ﬁrst vibrates in one
direction, then in another, the polarization is constantly changing. Remember
that one atom emits during 10—8 sec, and if one atom emits a certain polarization,
and then another atom emits light with a different polarization, the polarizations
will change every 10—8 see. If the polarization changes more rapidly than we can
detect it, then we call the light unpolarized, because all the effects of the polarization
average out. None of the interference effects of polarization would show up with
unpolarized light. But as we see from the deﬁnition, light is unpolarized only if
we are unable to ﬁnd out whether the light is polarized or not. 332 33—2 Polarization of scattered light The ﬁrst example of the polarization effect that we have already discussed is
the scattering of light. Consider a beam of light, for example from the sun, shining
on the air. The electric ﬁeld will produce oscillations of charges in the air, and mo
tion of these charges will radiate light with its maximum intensity in a plane normal
to the direction of vibration of the charges. The beam from the sun is unpolarized,
so the direction of polarization changes constantly, and the direction of vibration
of the charges in the air changes constantly. If we consider light scattered at 90°,
the vibration of the charged particles radiates to the observer only when the
vibration is perpendicular to the observer’s line of sight, and then light will be polarized along the direction of vibration. So scattering is an example of one means
of producing polarization. 333 Birefringence Another interesting effect of polarization is the fact that there are substances
for which the index of refraction is different for light linearly polarized in one
direction and linearly polarized in another. Suppose that we had some material
which consisted of long, nonspherical molecules, longer than they are wide, and
suppose that these molecules were arranged in the substance with their long axes
parallel. Then what happens when the oscillating electric ﬁeld passes through this
substance? Suppose that because of the structure of the molecule, the electrons
in the substance respond more easily to oscillations in the direction parallel to the
axes of the molecules than they would respond if the electric ﬁeld tries to push
them at right angles to the molecular axis. In this way we expect a different response
for polarization in one direction than for polarization at right angles to that direc
tion. Let us call the direction of the axes of the molecules the optic axis. When
the polarization is in the direction of the optic axis the index of refraction is
different than it would be if the direction of polarization were at right angles to it.
Such a substance is called birefringent. It has two refrangibilities, i.e., two indexes
of refraction, depending on the direction of the polarization inside the substance.
What kind of a substance can be birefringent? In a birefringent substance there
must be a certain amount of lining up, for one reason or another, of unsymmetrical
molecules. Certainly a cubic crystal, which has the symmetry of a cube, cannot be
birefringent. But long needlelike crystals undoubtedly contain molecules that
are asymmetric, and one observes this effect very easily. Let us see what effects we would expect if we were to shine polarized light
through a plate of a birefringent substance. If the polarization is parallel to the
optic axis, the light will go through with one velocity; if the polarization is per
pendicular to the axis, the light is transmitted with a different velocity. An inter
esting situation arises when, say, light is linearly polarized at 45° to the optic axis.
Now the 45° polarization, we have already noticed, can be represented as a super
position of the x and the y—polarizations of equal amplitude and in phase, as
shown in Fig. 33—2(a). Since the x and ypolarizations travel with different
velocities, their phases change at a different rate as the light passes through the
substance. So, although at the start the x and yvibrations are in phase, inside
the material the phase difference between x and yvibrations is proportional to the
depth in the substance. As the light proceeds through the material the polarization
changes as shown in the series of diagrams in Fig. 33—2. If the thickness of the
plate is just right to introduce a 90° phase shift between the x and ypolarizations,
as in Fig. 33—2(c), the light will come out circularly polarized. Such a thickness
is called a quarterwave plate, because it introduces a quartercycle phase difference
between the x and the ypolarizations. If linearly polarized light is sent through
two quarterwave plates, it will come out planepolarized again, but at right angles
to the original direction, as we can see from Fig. 33—2(e). One can easily illustrate this phenomenon with a piece of cellophane. Cello
phane is made of long, ﬁbrous molecules, and is not isotropic, since the ﬁbers lie
preferentially in a certain direction. To demonstrate birefringence we need a 33—3 CELLOPHANE \POLAROID/ Fig. 333. An experimental demon
stration of the birefringence of cellophane.
The electric vectors in the light are indi
cated by the dotted lines. The pass axes
of the polaroid sheets and optic axes of
the cellophane are indicated by arrows.
The incident beam is unpolarized. beam of linearly polarized light, and we can obtain this conveniently by passing
unpolarized light through a sheet of polaroid. Polaroid, which we will discuss
later in more detail, has the useful property that it transmits light that is linearly
polarized parallel to the axis of the polaroid with very little absorption, but light
polarized in a direction perpendicular to the axis of the polaroid is strongly
absorbed. When we pass unpolarized light through a sheet of polaroid, only that
part of the unpolarized beam which is vibrating parallel to the axis of the polaroid
gets through, so that the transmitted beam is linearly polarized. This same property
of polaroid is also useful in detecting the direction of polarization of a linearly
polarized beam, or in determining whether a beam is linearly polarized or not.
One simply passes the beam of light through the polaroid sheet and rotates the
polaroid in the plane normal to the beam. If the beam is linearly polarized, it will
not be transmitted through the sheet when the axis of the polaroid is normal to
the direction of polarization. The transmitted beam is only slightly attenuated
when the axis of the polaroid sheet is rotated through 90°. If the transmitted in
tensity is independent of the orientation of the polaroid, the beam is not linearly
polarized. To demonstrate the birefringence of cellophane, we use two sheets of polaroid,
as shown in Fig. 33—3. The ﬁrst gives us a linearly polarized beam which we pass
through the cellophane and then through the second polaroid sheet, which serves
to detect any effect the cellophane may have had on the polarized light passing
through it. If we ﬁrst set the axes of the two polaroid sheets perpendicular to each
other and remove the cellophane, no light will be transmitted through the second
polaroid. If we now introduce the cellophane between the two polaroid sheets, and
rotate the sheet about the beam axis, we observe that in general the cellophane
makes it possible for some light to pass through the second polaroid. However,
there are two orientations of the cellophane sheet, at right angles to each other,
which permit no light to pass through the second polaroid. These orientations in
which linearly polarized light is transmitted through the cellophane with no
effect on the direction of polarization must be the directions parallel and per
pendicular to the optic axis of the cellophane sheet. We suppose that the light passes through the cellophane with two different
velocities in these two different orientations, but it is transmitted without changing
the direction of polarization. When the cellophane is turned halfway between
these two orientations, as shown in Fig. 33—3, we see that the light transmitted
through the second polaroid is bright. It just happens that ordinary cellophane used in commercial packaging is
very close to a halfwave thickness for most of the colors in white light. Such a
sheet will turn the axis of linearly polarized light through 90° if the incident
linearly polarized beam makes an angle of 45° with the optic axis, so that the beam
emerging from the cellophane is then vibrating in the right direction to pass through
the second polaroid sheet. If we use white light in our demonstration, the cellophane sheet will be of the
proper halfwave thickness only for a particular component of the white light,
and the transmitted beam will have the color of this component. The color trans
mitted depends on the thickness of the cellophane sheet, and we can vary the
effective thickness of the cellophane by tilting it so that the light passes through the
cellophane at an angle, consequently through a longer path in the cellophane. As
the sheet is tilted the transmitted color changes. With cellophane of different
thicknesses one can construct ﬁlters that will transmit different colors. These
ﬁlters have the interesting property that they transmit one color when the two
polaroid sheets have their axes perpendicular, and the complementary color when
the axes of the two polaroid sheets are parallel. Another interesting application of aligned molecules is quite practical.
Certain plastics are composed of very long and complicated molecules all twisted
together. When the plastic is solidiﬁed very carefully, the molecules are all twisted
in a mass, so that there are as many aligned in one direction as another, and so
the plastic is not particularly birefringent. Usually there are strains and stresses
introduced when the material is solidiﬁed, so the material is not perfectly homo 33—4 geneous. However, if we apply tension to a piece of this plastic material, it is as
if we were pulling a whole tangle of strings, and there will be more strings preferen
tially aligned parallel to the tension than in any other direction. So when a stress
is applied to certain plastics, they become birefringent, and one can see the effects
of the birefringence by passing polarized light through the plastic. If we examine
the transmitted light through a polaroid sheet, patterns of light and dark fringes will
be observed (in color, if white light is used). The patterns move as stress is applied
to the sample, and by counting the fringes and seeing where most of them are, one
can determine what the stress is. Engineers use this phenomenon as a means of
ﬁnding the stresses in oddshaped pieces that are difficult to calculate. Another interesting example of a way of obtaining birefringence is by means
of a liquid substance. Consider a liquid composed of long asymmetric molecules
which carry a plus or minus average charge near the ends of the molecule, so that
the molecule is an electric dipole. In the collisions in the liquid the molecules
will ordinarily be randomly oriented, with as many molecules pointed in one direc
tion as in another. If we apply an electric ﬁeld the molecules will tend to line up,
and the moment they line up the liquid becomes birefringent. With two polaroid
sheets and a transparent cell containing such a polar liquid, we can devise an
arrangement with the property that light is transmitted only when the electric
ﬁeld is applied. So we have an electrical switch for light, which is called a Kerr cell. This effect, that an electric ﬁeld can produce birefringence in certain liquids,
is called the Kerr effect. 33—4 Polarizers So far we have considered substances in which the refractive index is different
for light polarized in different directions. Of very practical value are those crystals
and other substances in which not only the index, but also the coefﬁcient of ab
sorption, is different for light polarized in different directions. By the same argu
ments which supported the idea of birefringence, it is understandable that absorp
tion can vary with the direction in which the charges are forced to vibrate in an
anisotropic substance. Tourmaline is an old, famous example and polaroid is
another. Polaroid consists of a thin layer of small crystals of herapathite (a salt
of iodine and quinine), all aligned with their axes parallel. These crystals absorb
light when the oscillations are in one direction, and they do not absorb appreciably
when the oscillations are in the other direction. Suppose that we send light into a polaroid sheet polarized linearly at an
angle 0 to the passing direction. What intensity will come through? This incident
light can be resolved into a component perpendicular to the pass direction which
is proportional to sin 0, and a component along the pass direction which is pro
portional to cos 0. The amplitude which comes out of the polaroid is only the
cosine 0 part; the sin 0 component is absorbed. The amplitude which passes
through the polaroid is smaller than the amplitude which entered, by a factor
cos 0. The energy which passes through the polaroid, i.e., the intensity of the
light, is proportional to the square of cos 0. Cos 2 0, then, is the intensity transmitted
when the light enters polarized at an angle 0 to the pass direction. The absorbed
intensity, of course, is sin2 0. An interesting paradox is presented by the following situation. We know
that is is not possible to send a beam of light through two polaroid sheets with
their axes crossed at right angles. But if we place a third polaroid sheet between
the ﬁrst two, with its pass axis at 45° to the crossed axes, some light is transmitted.
We know that polaroid absorbs light, it does not create anything. Nevertheless,
the addition of a third polaroid at 45° allows more light to get through. The
analysis of this phenomenon is left as an exercise for the student. One of the most interesting examples of polarization is not in complicated
crystals or difficult substances, but in one of the simplest and most familiar of
situations—the reﬂection of light from a surface. Believe it or not, when light is
reﬂected from a glass surface it may be polarized, and the physical explanation of
this is very simple. It was discovered empirically by Brewster that light reflected 335 Fig. 334. Reﬂection of linearly po
larized light at Brewster’s angle. The
polarization direction is indicated by
dashed arrows; round dots indicate
polarization normal to the paper. Fig. 33—5. A molecule with a shape
that is not symmetric when reﬂected in a
mirror. A beam of light, linearly polarized
in the ydirection, falls on the molecule. from a surface is completely polarized if the reﬂected beam and the beam refracted
into the material form a right angle. The situation is illustrated in Fig. 33—4. If
the incident beam is polarized in the plane of incidence, there will be no reﬂection
at all. Only if the incident beam is polarized normal to the plane of incidence will
it be reﬂected. The reason is very easy to understand. In the reﬂecting material
the light is polarized transversely, and we know that it is the motion of the charges
in the material which generates the emergent beam, which we call the reﬂected
beam. The source of this socalled reﬂected light is not simply that the incident
beam is reﬂected; our deeper understanding of this phenomenon tells us that the
incident beam drives an oscillation of the charges in the material, which in turn
generates the reﬂected beam. From Fig. 33—4 it is clear that only oscillations normal
to the paper can radiate in the direction of reﬂection, and consequently the reﬂected
beam will be polarized normal to the plane of incidence. If the incident beam is
polarized in the plane of incidence, there will be no reﬂected light. This phenomenon is readily demonstrated by reﬂecting a linearly polarized
beam from a ﬂat piece of glass. If the glass is turned to present diﬂerent angles of
incidence to the polarized beam, sharp attenuation of the reﬂected intensity is
observed when the angle of incidence passes through Brewster’s angle. This
attenuation is observed only if the plane of polarization lies in the plane of incidence.
If the plane of polarization is normal to the plane of incidence, the usual reﬂected
intensity is observed at all angles. 33.5 Optical activity Another most remarkable effect of polarization is observed in materials
composed of molecules which do not have reﬂection symmetry: molecules shaped
something like a corkscrew, or like a gloved hand, or any shape which, if viewed
through a mirror, would be reversed in the same way that a lefthand glove reﬂects
as a righthand glove. Suppose all of the molecules in the substance are the same,
i.e., none is a mirror image of any other. Such’a substance may show an interesting
effect called optical activity, whereby as linearly polarized light passes through
the substance, the direction of polarization rotates about the beam axis. To understand the phenomenon of optical activity requires some calculation,
but we can see qualitatively how the effect might come about, without actually
carrying out the calculations. Consider an asymmetric molecule in the shape of
a spiral, as shown in Fig. 33—5. Molecules need not actually be shaped like a
corkscrew in order to exhibit optical activity, but this is a simple shape which we
shall take as a typical example of those that do not have reﬂection symmetry.
When a light beam linearly polarized along the ydirection falls on this molecule,
the electric ﬁeld will drive charges up and down the helix, thereby generating a
current in the ydirection and radiating an electric ﬁeld E, polarized in the ydirec—
tion. However, if the electrons are constrained to move along the spiral, they
must also move in the xdirection as they are driven up and down. When a current
is ﬂowing up the spiral, it is also ﬂowing into the paper at z = 21 and out of the
paper at z = 21 + A, if A is the diameter of our molecular spiral. One might
suppose that the current in the xdirection would produce no net radiation, since
the currents are in opposite directions on opposite sides of the spiral. However,
if we consider the xcomponents of the electric ﬁeld arriving at z = 22, we see
that the ﬁeld radiated by the current at z = 21 + A and the ﬁeld radiated from
2 = 21 arrive at 22 separated in time by the amount A/c, and thus separated in
phase by 71' + wA/c. Since the phase difference is not exactly 7r, the two ﬁelds
do not cancel exactly, and we are left with a small xcomponent in the electric ﬁeld
generated by the motion of the electrons in the molecule, whereas the driving
electric ﬁeld had only a ycomponent. This small xcomponent, added to the large
ycomponent, produces a resultant ﬁeld that is tilted slightly with respect to the
yaxis, the original direction of polarization. As the light moves through the material, the direction of polarization rotates about the beam axis. By drawing a
few examnles and cnnsiderina the currents that will he set in mntinn hv an incident electric ﬁeld, one can convince himself that the existence of optical activity and the
sign of the rotation are independent of the orientation of the molecules. Corn syrup is a common substance which possesses optical activity. The
phenomenon is easily demonstrated with a polaroid sheet to produce a linearly
polarized beam, a transmission cell containing corn syrup, and a second polaroid
sheet to detect the rotation of the direction of polarization as the light passes
through the corn syrup. 33—6 The intensity of reﬂected light Let us now consider quantitatively the reﬂection coeﬂicient as a function of
angle. Figure 33—6(a) shows a beam of light striking a glass surface, where it is
partly reﬂected and partly refracted into the glass. Let us suppose that the incident
beam, of unit amplitude, is linearly polarized normal to the plane of the paper.
We will call the amplitude of the reﬂected wave b, and the amplitude of the re
fracted wave a. The refracted and reﬂected waves will, of course, be linearly
polarized, and the electric ﬁeld vectors of the incident, reﬂected, and refracted
waves are all parallel to each other. Figure 33—6(b) shows the same situation, but
now we suppose that the incident wave, of unit amplitude, is polarized in the plane
of the paper. Now let us call the amplitude of the reﬂected and refracted wave
B and A, respectively. We wish to calculate how strong the reﬂection is in the two situations illus
trated in Fig. 33—6(a) and 33—6(b). We already know that when the angle between
the reﬂected beam and refracted beam is a right angle, there will be no reﬂected
wave in Fig. 33—6(b), but let us see if we cannot get a quantitative answer—an
exact formula for B and b as a function of the angle of incidence, i. The principle that we must understand is as follows. The currents that are
generated in the glass produce two waves. First, they produce the reﬂected wave.
Moreover, we know that if there were no currents generated in the glass, the in
cident wave would continue straight into the glass. Remember that all the sources
in the world make the net ﬁeld. The source of the incident light beam produces a
ﬁeld of unit amplitude which would move into the glass along the dotted line in
the ﬁgure. This ﬁeld is not observed, and therefore the currents generated in the
glass must produce a ﬁeld of amplitude —l which moves along the dotted line.
Using this fact, we will calculate the amplitude of the refracted waves, a and A. In Fig. 33—6(a) we see that the ﬁeld of amplitude b is radiated by the motion
of charges inside the glass which are responding to a ﬁeld a inside the glass, and
that therefore b is proportional to a. We might suppose that since our two ﬁgures
are exactly the same, except for the direction of polarization, the ratio B/A would
be the same as the ratio b /a. This is not quite true, however, because in Fig. 33—6(b)
the polarization directions are not all parallel to each other, as they are in Fig.
33—6(a). It is only the component of A which is perpendicular to B, A cos (i + r),
which is effective in producing B. The correct expression for the proportionality
is then b B
a (33.1) _ A cos (i + r) '
Now we use a trick. We know that in both (a) and (b) of Fig. 33—6 the electric
ﬁeld in the glass must produce oscillations of the charges which generate a ﬁeld of
amplitude — l, polarized parallel to the incident beam, and moving in the direction
of the dotted line. But we see from part (b) of the ﬁgure that only the component
of A that is normal to the dashed line has the right polarization to produce this
ﬁeld, whereas in Fig. 33—6(a) the full amplitude a is effective, since the polarization
of wave a is parallel to the polarization of the wave of amplitude —— 1. Therefore
we can write
Acos(i — r) _ —l
a _ —l , (33.2) since the two amplitudes on the left side of Eq. (33.2) each produce the wave of
amplitude — 1. 33—7 Fig. 33—6. An incident wave of unit
amplitude is reﬂected and refracted at a
glass surface. In (a) the incident wave is
linearly polarized normal to the plane of
the paper. In (b) the incident wave is
linearly polarized in the direction shown
by the dotted electric vector. Dividing Eq. (33.1) by Eq. (33.2), we obtain B _ cos (i + r) 3 — m ’ a result which we can check against what we already know. If we set (i + r) =
90°, Eq. (33.3) gives B = 0, as Brewster says it should be, so our results so far
are at least not obviously wrong. We haveassumed unit amplitudes for the incident waves, so that [B] 2/12 is
the reﬂection coeﬂicient for waves polarized in the plane of incidence, and b 2/ l 2
is the reﬂection coefﬁcient for waves polarized normal to the plane of incidence.
The ratio of these two reﬂection coefﬁcients is determined by Eq. (33.3). Now we perform a miracle, and compute not just the ratio, but each coeﬂicient
B 2 and lb! 2 individually! We know from the conservation of energy that the energy
in the refracted wave must be equal to the incident energy minus the energy in
the reﬂected wave, 1 — B2 in one case, 1 — b2 in the other. Furthermore, the
energy which passes into the glass in Fig. 33—6(b) is to the energy which passes
into the glass in Fig. 33—6(a) as the ratio of the squares of the refracted amplitudes,
A2/[al2. One might ask whether we really know how to compute the energy
inside the glass, because, after all, there are energies of motion of the atoms in
addition to the energy in the electric ﬁeld. But it is obvious that all of the various
contributions to the total energy will be proportional to the square of the amplitude
of the electric ﬁeld. Therefore we can write 1 — B2 _ ME.
1 ~ lbl2 _ lal2
We now substitute Eq. (33.2) to eliminate A /a from the expression above,
and express B in terms of b by means of Eq. (33.3): (33.4) cos2 (i + r)
1 _ bl2 cos2 (i — r) _ l (33 5
1— [1,42 ‘cos2(i—r)' ') This equation contains only one unknown amplitude, b. Solving for lb] 2, we obtain sin2 (1' — r) 2 _
[bl ‘ sin2 (i + r) (33.6) and, with the aid of (33.3),
tan2 (1' — r) tan2 (i + r) ' So we have found the reﬂection coefﬁcient lb]2 for an incident wave polarized
perpendicular to the plane of incidence, and also the reﬂection coefﬁcient IBI'2 for
an incident wave polarized in the plane of incidence! It is possible to go on with arguments of this nature and deduce that b is real.
To prove this, one must consider a case where light is coming from both sides of
the glass surface at the same time, a situation not easy to arrange experimentally,
but fun to analyze theoretically. If we analyze this general case, we can prove that
b must be real, and therefore, in fact, that b = =I= sin (i — r)/sin (i + r). It is
even possible to determine the sign by considering the case of a very, very thin
layer in which there is reﬂection from the front and from the back surfaces, and
calculating how much light is reﬂected. We know how much light should be
reﬂected by a thin layer, because we know how much current is generated, and we
have even worked out the ﬁelds produced by such currents. One can show by these arguments that 1312 = (33.7) sin (1' r) _ tan (i — r) m. These expressions for the reﬂection coefficients as a function of the angles of
incidence and refraction are called Fresnel’s reﬂection formulas. If we consider the limit as the angles i and r go to zero, we find, for the case of
normal incidence, that 82 z b2 m (i — r)2/(i + r)2 for both polarizations,
338 b = _ B = (33.8) since the sines are practically equal to the angles, as are also the tangents. But we
know that sin i/sin r = n, and when the angles are small, i/r 2: n. It is thus easy
to show that the coefficient of reﬂection for normal incidence is 2_ 2_("*1)2
B ‘b *m‘ It is interesting to ﬁnd out how much light is reﬂected at normal incidence
from the surface of water, for example. For water, n is 4/3, so that the reﬂection
coefﬁcient is (1/7)2 x 2%. At normal incidence, only two percent of the light
is reﬂected from the surface of water. 33—7 Anomalous refraction The last polarization effect we shall consider was actually one of the ﬁrst
to be discovered: anomalous refraction. Sailors visiting Iceland brought back to
Europe crystals of Iceland spar (CaCO 3) which had the amusing property of mak
ing anything seen through the crystal appear doubled, i.e., as two images. This
came to the attention of Huygens, and played an important role in the discovery
of polarization. As is often the case, the phenomena which are discovered ﬁrst are
the hardest, ultimately, to explain. It is only after we understand a physical concept
thoroughly that we can carefully select those phenomena which most clearly and
simply demonstrate the concept. Anomalous refraction is a particular case of the same birefringence that we
considered earlier. Anomalous refraction comes about when the optic axis, the
long axis of our asymmetric molecules, is not parallel to the surface of the crystal.
In Fig. 33—7 are drawn two pieces of birefringent material, with the optic axis as
shown. In the upper ﬁgure, the incident beam falling on the material is linearly
polarized in a direction perpendicular to the optic axis of the material. When this
beam strikes the surface of the material, each point on the surface acts as a source
of a wave which travels into the crystal with velocity vi, the velocity of light in
the crystal when the plane of polarization is normal to the optic axis. The wave
front is just the envelope or locus of all these little spherical waves, and this wave
front moves straight through the crystal and out the other side. This is just the
ordinary behavior we would expect, and this ray is called the ordinary ray. In the lower ﬁgure the linearly polarized light falling on the crystal has its
direction of polarization turned through 90°, so that the optic axis lies in the plane
of polarization. When we now consider the little waves originating at any point
on the surface of the crystal, we see that they do not spread out as spherical waves.
Light travelling along the optic axis travels with velocity vi because the polariza
tion is perpendicular to the optic axis, whereas the light travelling perpendicular
to the optic axis travels with velocity 22” because the polarization is parallel to the
optic axis. In a birefringent material 1)” ¢ vi, and in the ﬁgure v” < vi. A more
complete analysis will show that the waves spread out on the surface of an ellipsoid,
with the optic axis as major axis of the ellipsoid. The envelope of all these elliptical
waves is the wavefront which proceeds through the crystal in the direction shown.
Again, at the back surface the beam will be deﬂected just as it was at the front
surface, so that the light emerges parallel to the incident beam, but displaced from
it. Clearly, this beam does not follow Snell’s law, but goes in an extraordinary
direction. It is therefore called the extraordinary ray. When an unpolarized beam strikes an anomalously refracting crystal, it is
separated into an ordinary ray, which travels straight through in the normal man
ner, and an extraordinary ray which is displaced as it passes through the crystal.
These two emergent rays are linearly polarized at right angles to each other. That
this is true can be readily demonstrated with a sheet of polaroid to analyze the
polarization of the emergent rays. We can also demonstrate that our interpretation
of this phenomenon is correct by sending linearly polarized light into the crystal.
By properly orienting the direction of polarization of the incident beam, we can
make this light go straight through without splitting, or we can make it go through
without splitting but with a displacement. 33—9 Fig. 33—7. The upper diagram shows
the path of the ordinary ray through a doubly refracting crystal. The extraor
dinary ray is shown in the lower dia
gram. The optic axis lies in the plane of
the paper. 4\\
' Fig. 33—8. Two oppositely rotating
vectorsof equal amplitude add to produce
a vector in a ﬁxed direction, but with an
oscillating amplitude. Fig. 339. A charge moving in a
circle in response to circularly polarized
light. We have represented all the various polarization cases in Figs. 33—] and 33—2
as superpositions of two special polarization cases, namely x and y in various
amounts and phases. Other pairs could equally well have been used. Polarization
along any two perpendicular axes x’, y’ inclined to x and y would serve as well
[for example, any polarization can be made up of superpositions of cases (a) and
(e) of Fig. 33—2]. It is interesting, however, that this idea can be extended to other
cases also. For example, any linear polarization can be made up by superposing
suitable amounts at suitable phases of right and left circular polarizations [cases
(c) and (g) of Fig. 33—2], since two equal vectors rotating in opposite directions
add to give a single vector oscillating in a straight line (Fig. 33—8). If the phase of
one is shifted relative to the other, the line is inclined. Thus all the pictures of
Fig. 33—1 could be labeled “the superposition of equal amounts of right and left
circularly polarized light at various relative phases.” As the left slips behind the
right in phase, the direction of the linear polarization changes. Therefore optically
active materials are, in a sense, birefringent. Their properties can be described by
saying that they have different indexes for right and lefthand circularly polarized
light. Superposition of right and left circularly polarized light of diﬁerent intensi
ties produces elliptically polarized light. Circularly polarized light has another interesting property—it carries angular
momentum (about the direction of propagation). To illustrate this, suppose that
such light falls on an atom represented by a harmonic oscillator that can be dis
placed equally well in any direction in the plane xy. Then the xdisplacement of
the electron will respond to the E2 component of the ﬁeld, while the ycomponent
responds, equally, to the equal Ey component of the ﬁeld but 90° behind in phase.
That is, the responding electron goes around in a circle, with angular velocity w,
in response to the rotating electric ﬁeld of the light (Fig. 33—9). Depending on
the damping characteristics of the response of the oscillator, the direction of the
displacement a of the electron, and the direction of the force qu on it need not be
the same but they rotate around together. The E may have a component at right
angles to a, so work is done on the system and a torque T is exerted. The work done
per second is for. Over a period of time T the energy absorbed is mT, while TT is
the angular momentum delivered to the matter absorbing the energy. We see
therefore that a beam of right circularly polarized light containing a total energy
6 carries an angular momentum (with vector directed along the direction of prop
agation) S/w. For when this beam is absorbed that angular momentum is de
livered to the absorber. Lefthand circular light carries angular momentum of the
opposite sign, —8/w. 3310 ...
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 Spring '09
 LeeKinohara
 Physics

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