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the dotted line. Ủsing this fact, we will caleulate the amplitude of the refracted
waves, ø and 4.
In Eig. 33-6(a) we see that the field of amplitude ð is radiated by the motion
of charges Inside the glass which are responding to a field ø inside the glass, and
that therefore b is proportional to a. We might suppose that since our two fñgures
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. 35-6(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 Ö,
Acos (¿ + r), which is efective in producing Ø. The correct expression for the
proportionality is then
a Acos(i+r). 33.1)
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 —1, 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 4 that is normal to the dashed line has the right polarization
to produce this field, whereas in Eig. 33-6(a) the full amplitude ø is efective,
since the polarization of wave ø is parallel to the polarization of the wave of
amplitude —1. 'Therefore we can write
A cos (¿ — 7) — _1 (33.2)
since the two amplitudes on the left side of Eq. (33.2) each produce the wave of
amplitude —1.
Dividing Eq. (38.1) by Bq. (33.2), we obtain
B _ cos ứ + n: (33.3)
b — cos(—r}
--- Trang 582 ---
a result which we can check against what we already know. IÝ we set (¿-+z) = 909,
Edq. (33.3) gives =0, as Brewster says it should be, so our results so far are at
least not obviousÌly wrong.
We have assumed unit amplitudes for the incident waves, so that ||2/12 is
the reflection coefficient for waves polarized in the plane of ineidence, and |b|2/12
1s the refection coefficient for waves polarized normal to the plane of incidenee.
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 coefficlent
||? and |b|? individually! We know from the conservation of energy that the
energy in the reracted wave must be equal to the incident energy minus the energy
in the refected wave, 1 — ||? in one case, 1 — |b|? in the other. Purthermore,
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, |A|?/|a|?. 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 fñeld. 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 feld. 'Pherefore we can write
1—|BIÊ - |Al2
T— BE T lajP” (33.4)
We now substitute Eq. (33.2) to eliminate A/ø from the expression above,
and express Ö in terms of b by means of Eq. (33.3):
. co ự +r)
coS (¿ — r) — b (33.5)
1— ||? cosZ (2 — r)
This equation contains only one unknown amplitude, b. Solving for |b|2, we
obtain
b2 = Z8 ữ—") (33.6)
sin“ (2 + r)
and, with the aid of (33.3),
IBẸ = PHUE—T), (33.7)
tan (2 + r)
--- Trang 583 ---
So we have found the reflection coefficient |b|? for an incident wave polarized
perpendicular to the plane of ineidenee, and also the reflection coefficient |B|2
for an incident wave polarized in the plane of incidencel
Tt is possible to go on with arguments of this nature and deduce that 0 is real.
To prove this, one must consider a case where light is coming from both sides of
the glass surface at the same tỉme, 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 = +sin (¿ — r)/sin (¿ + r). It
is even possible to determine the sign by considering the case OŸ a very, Very
thin layer in which there is relection from the front and from the back surfaces,
and calculating how mụuch 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
,— SnU—r) u_ tan r). (33.8)
sin (¿+ r) tan (2+ 7)
These expressions for the relection coefficients as a function of the angles of
incidence and refraction are called Eresnel”s refection formulas.
Tf we consider the limit as the angles ? and z go to zero, we fñnd, for the case
of normal ineidenee, that 2 b2 (¿— r)2/(¡+r) for both polarizations, since
the sines are practically equal to the angles, as are also the tangents. But we
know that sin2/sin? = ø, and when the angles are small, ¿/r 2ø. It is thus easy
to show that the coefflicient of refection for normal incidenece is
B2ˆ—?— (n — Dã
(6+1)?
Tt is interesting to ñnd out how mụuch light is reflected at normal incidence
from the surface oŸ water, for example. Eor water, ø is 4/3, so that the reflection
coefficient is (1/7)2 ~ 2%. At normal incidence, only two percent of the light is
refected from the surface oŸ water.
33-7 Anomalous refraction
The last polarization elfect we shall consider was actually one of the first to be
discovered: anomalous refraction. Sailors visiting Iceland brought back to Burope
--- Trang 584 ---
crystals of Iceland spar (CaCOs) which had the amusing property of making
anything seen through the crystal appear doubled, I.e., as two images. 'Phis 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.