This preview shows pages 1–13. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 33 Iloflmitian from Surfaces 331 Reﬂection and refraction of light The subject of this chapter is the reﬂection and refraction of light—or electro
magnetic waves in general—at surfaces. We have already discussed the laws of
reﬂection and refraction in Chapter 35 of Volume 1. Here’s what we found out
there: 1. The angle of reﬂection is equal to the angle of incidence. With the angles deﬁned as shown in Fig. 33—1, (33.1) 2. The product n sm 0 is the same for the incident and transmitted beams
(Snell’s law).
n1 sin 0, = 112 sin 0,. (33.2)
3. The intensity of the reﬂected light depends on the angle of incidence and
also on the direction of polarization. For E perpendicular to the plane of
inCidence, the reﬂection coefficient R i is  2
RL 2 {1 _ Sln (6, 0;) I, H Sin2 (6, + Ht) (333) For E parallel to the plane of inCidence, the reﬂection coei‘fiCient RH is I, _ tan2 (0, — 0t) 4. For normal inCidence (any polarization, of course!),
IT n2 — n1 2
_, :2 _ . 3
I, (’12 + 711) (3‘ (Earlier, we used 1 for the inCident angle and r for the refracted angle Since we
can’t use r for both “refracted” and “reﬂected” angles, we are now using 6, =
incident angle, 6, = reﬂected angle, and 0, = transmitted angle.) Our earlier discussion is really about as far as anyone would normally need
to go With the subject, but we are going to do it all over again a different way
Why " One reason is that we assumed before that the indexes were real (no ab
sorption in the materials) But another reason is that you should kn0w h0w to
deal With what happens to waves at surfaces from the point of view of Maxwell’s
equations. We'll get the same answers as before, but now from a straightforward
solution of the wave problem, rather than by some clever arguments. We want to emphasize that the amplitude of a surface reﬂection is not a
property of the material, as is the index of refraction It is a “surface property,"
one that depends preCisely on how the surface is made. A thin layer of extraneous
junk on the surface between two materials of indices I11 and n2 will usually change
the reﬂection. (There are all kinds of pOSSlbliltICS of interference here~like the
colors of 011 ﬁlms Suitable thickness can even reduce the reﬂected amplitude to
zero for a given frequency; that’s how coated lenses are made.) The formulas
we Will derive are correct only if the change of index is sudden—within a distance
very small compared with one wavelength. For light, the wavelength is about
5000 A, so by a “smooth” surface we mean one in which the conditions change in 33—1 331 Reﬂection and refraction of
light 33—2 Waves in dense materials
33—3 The boundary conditions 33—4 The reﬂected and transmitted
waves 33—5 Reﬂection from metals 33—6 Total internal reﬂection Review. Chapter 35, Vol. I, Polarization SURFACE “2 Fig. 33—1.
of light waves at a surface.
directions are normal to the wave crests.) Reﬂection and refraction
(The wave WAVE caasrs
\ \ Fig. 33—2. For a wave moving in the
direction It, the phase at any point P is
(wt — k  r). going a distance of only a few atoms (or a few angstroms). Our equations will
work for light for highly polished surfaces. In general, if the index changes grad
ually over a distance of several wavelengths, there is very little reﬂection at all. 33—2 Waves in dense materials First, we remind you about the convenient way of describing a sinusmdal
plane wave we used in Chapter 36 of Volume 1. Any field component in the wave
(we use E as an example) can be written in the form E = EOeW—k'“, (33.6) where E represents the amplitude at the point r (from the origin) at the time t.
The vector k points in the direction the wave is travelling, and its magnitude [k] = k = 27r/>\ is the wave number. The phase veIOCity of the wave is m, = w/k,
for a light wave in a material of index n, up], = c/n, so
can
k — —ce (33.7) Suppose k is in the zdirection, then k  r is just kz, as we have often used it For k in any other direction, we should replace 2 by rk, the distance from the origin in the kdirection; that is, we should replace k2 by krk, which is Just k r. (See Fig. 33—2.) So Eq. (33.6) is a convenient representation of a wave in any direction.
We must remember, of course, that k‘r= k1x+kyy+kzz, where k,, k,,, and k; are the components of k along the three axes. In fact, we
pointed out once that (w, k,, ky, k2) is a four—vector, and that its scalar product
With (1, x, y, z) is an invariant. So the phase of a wave is an invariant, and Eq.
(33.6) could be written E = Eoelkm‘. But we don’t need to be that fancy now. For a sinusoidal E, as in Eq. (33.6), 6E/at is the same as in, and aE/ax is
— lsz, and so on for the other components. You can see why it is very convenient
to use the form in Eq. (33 6) when working with differential equations—differentia
tions are replaced by multiplications. One further useful pomt: The operation
V = (6/6x, 6/0y, 8/62) gets replaced by the three multiplications (—lk1, —ik,,,
—ikz). But these three factors transform as the components of the vector k, so
the operator V gets replaced by multiplication with —ik: 6
u ~—> 1w, at v —» —ik. (33.8) This remains true for any V operation—whether it is the gradient, or the diver
gence, or the curl. For instance, the zcomponent of V X E is
615, 613,. 6x 8y If both Ey and Ex vary as e77" ’, then we get
—zkxEy + zkyEr, which is, you see, the z—component of —Ik X E. So we have the very useful general fact that whenever you have to take the
gradient ofa vector that varies as a wave in three dimensmns (they are an important
part of physics), you can always take the derivations quickly and almost Without
thinking by remembering that the operation V is equivalent to multiplication by
—ik. 33—2 For instance, the Faraday equation 03
vxE——3? becomes for a wave
—ik X E = —in. This tells us that _ k X E 0.) B . (33.9) which corresponds to the result we found earlier for waves in free space—that B,
in a wave, is at right angles to E and to the wave direction. (In free space, w/k :
c.) You can remember the sign in Eq. (33 9) from the fact that k is in the direction
of Poynting’s vector S : enczE X B. If you use the same rule With the other Maxwell equations, you get again the
results of the last chapter and, in particular, that II
1
ii
I
l k ‘ k (33 10)
But since we know that, we won’t do it again. If you want to entertain yourself, you can try the following terrifying problem
that was the ultimate test for graduate students back in 1890: solve Maxwell’s
equations for plane waves in an anisotropic crystal, that is, when the polarization
P 18 related to the electric ﬁeld E by a tensor of polarizability. You should, of
course, choose your axes along the principal axes of the tensor, so that the relations
are simplest (then P, = aaEI, P,, = ahEu, and P2 = acEz), but let the waves
have an arbitrary direction and polarization. You should be able to ﬁnd the rela
tions between E and B, and how k varies with direction and wave polarization.
Then you Will understand the optics of an anisotropic crystal. It would be best
to start With the simpler case of a birefringent crystal—like calcite—for which
two of the polarizabilities are equal (say, oq, 2 ac), and see if you can understand
why you see double when you look through such a crystal If you can do that,
then try the hardest case, in which all three a’s are different. Then you Will know
whether you are up to the level of a graduate student of 1890. In this chapter,
however, we will consider only isotropic substances I I Fig. 33—3. The propagation vectors
I ‘ ~ ‘ . «  ‘ k, k’, and k” for the incident, reﬂected,
and transmitted waves. We know from experience that when a plane wave arrives at the boundary
between two different materials—say, air and glass, or water and Oil—there is a
wave reﬂected and a wave transmitted Suppose we assume no more than that and
see what we can work out. We choose our axes With the yzplane in the surface
and the xyplane perpendicular to the incident wave surfaces, as shown in Fig. 33—3. 33—3 The electric vector of the incident wave can then be written as
E, = Eoezm‘”k '). (33.11)
Since k is perpendicular to the zaxis, k  r = kxx + kyy. (33 12)
We write the reﬂected wave as E. = E{,e““"‘_""”, (33.13) so that its frequency is w', its wave number is k’, and its amplltude 1s E6. (We
know, of course, that the frequency is the same and the magnitude ofk 15 the same
as for the incident wave, but we are not going to assume even that. We W111 let 1t
come out of the mathematical machrnery.) F1nally, we wr1te for the transmltted
wave, E) = E6’e“‘“"’_"""). (33.14) We know that one of Maxwell’s equations gives Eq (33.9). so for each of the
waves we have I N
13.: 5‘ X E" B. = {5 X,E', B, = Lia. (33.15)
(U CO CO Also, 1f we call the 1ndexes of the two media ml and 112, we have from Eq. (33.10) 22
k2 = = (3316) C2 Since the reﬂected wave 1s 1n the same material, then [2 2
k’2 = (33.17)
whereas for the transmitted wave,
H2 2
W = E962”? (33.18) 33—3 The boundary conditions All we have done so far is to describe the three waves; our problem now 18
to work out the parameters of the reﬂected and transmitted waves in terms of
those of the incident wave. How can we do that? The three waves we have de
scribed satisfy Maxwell’s equations 1n the uniform mater1al, but Maxwell’s equa
tions must also be satisﬁed at the boundary between the two d1ﬂerent materials.
So we must now look at what happens right at the boundary. We w1ll ﬁnd that
Maxwell‘s equations demand that the three waves ﬁt together in a certam way. As an example of what we mean, the ycomponent of the electr1c ﬁeld E must
be the same on both s1des of the boundary. This is required by Faraday’s law, _ VXE=—w
Fig. 33—4. A boundary condition at EN = 5,1 is obtained from fr E ds = O. , (33.19) as we can see in the following way. ConSIder a little rectangular loop I‘ which
straddles the boundary, as shown in F1g 33—4. Equation (33.19) says that the line
1ntegra1 of E around I‘ is equal to the rate of change of the ﬂux of B through the loop: _
6
fEds = ———/B'nda.
1‘ 6! Now imagine that the rectangle is very narrow, so that the loop encloses an 1n
ﬁnitesimal area. If B remams ﬁmte (and there‘s no reason 1t should be 1nﬁn1te
at the boundary!) the ﬂux through the area is zero So the l1ne 1ntegral of E must 33—4 be zero. If EU] and EH2 are the components of the ﬁeld on the two sides of the
boundary and if the length of the rectangle 1s I, we have E111] — Eu‘ll : 0
or
Eu] = E.,2, (33.20) as we have said. This gives us one relation among the ﬁelds of the three waves. The procedure of working out the consequences of Maxwell’s equations at
the boundary IS called “determining the boundary cond1trons.” Ordinarily, it IS
done by ﬁnding as many equations like Eq. (33 20) as one can, by making argu—
ments about little rectangles hke F in Flg. 334, or by using little gaussian surfaces
that straddle the boundary Although that is a perfectly good way of proceeding,
it gives the impression that the problem of dealing With a boundary 15 different
for every different physrcal problem For example, in a problem of heat ﬂow across a boundary, how are the tem
peratures on the two SldCS related? Well, you could argue, for one thing, that the
heat flow In the boundary from one stde would have to equal the ﬂow away from
the other side. It IS usually possible, and generally quite useful, to work Out the
boundary conditions by making such physrcal arguments. There may be times,
however, when in working on some problem you have only some equations, and
you may not see right away what physical arguments to use. So although we are
at the moment Interested only in an electromagnetic problem, where we can make
the physical arguments“ we want to show you a method that can be used for any
problem—at general way of ﬁnding what happens at a boundary directly from the
differential equations We begin by writing all the Maxwell equations for a dlelectric—and this time
we are very specrﬁc and write out explicitly all the components: VE:—Ff
6I)
{)E 8E, 0E aP OB, 6P
J m; a: : _ ,4 in ,J .2
6“<le + By + 8:) (ﬁx + 6y + 02) (33 1)
QB
VXE"—a
OE: _ “34 _ _ 9?: (3% 22a)
0)) ('32 — (it “
0E 0E, OBJ
Vt' _ 7r :2 _ .7; 33.22
OZ 6A a! ( b)
013, 81? 0t?
, ' _ v 2 _ , .i 3 22
0x 6y (1; (3 c)
V ‘ B ’7 0
£151 + .1er 01% : 0 (33 23)
0A d)’ dz
2 . I 613 (1E,
( ‘ X B x (‘1) + , 03, 68, 1 OF 315,
“ ‘—~7 =v # V 3.24,
‘ (av 62) e” at + at (3 d)
9 03 OB~ l an 6E”
.~ I" i ~ : 7 , ﬁi .2
( (02 0x) 60 61 + at (33 4b)
) 03,, at; 1 ap 6E:
.~ _ 30 _ , “z W 3 .2
‘ (Ox 6y) e0 61 + at (3 4C) 33—5 REGION 1 REGION 3' REGION 2 The ﬁelds in the transition Fig. 33—5.
region (3) between two different ma
terials in regions (1) and (2). Now these equations must all hold in region 1 (to the left of the boundary)
and in region 2 (to the right of the boundary). We have already written the solu
tions in regions 1 and 2. Finally, they must also be satisﬁed In the boundary, which
we can call region 3. Although we usually think of the boundary as being sharply
discontinuous, in reality it is not. The physical properties change very rapidly
but not inﬁnitely fast. In any case, we can imagine that there is a very rapid, but
continuous, transition of the index between region 1 and 2, in a short distance we
can call region 3. Also, any ﬁeld quantity like P1, or E,,, etc., Will make a Similar
kind of tranSition in region 3. In this region, the diﬁerential equations must still
be satisﬁed, and it is by following the differential equations in this region that we
can arrive at the needed “boundary conditions.” For instance, suppose that we have a boundary between vacuum (region 1)
and glass (region 2). There IS nothing to polarize in the vacuum, so P, = 0.
Let‘s say there is some polarization P2 in the glass. Between the vacuum and the
glass there is a smooth, but rapid, transition If we look at any component of
P, say P7,, it might vary as drawn in Fig. 33—5(a). Suppose now we take the ﬁrst
of our equations, Eq (33.21). It involves derivatives of the components of P with
respect to x, y, and z. The y and zderivatives are not interesting; nothing spec
tacular is happening in those directions. But the xderivative of PI Will have some
very large values in region 3. because of the tremendous slope of PI. The derivative
()PJ/ax Will have a sharp spike at the boundary, as shown in Fig. 33—5(b). If we
imagine squashing the boundary to an even thinner layer, the spike would get
much higher If the boundary is really sharp for the waves we are interested in,
the magnitude of BPI/ax in region 3 Will be much, much greater than any contribu—
tions we might have from the variation ofP in the wave away from the boundary——
so we ignore any variations other than those due to the boundary. Now how can Eq. (33 21) be satisﬁed if there is a whopping big spike on the
righthand side? Only if there is an equally whopping big spike on the other side.
Something on the lefthand Side must also be big. The only candidate is aE,/ax,
because the variations with y and z are only those small effects in the wave wejust
mentioned. So —eo(aE/6x) must be as drawn in Fig. 33—5(c)—just a copy of
(am/ax. We have that €9£e=_apx. :3
06x 6x i If we integrate this equation With respect to x across region 3, we conclude that 60(Ex2 — Em) = _(Px2 In other words, thejump in 60E, in gOing from region 1 to region 2 must be equal
to the jump in —Px.
We can rewrite Eq. (33.25) as EoEw 4‘ P352 = 60E“ ‘i’ P11, (3326) which says that the quantity (60E, + Pr) has equal values in region 2 and region 1.
People say: the quantity (60E,r + P,) is COHIII’ILIOHS‘ across the boundary. We have,
in this way, one of our boundary conditions. Although we took as an illustration the case in which P1 was zero because
region 1 was a vacuum, it is clear that the same argument applies for any two
materials in the two regions, so Eq. (33.26) is true in general. Let’s now go through the rest of Maxwell’s equations and see What each of
them tells us. We take next Eq. (33.22a). There are no xderivatrves, so it doesn’t
tell us anything. (Remember that the ﬁelds themselves do not get espeCially large
at the boundary; only the derivatives With respect to x can become so huge that
they dominate the equation.) Next, we look at Eq. (33 22b). Ah' There is an
xderivative! We have aEz/ax on the lefthand Side. Suppose it has a huge de
rivative But wait a moment! There IS nothing on the righthand side to match it
With; therefore E3 cannot have any jump in gomg from region 1 to region 2.
[If it did, there would be a spike on the left of Eq. (33.22a) but none on the right, 33~6 and the equation would be false ] So we have a new condition: E:2 = Ezl. (33.27)
By the same argument, Eq (33.22c) gives
Ell/2 : EUl. This last result is just what we got in Eq. (33 20) by a line integral argument.
We go on to Eq. (33 23) The only term that could have a spike is (am/ax.
But there’s nothing on the right to match it, so we conclude that 8,2 = 8,1. (33.29) On to the last of Maxwell’s equations! Equation (33 24a) gives nothing,
because there are no AdCl‘lVElthCS Equation (33 23b) has one, —c2 6B2/6x, but
again, there is nothing to match it with. We get B22 = B“. (33.30)
The last equation is quite similar, and gives
31/2 : Byl The last three equations gives us that BZ = 3;. We want to emphasize,
however, that we get this result only when the materials on both Sides of the
boundary are nonmagnetic—or rather, when we can neglect any magnetic effects
of the materials. This can usually be done for most materials, except ferromagnetic
0;;‘65 (We Will treat the magnetic properties of materials in some later chapters.) E Our program has netted us the six relations between the ﬁelds in region 1 and
those in region 2. We have put them all together in Table 33—1. We can now use
them to match the waves in the two regions. We want to emphasize, however, that
the idea we have just used Will work in any phySICal situation in which you have
dilTerential equations and you want a solution that crosses a sharp boundary
between two regions where some property changes. For our present purposes,
we could have easin derived the same equations by usmg arguments about the
ﬂuxes and circulations at the boundary. (You might see whether you can get the
same result that way.) But now you have seen a method that will work in case you
ever get stuck and don't see any easy argument about the phySlCS of what IS happen
ing at the boundary—you can Just work With the equations. 33—4 The reﬂected and transmitted waves Now we are ready to apply our boundary conditions to the waves we wrote
down in Section 33—2. We had: 5, = Eoe'(W’—’tr“—W’, (33 32) ET : Eﬁertw'l—Agzwki’ﬂ), El : E;;y<w”'~ké’xké’v>, (33.34) B] : 52:79, (33.35) Br : Egg/E, (33.36)
0.) Bi : I: 6:; 9. (33 37) We have one further bit of knowledge: E is perpendicular to its propagation
vector k for each wave. 3 3—7 Table 33—1
Boundary conditions at the surface of :
dielectric
(GOEi + Pl); : (60152 + P2);
(El)7/ : (E2)rl
(E i)z = (E2):
BI L B; (The surface is in the yzplane) The results Will depend on the direction of the Evector (the “polarization”)
of the incoming wave. The analySIS is much Simpliﬁed if we treat separately the case
of an inCident wave With its E—vector parallel to the “plane of incidence" (that is,
the xyplane) and the case of an inCIdent wave with the Evector perpendicular to
the plane of incidence. A wave of any other polarization is just a linear combina
tion of two such waves. In other words, the reﬂected and transmitted intenSIties
are different for different polarizations, and it is easiest to pick the two Simplest
cases and treat them separately. We will carry through the analysis for an incoming wave polarized per
pendicular to the plane of inCidence and thenjust give you the result for the other.
We are cheating a little by taking the simplest case, but the prinCiple is the same
for both. So we take that E, has only a zcomponent, and Since all the Evectors
are in the same direction we can leave off the vector signs. So long as both materials are isotropic, the induced oscillations of charges in
the material will also be in the z—direction, and the E—ﬁeld of the transmitted and
radiated waves will have only z—components. So for all the waves, Ex and E1,
and Pr and Pg are zero. The waves will have their E and Bvectors as drawn in
Fig. 33—6 (We are cutting a corner here on our original plan of getting everything
from the equations. This result would also come out of the boundary conditions, Fig 334,. polarization of the re_ but we can save a lot of algebra by usmg the phySical argument When you have
ﬂecfed and transmitted waves when the some spare time, see if you can get the same result from the equations. It is clear
Eﬂeld of the incident wave is perpendicu that what we have said agrees With the equations; it lSJUSt that we have not shown
lar to the plane of incidence. that there are no other pOSSibilities.) Now our boundary conditions, Eqs. (33 26) through (33.31), give relations
between the components of E and B in regions 1 and 2. For region 2 we have only
the transmitted wave, but in region 1 we have two waves. Which one do we use?
The ﬁelds in region 1 are, of course, the superposition of the ﬁelds of the inCident
and reﬂected waves. (Since each satisﬁes Maxwell’s equations, so does the sum.)
So when we use the boundary conditions, we must use that 'E12E1+Ers E2:Ets
and Similarly for the B’s. \ For the polarization we are considering, Eqs. (33.26) and (33.28) give us no
new information; only Eq (33.27) is useful. It says that El + Er : E;
at the boundary, that is, for x = 0. So we have that
Eoeuwtwkuy) + Ebemm’l—kuu) : E(;)/gi(t1i"l—/,1r[rll)’ which must be true for all t and for ally. Suppose we look ﬁrst at y = 0. Then we
have t ’1 i”!
E06”) ‘1” E68”) : Eifem This equation says that two oscﬂlating terms are equal to a third oscﬂlation.
That can happen only if all the oscillations have the same frequency. (It is im
poss1ble for three—0r any number~of such terms With different l‘requenCIes to
add to zero for all times.) So to” = w’ = 0.). (33.39) As we knew all along, the frequenCies of the reﬂected and transmitted waves are
the same as that of the inCident wave. We should really have saved ourselves some trouble by putting that in at the
beginning, but we wanted to show you that it can also be got out of the equations.
When you are doing a real problem, it is usually the best thing to put everything you
know into the works right at the sta rt and save yourself a lot of trouble. By deﬁnition, the magnitude of k is given by k2 = n2w2/c2, so we have also that
k/IZ kl2 k2
err = W = a (33.40)
n3 n? nf 33—8 Now look at Eq. (33.38) for I = 0. Usmg again the same kind of argument
we have Just made. but this time based on the fact that the equation must hold
for all values of y, we get that k1] : k;, 2 kg. (33.41)
From Eq. (33.40), k’2 2 k2, so
k;2 + kff = k: + k3.
Combining this with Eq. (33.41), we have that
kll : kf, or that k’, = ikx. The positive Sign makes no sense; that would not give a
reﬂected wave, but another incident wave, and we said at the start that we were
solving the problem of only one incident wave. So we have M = —k.. (33 42) The two equations (33.41) and (33.42) give us that the angle of reﬂection is equal
to the angle of inCidence, as we expected. (See Fig. 33—3 ) The reﬂected wave is E, = (,e'<w‘—kzx+kyy>. (33.43) For the transmitted wave we already have that kg; = k.
and k”2 k2 4* = 4 .4 "g "f , (33 4)
so we can solve these to ﬁnd k;’. We get 2
kg? = W — kg,” = g; k2 — k3. (33.45) Suppose for a moment that ml and I13 are real numbers (that the imaginary
parts of the indexes are very small). Then all the k’s are also real numbers, and
from Fig. 33—3 we ﬁnd that k, _ . k’y’ _ .
7; — Sin 0,, F, — Sin 0). (33.46)
From (33.44) we get that
112 sin 6; = n1 sin 0,, (33.47) which is Snell’s law of refraction—again, something we already knew. If the
indexes are not real. the wave numbers are complex, and we have to use Eq. (33.45).
[We could still deﬁne the angles 6, and 0) by Eq. (33.46), and Snell’s law, Eq. (33.47),
would be true in general. But then the “angles” also are complex numbers, thereby
losing their simple geometrical interpretation as angles. It 15 best then to describe
the behaVior of the waves by their complex km or k’,’ values ] So far, we haven’t found anything new. We havejust had the Simpleminded
delight of getting some obvious answers from a complicated mathematical ma
chinery. Now we are ready to ﬁnd the amplitudes of the waves which we have
not yet known. Using our results for the (0’5 and k’s, the exponential factors in
Eq. (33.38) can be cancelled, and we get E, + E{, 2 E6’. (33 48) Since both Et’, and E{)' are unknown, we need one more relationship. We must
use another of the boundary conditions. The equations for E, and E, are no help,
because all the ES have only a zcomponent So we must use the conditions on
B. Let’s try Eq. (33 29): BIZ = le 3 3—9 Fig. 33—7.
when the Eﬁeld of the incident wave is
parallel to the plane of incidence. Polarization of the waves From Eqs. (33.35) through (33.37). Recalling that w” = w’ w and kj’,’ = kl, = k,,, we get that H / w II
E0 ‘l' EU _ 0 But this isjust Eq. (33 48) all over again‘ We’vejust wasted time getting something
we already knew. We could try Eq. (33.30), Bzg 2
So there’s only one equation left: Eq. (33.31), 8,,2 : B21, but there are no zcomponents of B‘
B“. For the three waves. /  I/
B _ B i kLE, B g k, E,
111 H _ ‘_" ’ yr _ _ TT/i ’ y! _ _ T /7 ‘
(.0 (,0 0.) (33.49) Putting for E,, E,, and E, the wave expression for x = 0 (to be at the boundary),
the boundary condition is
k k5; k t—k iv ’i—k’i ”i—A”
i Enema) V11) + ‘7 E‘ljeuu) yJ) : _/7 Ezjleuw U ll).
w w (.0 Again all w’s and ky’s are equal, so this reduces to
kIE” + kLEE, = k;’E(,’. (33.50) This gives us an equation for the ES that is different from Eq. (33 48). With the two, we can solve for E6 and E6’. Remembering that k: : —k,, we get
kg, — kﬁ/
Et’) 2 Riki/a, Em (33 51)
2k
" = IA 33. 2
E0 kr + kit! EO' ( 5 ) These, together With Eq. (33.45) or Eq. (33 46) for kf’, give us what we wanted to
know. We will discuss the consequences of this result in the next section. If we begin With a wave polarized with its EVector parallel to the plane of
inCIdence, E will have both x and ycomponents, as shown in Fig. 33—7. The
algebra is straightforward but more complicated (The work can be somewhat
reduced by expressmg things in this case in terms of the magnetic ﬁelds, which are
all in the zdirection.) One ﬁnds that 2 _ 2 //
lEril = IEol <33 53>
l’lg r "1 x
and
{Em : i£”1”~’lf,£ lEUi. (33 54) 2 2
"2k; + (11kg Let’s see whether our results agree with those we got earlier Equation (33 3)
is the result we worked out in Chapter 35 of Volume I for the ratio of the intensity
of the reﬂected wave to the intenSity of the 1nc1dent wave Then, however, we were
considering only real indexes For real indexes (and k’s), we can write k, = kcos 0, : :0? cos 0,, kg = k” cos 6; = 9% cos 6;
Substituting in Eq. (33.51), we have 1% ﬂLcos 6, — *1; cos 6; (33.55) _. iiiiiii 7 , E0 n1 cos 6, + 112 cos 0, 33—10 which does not look the same as Eq. (33.3). It will, however, if we use Snell’s law to get rid of the n’s. Setting n2 2 n1 sin HL/sin 6,, and multiplying the numerator
and denominator by sm 6!, we get 6 cos 0, sin 6; — sin 0, cos 6; E0 cos 6, sm 6; —l— Sin 0, cos 0, The numerator and denominator are just the sines of (0, — 6t) and (671 —l— 0,);
we get §§ _ 3195— at)
E0 _ Sm (gitl' 0:) (33.56) Since EX, and E0 are in the same material, the intensities are proportional to the
squares of the electric ﬁelds, and we get the same result as before. Similarly, Eq.
(33.53) is the same as Eq. (33.4). For waves which arrive at normal incidence, 0, = 0 and 6, = 0. Equation
(33.56) gives 0/0, which is not very useful. We can, however, go back to Eq. (33.55), which gives
Q 2 (ﬂy = <L:Q>2.
1, Eu "1 + "2 This result, naturally, applies for “either” polarization, since for normal incidence
there is no speCial “plane of incidence.” (33.57) 33—5 Reﬂection from metals We can now use our results to understand the interesting phenomenon of
reﬂection from metals. Why is it that metals are shiny? We saw in the last chapter
that metals have an index of refraction which, for some frequenc1es, has a large
imaginary part. Let’s see what we would get for the reﬂected intenSity when light
shines from air (with n = 1) onto a material with n = —im. Then Eq. (33.55)
gives (for normal incidence) E6 _ 1 + [H1
F0 — li— in] . For the intensity of the reﬂected wave, we want the square of the absolute values
of E6 and EU: Q = iEg2 _Ll+in12
I. lE0l2 _ 1 —in112’
01‘
2
_1: 2 Lily. I 1 (33 58)
1: 1 + n? For a material With an index which is a pure imaginary number, there is 100 per
cent reﬂection' Metals do not reﬂect 100 percent, but many do reﬂect Visible light very well.
In other words, the imaginary part of their indexes is very large But we have seen
that a large imaginary part of the index means a strong absorption. So there is a
general rule that if any material gets to be a very good absorber at any frequency.
the waves are strongly reﬂected at the surface and very little gets inSide to be ab
sorbed You can see this effect with strong dyes Pure crystals of the strongest
dyes have a “metallic” shine. Probably you have noticed that at the edge of a bottle
of purple ink the dried dye Will give a golden metallic reﬂection, or that dried red
ink Will sometimes give a greenish metallic reﬂection. Red ink absorbs out the
greens of transmitted light, so if the ink is very concentrated, it will exhibit a strong
surface reflection for the frequencies of green light. You can easin show this eﬂect by coating a glass plate with red ink and
letting it dry. If you direct a beam of white light at the back of the plate, as shown
in Fig. 33—8, there Will be a transmitted beam of red light and a reﬂected beam of
green light. 33—11 GLASS PLATE DRIED RED INK Fig. 33—8. A material which absorbs
light strongly at the frequency (.0 also
reﬂects light of that frequency. n2=0 n3=n‘.‘. If there is a small gap, Fig. 33—l0.
internal reﬂection is not "total"; a trans—
mitted wove appears beyond the gap. Fig. 33—9. Total internal reﬂection. 336 Total internal reﬂection If light goes from a material like glass, with a real index ll greater than 1.
toward, say, air. with an index n2 equal to l, Snell‘s law says that sm 0; = )1 sin 0L. The angle 05 of the transmitted wave becomes 90" when the incident angle 0. is
equal to the “critical angle" 0, given by nsm a0 = 1. (33.59) What happens for (91 greater than the critical angle" You know that there is total
internal reﬂection. But how does that come about" Let’s go back to Eq (33.45) which gives the wave number k’,’ for the trans
mitted wave. We would have k5’“ = 25 kg.
Now k,, = k sin 61 and k = wn/c, so
2
kg” = E:1)?“ ~ n2 Slnz 61). If n Sln BL 18 greater than one, kg’2 18 negative and k'r’ is a pure imaginary, say
ilk]. You know by now what that means' The “transmitted” wave (Eq. 33.34)
Will have the form Et 2 Ege::k[xei(ul—I.,ly). The wave amplitude either grows or drops off exponentially With increasing x.
Clearly, what we want here is the negative Sign. Then the amplitude of the wave
to the right of the boundary will go as shown in Fig. 33—9. Notice that A, IS of
the order w/c—which is A”. the free—space wavelength of the light. When light is
totally reﬂected from the mSide of a glassair surface. there are ﬁelds in the air,
but they extend beyond the surface only a distance of the order of the wavelength
of the light We can now see how to answer the followmg question: If a light wave in glass
arrives at the surface at a large enough angle, it IS reﬂected, if another piece of
glass is brought up to the surface (so that the “surface” in effect disappears) the
light is transmitted. Exactly when does this happen? Surely there must be con
tinuous change from total reﬂection to no reﬂection' The answer, of course, is
that if the air gap is so small that the exponential tail of the wave in the air has an
appreciable strength at the second piece of glass, it Will shake the electrons there
and generate a new wave, as shown in Fig. 33—10. Some light Will be transmitted.
(Clearly, our solution is incomplete, we should solve all the equations again for a
thin layer of air between two regions of glass.) 33—12 (0) TRANSMITTER H) B DETECTOR (b) DETECTOR ii) ml Bl (c) A"
TRANSMTTTER DETECTOR DETECTOR TRANSMITTER DETECTOR
Fig. 33—1 1. A demonstration of the penetration of internally reﬂected waves. This transmission effect can be observed with ordinary light only if the air
gap is very small (of the order of the wavelength of light, like 10’5 cm), but it is
easily demonstrated with threecentimeter waves. Then the exponentially de
creasing ﬁeld extends several centimeters. A microwave apparatus that shows the
effect is drawn in Fig. 33—11 Waves from a small threecentimeter transmitter are
directed at a 45° prism of paraffin. The index of refraction of parafﬁn for these
frequencies is 1.50, and therefore the critical angle is 415°. So the wave is totally
reﬂected from the 45° face and is picked up by detector A, as indicated in
Fig. 33—ll(a). If a second parafﬁn prism is placed in contact with the ﬁrst, as
shown in part (b) of the ﬁgure, the wave passes straight through and is picked up
at detector B. If a gap of a few centimeters is left between the two prisms, as in
part (0). there are both transmitted and reﬂected waves. The electric ﬁeld outSide
the 45° face of the prism in Fig. 33—11(a) can also be shown by bringing detector
B to within a few centimeters of the surface. 33—13 DETECTOR ...
View
Full
Document
This note was uploaded on 06/18/2009 for the course PHYSICS none taught by Professor Leekinohara during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LeeKinohara
 Physics

Click to edit the document details