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Eaq. (31.23) should be ) AzqeE2,u. Our energy equation now looks like
ðă? = aE? + 2aE,E„ + N AzqeE,0. (31.24)
The #2 terms cancel, and we have
2ăl,E„ =—N Azqef.0. (31.25)
W©e now go back to Ed. (31.19), which tells us that for large z
đ„ = _XÂzứ 0(ret by z/c) (31.26)
--- Trang 551 ---
(recalling that ạ = W A2). Putting Eq. (31.26) into the left-hand side of (31.25),
W© getE
NWAzdqe—————
2v S—— E;(at 2) - 0(ret by z/c).
However, #2;(at z) is J;(at atoms) rebarded by z/c. Since the average is inde-
pendent of time, it is the same now as retarded by z/e, or is ;(at atom$) - 0,
the same average that appears on the right-hand side of (31.25). The two sides
are therefore equal if
Ý“= 1, Or œ = cục. (31.27)
We have discovered that 1Ý energy is to be conserved, the energy carried ïn an
electric wave per unit area and per unit time (or what we have called the imensify)
must be given by cocE2. IÝ we call the intensity Š, we have
s— 1ntensity =
Ss= Or = cạụcE2, (31.28)
energy/area/time
where the bar means the #ữne a0eragc. We have a nice bonus result om our
theory of the refractive indexl
31-6 Diffraction of light by a screen
lt is now a good time to take up a somewhat diferent matter which we can
handle with the machinery of this chapter. In the last chapter we said that
when you have an opaque screen and the light can come through some holes, the
distribution of intensity—the difraction pattern——could be obtained by imagining
instead that the holes are replaced by sources (oscillators) uniformly distributed
over the hole. In other words, the difracted wave is the same as though the hole
were a new source. We have to explain the reason for that, because the hole is, of
course, just where there are øø sources, where there are ?ø accelerating charges.
Let us first ask: “What 7s an opaque screen?” Suppose we have a completely
opaque screen bebween a source Š and an observer at P, as in Fig. 3I-6(a). Tf the
screen is “opaque” there is no field at P. Why is there no field there? According
to the basic principles we should obtain the field at as the field ; of the
source delayed, plus the field from all the other charges around. But, as we
have seen above, the charges in the screen will be set in motion by the field 2,
--- Trang 552 ---
X E=E, E=0 ,
Opaque screen
s E=E:; E=E: + Euai PP
xzhole
J—wall
S ~—plug P
x °
E=E. F =Es + Ej + EDi,g = Ô
Fig. 31-6. Diffraction by a screen.
and these motions generate a new field which, if the screen is opaque, must
czactlU cancel the field 2 on the back side of the screen. You say: “What a
miracle that it balances ezøctl Suppose it was not exactly right!” T it were
not exactly right (remember that this opaque screen has some thickness), the
field toward the rear part of the screen would not be exactly zero. So, not
being zero, it would set into motion some other charges in the material of
the screen, and thus make a little more field, trying to get the total balanced
out. So if we make the screen thick enough, there is no residual feld, because
there is enough opportunity to fñnally get the thing quieted down. In terms
of our formulas above we would say that the screen has a large and imaginary
Index, so the wawve is absorbed exponentially as it goes through. You know,
of course, that a thin enough sheet of the most opaque material, even gold, 1s
transparent.
Now let us see what happens with an opaque screen which has holes in it, as
in Eig. 3I-6(b). What do we expect for the fñeld at P? 'The field at P can be
represented as a sum of two parts—the field due to the source Š plus the field
due to the wall, i.e., due to the motions of the charges in the walls. We might
expect the motions of the charges in the walls to be complicated, but we can fnd
out :0ha‡ ftelds the produce in a rather simple way.
--- Trang 553 ---
Suppose that we were to take the same screen, but plug up the holes, as
indicated in part (c) of the fñgure. We imagine that the plugs are of exactly
the same material as the wall. Mind you, the plugs go where the holes were In
case (b). Now let us calculate the fñeld at P. The field at P is certainly zero in
case (©), but it is aiso equal to the ñeld from the source plus the feld due to
all the motions of the atoms in the walls and in the plugs. We can write the
following equations:
Case (h): đà p—= Hs + F2wall›
Case (c): FEÒp=0=E,+E\ạ+ Ebiug
where the primes refer to the case where the plugs are in place, but 2 1s, of
course, the same in both cases. Now if we subtract the two equations, we get
đạt p= (Evan - van) - EDnng:
Now ïf the holes are not too smaill (say many wavelengths across), we would not
expect the presence of the plugs to change the fields which arrive at the walls
except possibly for a little bit around the edges of the holes. Neglecting this
small efect, we can set #⁄van = # „¡ị and obtain that
đài p — —Ebiug
We have the result that the field at ÐP hen there œre holes ìn a sereen (case
b) is the same (except for sign) as the field that is produced by ứhat part of a
complete opaque wall which is located there the holes are! (The sign 1s not too
interesting, since we are usually interested in intensity which is proportional to
the square of the field.) It seems like an amazing backwards-forwards argument.
It is, however, not only true (approximately for not too small holes), but useful,
and is the justification for the usual theory of diÑraction.
'The field TỚNG 1s computed in any particular case by remembering that the
motion of the charges euerhere in the sereen is just that which will cancel out
the fñeld #⁄¿ on the back of the screen. OÔnece we know these motions, we add the
radiation fields at ? due just to the charges in the plugs.
W© remark again that this theory of difraction is only approximate, and will
be good only if the holes are not too smaill. Eor holes which are too small the
Tinng term will be small and then the diference between #2 „ and #¡ (which