oing work. Since there are N Az atoms in a unit area, the last term in Eq.(31.23) aE=aE?+ 2aEs Ea t Naz qe Esu (3124 The E terms cancel, and we have 2aEE=N△zqE2 We now go back to Eq (30.19), which tells us that for large z ea N△zqe (ret by z/c) ( recalling that n=N Az). Putting Eq.(3126)into the left-hand side of (31 25) NAzqe e(at z)v(ret by z/c) However, Es(at z)is Es(at atoms)retarded by z/c. Since the average pendent of time, it is the same now as retarded by z/c, or is Es(at atom).v, the same average that appears on the right-hand side of (31. 25). The two sides are efore equal if (3127) We have discovered that if energy is to be conserved, the energy carried in an elec- tric wave per unit area and per unit time (or what we have called the intensity) must be given by EoCEZ. If we call the intensity S,we have CE2 nergy/area/time opaque screen where the bar means the time average. We have a nice bonus result from our of the refractive index! S EEs E=es 31-6 Diffraction of light by a screen It is now a good time to take up a somewhat different matter which handle with the machinery of this chapter. In the last chapter we said that when c you have an opaque screen and the light can come through some holes, the distribu tion of intensity-the diffraction pattern- could be obtained by imagining instead that the holes are replaced by so (oscillators)uniformly distributed over the wofl+Eplug"o hole. In other words, the diffracted 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 no sources, where there are no accelerating charges Diffraction by a screen Let us first ask: "What is an opaque screen? "Suppose we have a completely paque screen between a source S and an observer at P, as in Fig. 31-6(a).If the creen 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 P as the field es 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 Ea, and these motions generate a new field which, if the screen is opaque, must exactly cancel the field Es on the back side of the screen. You say: What a miracle that it bal ances exactly! Suppose it was not exactly right! " If it were not exactly right(re- member that this opaque scree has some thickness ), the field toward the 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 field, because there is enough opportunity to finally get the thing quieted down. In terms of our formulas above we would say that the