For our problem, the driving force comes from the electric field of the wave from the source, so we should us f= gees- eee, where qe is the electric charge on the electron and for Es we use the expression Es= eoew from(31. 10). Our equation of motion for the electron is then geese We have solved this equation before, and we know that the solution is (31.14) where, by substituting in(31. 13), we find that geO (31.15) geO (31.16) We have what we needed to know-the motion of the electrons in the plate. And it is the same for every electron, except that the mean position(the "zero"of the motion)is, of course, different for each electron Now we are ready to find the field ea that these atoms produce at the point P, because we have already worked out(at the end of Chapter 30)what field is pro- duced by a sheet of charges that all move together. Referring back to Eq. (30.19), we see that the field ea at P is just a negative constant times the velocity of the charges retarded in time the amount z/c. Differentiating x in Eq. ( 31. 16) to get the velocity, and sticking in the retardation [or just putting xo from (31.15) into(30. 18)] yield eoc ia qeEo wo(4-2/e) (31.17 Just as we expected, the driven motion of the electrons produced an extra wave which travels to the right(that is what the factor eiw -z/e)says), and the amplitude of this wave is proportional to the number of atoms per unit area in the plate (the factor n)and also proportional to the strength of the source field(the factor Eo). Then there are some factors which depend on the atomic properties (qe, m, and wo), as we should expect n. The two expressions will, in fact, be identical i rial wha,o The most important thing, however, is that this formula(31. 17) for Ea looks very much like the expression for ea that we got in Eq. (31.8)by saying that the original wave was delayed in passing through a mat (n-1)△z=-m (31.18) Notice that both sides are proportional to Az, since n, which is the number of atoms per unit area, is equal to N AZ, where N is the number of atoms per unit volume of the plate. Substituting N AZ for n and cancelling the Az, we get our main ul, a formula for the index of refraction in terms of the properties of the atoms of the material-and of the frequency of the light 1+ (31.19) This equation gives the"explanation "of the index of refraction that we wished to obt 31-5