Dividing Eq(33. 1)by Eg(33. 2), we obtain B a result which we can check against what we already know. If we set(i + r) 90, Eq. (33. 3)gives B=0, as Brewster says it should be, so our results so far are at least not obviously wrong We have assumed unit amplitudes for the incident waves, so that B 2/12 the reflection coefficient for waves polarized in the plane of incidence, and b /1 is the reflection coefficient for waves polarized normal to the plane of incidence 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 coefficient Band bl individually! We know from the conservation of energy that the energy in the refracted wave must be equal to the incident energy minus the energy in the reflected wave, 1-B2in one case, 1- blin the other.Furthermore,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 JA 2/al. 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 field. 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 field. Therefore we can write 1=BF=2 334) We now substitute Eq (33. 2)to eliminate A/a from the expression above d express B in terms of b by means of Eq. (33.3) 1-{b2 (i+r) This equation contains only one unknown amplitude, b. Solving for b 2, we obtain bl 336) and, with the aid of (33. 3), tan (33.7 So we have found the reflection coefficient b2 for an incident wave polarized perpendicular to the plane of incidence, and also the reflection coefficient B 2for an incident wave polarized in the plane of incidence It is possible to go on with arguments of this nature and deduce that b is real To prove this, one must consider a case where light is coming from both sides of the glass surface at the same time, 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(i-r)/sin(i +r). It is ven possible to determine the sign by considering the case of a very, very thin layer in which there is reflection from the front and from the back surfaces, and calculating how much light is reflected. We know how much light should be eflected by a thin layer, because we know how much current is generated and we have even worked out the fields produced by suc One can show by these arguments tha b=sin(i-r, B tan(i (i+ These expressions for the reflection coefficients as a function of the angles of incidence and refraction are called Fresnels reflection formulas If we consider the limit as the angles i and r go to zero we find for the case of normal incidence, that B2 b2s(i-r)/(i+r)2 for both polarizations