FROM A REGULARLY STRATIFIED MEDIUM In this we assume un= vn+1/vn, so that 497 l+1 + ?vn=0...........5,.......(31) The solution of (31) is where p + q = r2-t2-!, pq = t2, .....................(32) and H, K are arbitrary constants. Accordingly BVi'i+i i ZfV>7i+i -"• f-i r -fi- (/ ,__. -J/ — ---i.________ -i___ ^QQ 1 n Hpn + Kqn '.........................^oa' in which there is but one constant of integration effectively. This constant may be determined from the case of n = l, for which tt1 = 7'9—1. By means of (32) we get (p + l)H+(q + l) K=0, so that ^ — L^^£_^yLt_t/_.^J > ..................(34) and or since by (32) ?>a = (p fa = __I-__ ? ...........................(35) where A/(-) = 6, A/(-—Wrt......................(36) V \gy V Vp + 1/ In order to find tym we may put n= 1 in (17); and by use of (29), with m substituted for n, we get T and on reduction with use of (35), (32), »--*afa-"^-.........................<37> By putting TO = 0, we see that the upper sign is to be taken. The expressions thus obtained are those of Stokes: 0m_____ym ____________ f^R\ The connexion between a, b and r, t is established by setting, m = 1. Thus b — 6—1 a — or*1 ft6 — or1 b~1 In Stokes' problem, where r, t, (j>, ty represent intensities, a and b are real. If there is no absorption, r + t = 1, so that a — 1, 6 — 1 are vanishing quantities. In this case r____t 1 b^l~a-l~a-l + b-l' E. vi. 32not introducing other than integral values of m and n. If we make m = 1 in (17),