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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 = 7'91.    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  61    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),