22 M‘CreLttanp—The Energy of Secondary Radiation. 
In our problem we know that thick plates absorb the radiation completely, 
so that we are not concerned at present with the periodic solution corresponding 
to negative values of 
a — 0’. 
For positive values we must have 
K 
b <a, or <a) 5); or «<1. 
The energy set free as secondary radiation from any element of volume 
cannot therefore exceed the incident energy absorbed by the element. 
For values of « less than unity we have 
R = fig RPaP a, MUploae 
and. HERE 
r — Be-* Pale on, Bieve-” 
Taking, in the first place, an infinitely thick plate, which, for our purpose, 
need only be a plate thick enough to transmit no radiation, we have 
TN = 0, 153) = 0, and Al = Ite 2 
and from equation (1) we get 
Ley ees o 
B= UERe ZR. 
Therefore pie 
R = Bg PAO ® 
and 
Now p is equal to 
a when z=0; 
_@= Ja — 
(p= b ° 
Substituting for a and 2, 
K 
1-=-—/]l-« 
LS z ) 
K 
Y 
or 4p 
