210  Mr.  N.  K.  Campbell  on  the 
AB  of  the  slab  (fig.  13).     Take  0  as  pole  and  ON,  the 
normal,  as  axis,  of  polar  coordinates.     The  rays   from  this 
FiR-.  13. 
/I/R 
//////////////////I '//////////////// 
Slab 
Io 
element  will  cause  ionization  —dB.dl  dv  in  any  small  volume 
r 
of  air  dv,  distant  r  from  0,  so  long  as  r—  I  sec  6  is  not  greater 
than  a  —  pi  sec  6*;  if  r  is  greater  than  this  there  will  be  no 
ionization.     Now,  in  an  element  of  volume  r2  sin  6  dr  d9  d(j> 
there  will  be  ionization  I0  sin  6  dr  d6  dcf>  dS  dl.     This  has  to 
be  integrated  within  the  limits  indicated.     <f)  can  range  from 
0  to  27r,  6  from  0  to  the  value  given  by  a  —pi  sec  0  =  0,  or 
cos-1—,    r    from    dsecO   to    a  —  (p  — 1)1  sec  0 ;    hence    the 
ionization  is 
J 
Jo  J 
*a—(p— 1)1  sec 
pi 
dO  sin  6  .  rfS  £# 
cos-i — 
^  £  sec  i 
=  2ttI0  i  "  (a-pl  sec  0)  sin  0  a'0  r/S  dl 
Jo 
=  2wI0dSdZ-{a-/d/+pZlog^\.     ...     (2) 
To  get  the  effect  of  the  whole  slab  we  must  integrate  this 
with  regard  to  S,  the  surface,  and  I,  the  depth.  We  thus 
find  as  the  whole  effect 
a 
2ttI0S   Y  \a-pl  +  pl\ogep~yil  =  l0  2ttS 
a- 
v    •  (3) 
for  if  pl  is  greater  than  a,  no  rays  emerge  from  the  surface. 
*  It  is  assumed  here  that  the  rays  are  emitted  with  equal  intensities 
in  all  directions.  This  seems  the  most  probable  distribution,  but  Bragg 
inserts  an  obliquity  factor  cos  0.  I  am  unable  to  find  a  justification  for 
such  a  procedure. 
