54  Mr.  L.  R.  Ingersoll  on  the  Faraday  and 
Let  V  represent  the  intensity,  for  any  chosen  wave-length, 
of  the  beam  o£  natural  light  incident  on  the  polarizer.  Then 
the  reflected  beam  will  be  made  up  of  two  parts, — 
l  K:  V  (polarized  in  the  plane  of  incidence)  ; 
and 
\  «K:  V  (polarized   perpendicular  to    the    plane  of 
incidence), 
where  Kj  is  a  factor  which  takes  account  of  the  percentage 
of  light  reflected  by  the  surfaces,  and  a  is  a  constant  to  be 
determined  later. 
On  emerging  from  the  analyser,  whose  plane  of  incidence 
makes  an  angle  a  with  that  of  the  polarizer,  the  intensity  is 
I  =  lK1K2r  cos2a-f-JrK1K2I/&  sin2* 
+ 1  KiK2  F  a  sin2  a  -f  1 K^  I'  a  b  cos2  a, 
where  b  and  K2  are  constants  for  the  analyser  corresponding 
to  a  and  K:  for  the  polarizer.  In  general  a  and  b  will  be 
small  and  their  product  may  be  neglected  ;  hence,  dropping 
the  last  term  and  collecting, 
I=J-K1K2I'{cos2a  +  (a-l-6)  sin2a}. 
A  small  rotation  8  will  cause  a  change, 
dl  =  —  SK]K2I'  (1—a  —  b)  sin  a  cos  a. 
Solving, 
~  dIcos2a+(a  +  &)  sin2  a 
I      (i  —  a  —  b)  sin  2a    ' 
or,  if  a  =  45°, 
-  I  1-a-b' 
The  constants  a  and  b  are  seen  to  represent  the  ratios  of 
intensities  of  the  light  polarized  perpendicular  to  and  in  the 
plane  of  incidence,  for  both  polarizer  and  analyser.  They 
may  be  determined  with  the  aid  of  a  nicol  or  other  com- 
pletely polarizing  agent  placed  at  N,  fig.  1.  If  its  principal 
plane  is  turned  at  right  angles  to  that  of  the  polarizer,  the 
intensity  of  the  light  on  emerging  from  the  analyser  will  be 
Ix = l  a  K^Ko  V  sin2  a  +  \  ab  KiK2K3  I'  cos2 «, 
K3  allows  for  absorption  in  the  nicol. 
-ossed  with  the  analyser  instead  it  will  be 
I2= J- b K1K2K3 V  sin2 a+l<&  K^oK, I'  cos2 «. 
