﻿Motion 
  of 
  the 
  Earth 
  and 
  JEther. 
  21 
  

  

  Now 
  this 
  can 
  be 
  true 
  only 
  if 
  the 
  images 
  are 
  symmetrical 
  with 
  

   respect 
  to 
  L, 
  or 
  if 
  each 
  wave-front 
  is 
  absolutely 
  uniform 
  in 
  

   character. 
  But 
  in 
  dealing 
  with 
  wave-fronts 
  of 
  light 
  in 
  con- 
  

   nexion 
  with 
  interference 
  we 
  must 
  be 
  careful 
  to 
  conjoin 
  the 
  

   effects 
  of 
  parts 
  only 
  of 
  identical 
  origin, 
  as, 
  for 
  instance, 
  that* 
  

   of 
  C 
  with 
  that 
  of 
  E, 
  or 
  that 
  of 
  0' 
  with 
  that 
  of 
  7 
  or 
  that 
  of 
  L 
  

   in 
  CD' 
  with 
  that 
  of 
  L 
  in 
  EF. 
  

  

  But 
  in 
  the 
  actual 
  experiments 
  it 
  is 
  practically 
  impossible 
  to 
  

   secure 
  that 
  the 
  two 
  images 
  intersect 
  in 
  a 
  point 
  that 
  corre- 
  

   sponds 
  to 
  itself 
  as 
  L 
  does. 
  In 
  general 
  we 
  must 
  assume 
  that 
  

   the 
  two 
  images 
  CD' 
  and 
  EF 
  intersect 
  under 
  the 
  conditions 
  

   represented 
  in 
  the 
  next 
  diagram 
  (fig. 
  3), 
  where 
  in 
  EF 
  and 
  

   CK 
  in 
  CD' 
  are 
  images 
  of 
  the 
  same 
  point 
  in 
  the 
  original 
  wave- 
  

   front. 
  

  

  Let 
  00' 
  = 
  c 
  ; 
  then 
  if 
  T 
  is 
  any 
  point 
  at 
  distance 
  x 
  from 
  

  

  Fi°\ 
  3. 
  

  

  in 
  EF, 
  the 
  corresponding 
  point 
  U 
  in 
  CD' 
  is 
  at 
  distance 
  

   c 
  + 
  x 
  from 
  0. 
  Let 
  us 
  bisect 
  FOD' 
  by 
  AB, 
  and 
  find 
  the 
  

   value 
  of 
  x 
  which 
  determines 
  a 
  pair 
  of 
  corresponding 
  points 
  

   so 
  that 
  they 
  are 
  equidistant 
  from 
  a 
  point 
  whose 
  polar 
  coordi- 
  

   nates 
  relative 
  to 
  and 
  OB 
  are 
  r 
  and 
  0. 
  Denote 
  the 
  anode 
  

   FOB 
  by 
  a 
  ; 
  then 
  the 
  equality 
  of 
  PT 
  and 
  PU 
  gives 
  

  

  r 
  2 
  + 
  x 
  2 
  -2rxcos(6 
  + 
  *) 
  = 
  r 
  2 
  +(c 
  + 
  x) 
  2 
  -2r(c 
  + 
  x)cos{0-u), 
  (1) 
  

  

  .*. 
  2x(c 
  — 
  2r 
  sin 
  6 
  sin 
  a) 
  — 
  2rc 
  cos 
  [6 
  — 
  a) 
  + 
  c 
  2 
  = 
  ; 
  

   and 
  treating 
  a 
  as 
  a 
  small 
  angle 
  

  

  2x(c 
  — 
  2rot 
  sin 
  6) 
  — 
  2rc 
  cos 
  6 
  — 
  2rca 
  sin 
  6 
  f 
  c 
  l 
  = 
  nearly. 
  

   Denote 
  r 
  cos 
  6 
  by 
  q, 
  and 
  r 
  sin 
  by 
  p, 
  and 
  then 
  

  

  2x 
  (c 
  — 
  2pa) 
  +c(c—2pa) 
  —2cg 
  = 
  0. 
  ... 
  (2) 
  

   In 
  this 
  equation 
  first 
  suppose 
  that 
  

  

  c 
  — 
  2px 
  = 
  0, 
  

   then 
  either 
  c 
  = 
  or 
  ^ 
  = 
  0. 
  If 
  c 
  happens 
  to 
  be 
  zero, 
  then 
  

  

  