﻿Experiment 
  relating 
  to 
  the 
  Drift 
  of 
  the 
  JEther. 
  33 
  

  

  question 
  is, 
  Are 
  there 
  loci 
  where 
  the 
  maximals 
  for 
  all 
  the 
  

   various 
  component 
  slits 
  coincide 
  ? 
  

  

  The 
  general 
  formulae 
  developed 
  above 
  involve 
  the 
  angle 
  6, 
  

   — 
  the 
  angle 
  which 
  the 
  incident 
  light 
  makes 
  with 
  the 
  first 
  

   mirror. 
  If 
  a 
  point 
  on 
  the 
  flame 
  makes 
  an 
  angle 
  with 
  the 
  

   datum 
  line, 
  it 
  will 
  be 
  sufficient 
  to 
  replace 
  6 
  by 
  } 
  + 
  6 
  in 
  

   the 
  various 
  formulae 
  in 
  order 
  to 
  obtain 
  the 
  result 
  for 
  this 
  part 
  

   of 
  the 
  flame. 
  

  

  The 
  direction 
  of 
  drift 
  now 
  makes 
  an 
  angle 
  a—0 
  1 
  with 
  the 
  

   line 
  from 
  the 
  slit 
  to 
  the 
  centre 
  of 
  the 
  plate. 
  Hence 
  

  

  Vsm0=Usin(«-0i) 
  

  

  = 
  vcos 
  0j 
  — 
  Msin 
  L 
  . 
  

  

  The 
  position 
  of 
  the 
  white 
  point 
  — 
  i. 
  e. 
  where 
  the 
  central 
  

   maximal 
  cuts 
  the 
  first 
  mirror, 
  is 
  given 
  by 
  

  

  ^ 
  V 
  2 
  -iu 
  2 
  

  

  Oi 
  only 
  enters 
  through 
  iv. 
  Hence 
  when 
  f 
  is 
  small 
  x 
  is 
  inde- 
  

   pendent 
  of 
  0, 
  — 
  that 
  is, 
  the 
  central 
  maximal 
  cuts 
  the 
  plane 
  of 
  

   the 
  first 
  mirror 
  in 
  a 
  point 
  which 
  is 
  the 
  same 
  for 
  all 
  points 
  of 
  

   the 
  plane. 
  

  

  Next 
  the 
  common 
  direction 
  of 
  the 
  maximals 
  is 
  given 
  by 
  (17) 
  . 
  

  

  In 
  this 
  we 
  put 
  B\ 
  + 
  for 
  0. 
  Now 
  

  

  Vcos 
  (0 
  + 
  0i) 
  — 
  m=Vcos 
  0cos 
  0i— 
  Vsin0! 
  sin 
  0— 
  u, 
  

  

  where 
  the 
  source 
  is 
  fixed 
  to 
  the 
  apparatus, 
  

  

  V 
  sin 
  Q 
  = 
  v 
  cos 
  X 
  — 
  u 
  sin 
  X 
  ; 
  

  

  whence 
  

  

  V 
  cos 
  (0, 
  + 
  0)— 
  w 
  = 
  cos0 
  1 
  |V 
  cos 
  — 
  «cos 
  X 
  — 
  vsin 
  0,}, 
  

  

  Vsin 
  (0 
  t 
  + 
  0)— 
  v=sin 
  0!-|Vcos 
  — 
  ?* 
  cos 
  Qi—vsmQ^. 
  

  

  Hence 
  

  

  _ 
  V 
  2 
  sin 
  (0x 
  + 
  x) 
  — 
  v 
  2 
  sin 
  % 
  cos 
  t 
  + 
  2uv 
  cos 
  % 
  cos 
  X 
  — 
  u 
  2 
  cos 
  x 
  sm 
  #i 
  

  

  COt 
  *" 
  " 
  (V*-*»)cos(0 
  1 
  + 
  x 
  ) 
  

  

  or 
  

  

  • 
  , 
  t 
  , 
  a 
  . 
  , 
  V 
  2 
  sin 
  0! 
  4- 
  2uv 
  cos 
  X 
  — 
  u 
  2 
  sin 
  0! 
  cos 
  % 
  

  

  cotiIr 
  = 
  tan(Y 
  + 
  1 
  )+ 
  • 
  *™ 
  n 
  1 
  . 
  / 
  , 
  /, 
  > 
  

  

  r 
  V/V 
  ' 
  \ 
  2 
  —v 
  2 
  cos 
  (% 
  + 
  0i) 
  

  

  V/V 
  l/ 
  \ 
  J 
  — 
  v 
  2 
  cos(% 
  + 
  0i) 
  

  

  When 
  f 
  2 
  is 
  small 
  

  

  ^=i*— 
  (% 
  + 
  0i). 
  

   Phil. 
  Mag. 
  S. 
  6. 
  Vol. 
  3. 
  No. 
  13. 
  Jan. 
  1902. 
  D 
  

  

  