﻿Focal 
  Plane 
  of 
  a 
  Telescope 
  with 
  Circular 
  Aperture. 
  17 
  

  

  As 
  the 
  point 
  lies 
  very 
  near 
  the 
  periphery, 
  S 
  is 
  a 
  small 
  

   quantity 
  and 
  k 
  very 
  near 
  nnity 
  ; 
  whence, 
  if 
  p 
  be 
  expressed 
  in 
  

   terms 
  of 
  </>, 
  <£ 
  is 
  nearly 
  a 
  right 
  angle 
  for 
  points 
  of 
  the 
  

   periphery 
  AB 
  included 
  within 
  the 
  circle. 
  

  

  Let 
  

  

  </> 
  = 
  2 
  - 
  ^' 
  

  

  then 
  

  

  l-Psm 
  2 
  c/> 
  - 
  \J 
  k' 
  2 
  \ 
  k'^-k 
  2 
  ^ 
  nearly, 
  

  

  Denoting 
  the 
  angle 
  subtended 
  by 
  AB 
  at 
  the 
  centre 
  of 
  the 
  

   circular 
  image 
  by 
  4^ 
  b 
  we 
  obtain 
  by 
  the 
  formula 
  already 
  

  

  given 
  

  

  Expressing 
  k' 
  in 
  terms 
  of 
  a 
  and 
  8, 
  we 
  obtain 
  

  

  3 
  4a 
  (a 
  — 
  £)* 
  

  

  Practically, 
  a 
  is 
  very 
  large 
  compared 
  to 
  # 
  1; 
  and 
  t/tj 
  is 
  a 
  

   very 
  small 
  angle, 
  so 
  that 
  we 
  may 
  safely 
  neglect 
  I 
  — 
  and 
  put 
  

  

  2 
  _ 
  ^l 
  O 
  

  

  Yl 
  "~4a(a-S)' 
  

   and 
  for 
  points 
  on 
  the 
  periphery 
  AB 
  

  

  p 
  = 
  (2a-S) 
  V/c^+l^. 
  

   Putting 
  

  

  f 
  = 
  \ 
  / 
  4a{a 
  — 
  6)yjr 
  7 
  f 
  x 
  = 
  \/4a(a 
  — 
  8)^ 
  = 
  ct*! 
  2 
  — 
  8 
  2 
  nearly, 
  

   we 
  have 
  

  

  P 
  = 
  v^ 
  2 
  TF 
  ? 
  

  

  a 
  7 
  ^ 
  = 
  -df 
  = 
  

  

  \/4a(a 
  — 
  6) 
  

  

  Returning 
  to 
  formula 
  (II.) 
  we 
  find 
  the 
  intensity 
  at 
  due 
  

   to 
  the 
  sector 
  OAB, 
  

  

  L 
  

  

  = 
  ^f 
  a 
  (l-Jo 
  9 
  (p)-Ji 
  2 
  W)^ 
  

  

  TTJO 
  

  

  7T 
  7T 
  1 
  

  

  where 
  2a 
  is 
  the 
  angle 
  subtended 
  by 
  AB 
  at 
  0. 
  

  

  Phil. 
  Mag. 
  S. 
  5. 
  Vol. 
  45. 
  No. 
  272. 
  Jan. 
  1898. 
  G 
  

  

  