﻿of 
  correcting 
  Telescopic 
  Objectives. 
  479 
  

  

  Problem 
  (4). 
  To 
  correct 
  simultaneously 
  for 
  spherical 
  aber- 
  

   ration 
  and 
  for 
  sine-error; 
  in 
  other 
  -words, 
  to 
  find 
  the 
  two 
  

   " 
  aplanatic 
  points. 
  - 
  " 
  The 
  position 
  of 
  these 
  is 
  well-known, 
  so 
  

   that 
  the 
  exercise 
  affords 
  a 
  useful 
  check 
  on 
  the 
  foregoing 
  

   formulae. 
  Equate 
  the 
  expressions 
  (1) 
  and 
  (2) 
  to 
  zero, 
  

   and 
  solve 
  simultaneously, 
  with 
  P 
  = 
  4*1603, 
  Q=— 
  6*6683, 
  

   S= 
  — 
  18-4354, 
  and 
  # 
  = 
  0; 
  the 
  solutions 
  are 
  ^=4*8426, 
  

   «i= 
  12-1932, 
  c 
  2 
  = 
  7-3506, 
  or 
  u 
  x 
  =— 
  7*3506, 
  ^=-7*3506, 
  

   c 
  2 
  = 
  — 
  12*1932. 
  These 
  tally 
  with 
  the 
  familiar 
  solutions 
  

   w 
  2 
  ; 
  =c 
  2 
  =jt? 
  1 
  /(N 
  — 
  1); 
  Ui 
  — 
  Ci=— 
  pi/(N 
  — 
  1). 
  In 
  one 
  case 
  the 
  

   rays 
  enter 
  the 
  first 
  face 
  normally, 
  in 
  the 
  other 
  they 
  leave 
  

   the 
  second 
  face 
  normally. 
  The 
  cases 
  are 
  both 
  useless, 
  so 
  far 
  

   as 
  telescopic 
  work 
  is 
  concerned. 
  

  

  I 
  take 
  next 
  a 
  few 
  examples 
  on 
  the 
  doublet. 
  The 
  coefficients 
  

   are 
  now 
  :— 
  A 
  = 
  1*22153, 
  B 
  = 
  2*22153, 
  = 
  -1*68475, 
  

   D 
  = 
  -3'44306, 
  E=— 
  3*36951, 
  F=4*8775i, 
  Gr= 
  -3*73215, 
  

   H= 
  4-83548, 
  K 
  = 
  20 
  77590, 
  L 
  = 
  1*81770, 
  P 
  = 
  1*72153, 
  

   Q=-2*72153, 
  R=-2*43875, 
  S 
  = 
  -2*66771. 
  

  

  The 
  latus 
  rectum 
  of 
  all 
  parabolas 
  of 
  class 
  I 
  is 
  1/(1*22153); 
  

   II, 
  -1/(2*10720;; 
  III, 
  -1/(0*103042); 
  LV, 
  -1/^0*099481) 
  * 
  

   V, 
  -1/(1-19840). 
  

  

  Problem 
  (5). 
  With 
  parallel 
  light 
  and 
  a 
  cemented 
  lens, 
  

   what 
  carves 
  give 
  (algebraic-ally) 
  least 
  aberration 
  ? 
  Solve 
  

   (3) 
  with 
  ?/ 
  1 
  = 
  0, 
  ?=0 
  :<*!= 
  1-5277, 
  c 
  2 
  = 
  c 
  3 
  = 
  -3*3149, 
  c 
  4 
  =c 
  3 
  

   + 
  2*4315 
  = 
  0-8834; 
  the 
  aberration 
  is 
  -1*0330. 
  

  

  Problem 
  (6). 
  Given 
  1^ 
  = 
  and 
  # 
  = 
  0, 
  find 
  the 
  aberration 
  

   for 
  c 
  2 
  = 
  — 
  4*5. 
  Parabola 
  (I) 
  gives 
  : 
  excess 
  aberration 
  above 
  

   — 
  1*0330 
  = 
  1*22153 
  x 
  (square 
  of 
  excess 
  curvature 
  above 
  

   — 
  3*3149), 
  so 
  that 
  the 
  aberration 
  -=— 
  1*0330 
  + 
  1*7155 
  

   = 
  0*6824. 
  

  

  Problem 
  (7). 
  Given 
  i^ 
  = 
  and 
  g 
  = 
  0, 
  what 
  curves 
  will 
  

   remove 
  spherical 
  aberration 
  ? 
  We 
  must 
  have 
  1*0330 
  = 
  

   l-22153(c 
  1 
  -l-5277) 
  2 
  , 
  or 
  Cl 
  = 
  0-60806 
  or 
  2*4473. 
  Then, 
  as 
  

   usual, 
  deduct 
  4*8426 
  for 
  c 
  2 
  , 
  and 
  add 
  2*4315 
  to 
  c 
  3 
  for 
  c 
  4 
  . 
  

  

  Problem 
  (8). 
  Repeat 
  (5), 
  with 
  # 
  = 
  0*1. 
  The 
  vertex 
  of 
  the 
  

   parabola 
  is 
  now 
  given 
  by 
  c 
  1 
  = 
  1*6656 
  instead 
  of 
  1*5277; 
  

   indeed, 
  every 
  0*1 
  in 
  y 
  means 
  0*1379 
  extra 
  in 
  c 
  Y 
  ; 
  so, 
  an 
  axial 
  

   gap 
  (g= 
  —0*1) 
  would 
  mean 
  c 
  1 
  = 
  l*3898. 
  With 
  wi=0, 
  # 
  = 
  0*1, 
  

   and 
  (.'! 
  = 
  1*6656, 
  the 
  aberration 
  is 
  0*4897. 
  As 
  this 
  is 
  positive, 
  

   no 
  choice 
  of 
  curves 
  with 
  air-gap 
  0*1 
  (and 
  parallel 
  rays) 
  can 
  

   give 
  freedom 
  from 
  aberration. 
  Yet 
  for 
  a 
  suitable 
  value 
  of 
  

   i«j 
  this 
  could 
  be 
  done. 
  

  

  Problem 
  (9). 
  Over 
  what 
  range 
  of 
  values 
  of 
  u^ 
  could 
  it 
  be 
  

   done? 
  To 
  answer 
  tins, 
  we 
  require 
  parabola 
  III. 
  Its 
  latus 
  

   rectum 
  has 
  been 
  given 
  above; 
  its 
  vertex 
  is 
  found 
  by 
  solving 
  (3) 
  

   and 
  (5) 
  with 
  ^ 
  = 
  0*1; 
  namely, 
  ^ 
  = 
  0*24933, 
  aberration 
  = 
  0'6965. 
  

  

  2 
  L 
  2 
  

  

  