:i-2 



ON ABERRATION AT A CURVED SURFACE. 



solid, of which AC is a section in the plane of refraction, 

 QCy being its axis. The refractive index = m. 



Here QC is given. 



Also CD and DA co-ordinates of the point A. 



Hence, also, the normal AE and subnormal DE may be 

 found. 



Let kq be the refracted ray required, cutting Q,Cg in 

 Now sin inc. : sin E : : QE : QA. 

 sin E : sin refr. : : qA. : q~E. 



.-. sin inc. : sin refr. : : m : 1 : : 4-., : — -. 

 qh QE 



. . *=, = — q,,- = c (a known quantity). 

 .-. qk*=c t .qE?. 

 i.e. qW + EA 2 + 2?E . ED = c* . qW. 



.-. (c 2 — 1) q W — 2ED . qE = EA 2 . 



_, a 2. ED EA 2 



? E ~ 72 7 • ? E = 



c-— 1 

 ED 



c" — 1 

 ED 2 

 (C - l) 2 



(c 2 — 1)EA* 



.■J «E ■— - 



L ' c 2 — 1 J (c 2 — l) 2 (c 2 — 1)= 



•'• ? E = jztj { ED ± ^ { E D ' J + (6>s ~ 1)E A! )" QEJ - 



Let AC represent a section of a spherical surface (rad = EC) . 

 The lines CD, AD, DE are given in the tallies. Whence 

 find QA and QE ; Let m = |. 



Ex. 1. Let AC = 40° AD = -64279 (by tables.) 



QC = 2 inches .-.AD 2 = -4131789841 

 CE = 1 inch ED = -70004 (bj tables.) 



QE3 inches .-. QA = 2-324598 



.-. c m 1-102299 



.-. c 2 = 135093S9654U1 

 1*73442 



.•, »E ! 1 m 4-94220 inches 



1 350928905401 



.-. qC = 5 94220 inches. 



