42 
MR. T. Y. BAKER AND PROF. L. N. G. FILON : LONGITUDINAL 
If M, = La u /a 2 , which we shall call the ray magnification , we have 
M, = nfi 2 lnfi Q = n 2 x 2 /n Q x 0 . (14) 
The transverse magnification Mj is given by 
M, = i 2 /io = x 2 /x 0 = nJNLjn* .(15) 
We can also express (13) in another well-known form, namely, 
n 0 /x 2 -n 2 /x o = l [f], . 
whence, using (14), we obtain 
^0 = n 2 f\ (l 
x 2 = w 0 /i (l M x ) ) 
Again, from (9), (8) and (10) 
X 2 = 2 ? 2 /sin a 2 = n 0 x 0 sin a ( Jn 2 sin a 2 = n () f l (l — M,) sin a./M, sin a 2 
= x 2 sin a,,/M, sin a 2 ,. 
and the longitudinal aberration 
(16) 
(17) 
(18) 
Acc 2 = X 2 — 2 f 2 = a; 2 (sin a 0 /M, sin a 2 —l).(19) 
The corresponding longitudinal aberration in the equivalent Gaussian system is 
found from 
A $2 = A x 2 /n 2 - {n 0 fjn 2 ) (l -M,) (sin a n /M, sin a 2 -l) .(20) 
Now from (ll) 
sin a 2 = sin a 0 cos Go cos i/r 2 + sin \/r u cos a u cos fi 2 —sin \Js 2 cos a 0 cos Go 
+ sin a 0 sin Go sin \fs 2 , 
whence, using (8), (9) and (10), 
sin a 2 /sin a 0 = {1 — {xjrfi sin 2 a 0 }® {1 — (n^xjn^fi 2 sin 2 a 0 }* 
+ (*oM) { 1 -sin 2 aoP { l-{n 0 xfii 2 rfi sin 2 a 0 p 
— {n 0 x 0 /n 2 r x ) {1 -sin 2 a 0 } 4 {1 — (a 0 /n) 2 sin 2 a 0 } 4 
+ (n 0 x 0 2 /n 2 r 1 2 )sm 2 a 0 , 
and developing this in ascending powers of sin a 0 , we obtain, retaining only terms of 
fourth degree 
sin « 2 /sin a 0 = 1 +x 0 (n 2 —n 0 )/n 2 r i—^-P sin 2 a 0 
—iP sin 4 a 0 {{n 2 + n 0 ) 2 x 2 + n 2 (n-a: 0 ) ( n 2 r^+n 0 x 0 ))ln 2 r x 2 , 
where 
P = (1 -n 0 /n 2 ) (x 0 /ri) (l+xjn) (1 -n^xjn^i). 
