SPHERICAL ABERRATION FOR A SYMMETRICAL OPTICAL SYSTEM. 
53 
§ 8. Nature of the Quantities A, B, C, E in the General Case of any System. 
In the case of the single refracting surface we found that A, B, C were polynomials 
of degrees 4, 3, 2 in M, and that E was identically zero. 
In addition, for such a surface, equations (23), (25) and (27) show that the 
coefficient of M , 1 in A is nfn % times the coefficient of Mj 3 in B or B. This may be 
otherwise stated in the form :— 
Aj—noMjBj/^2) he., Aj —MjBj.(I) 
is of the fourth degree only in appearance and reduces to an expression of the third 
degree in M x or M a . 
The same holds good for A x — MjBj, so this result is independent of whether the 
sine or tangent is taken as argument. The same remark applies to all the results of 
the present section and to the other invariant relations shortly to he proved. We 
may therefore conveniently state it here once for all. 
If we now refer to the equations (47), (48), and remember that in any combination— 
M ls // ! =M 3 //„-l// 1 .(52) 
and 
l//iM ls = 1 //, s M,-1//,.(53) 
with the corresponding equations 
n u M r ,/fs = nM,/f I3 - njf .( 54 ) 
n Jf [ ®b.3 — >l "Jf 13 n Jfz . (55) 
and the obvious conditions 
M, :; = M,M 3 ; M 13 = MjMg, 
we note first that, if A 3 is a quartic in M :! it is also a quartic in M 13 or M 13 , and that 
if A 2 is a quartic in M l5 it becomes, on multiplication by M 3 2 M 3 2 , a quartic in M 13 
or M 13 , since M 3 = n.Mjn 4 , and therefore M 1 r M 3 2 M 3 2 = ni~M l JM : f r /n i \ which makes 
every term in AjM^M.- 2 a quartic in M 13 , because M 3 is a linear function of M 13 and 
4 —r is here zero or positive. 
(47) then shows that A 13 will be a quartic function of M 13 , if A x and A 3 are quartic 
functions of M, and M 3 respectively. But we know this to be the case for a single 
refracting surface. Hence it holds good of any system compounded of such surfaces. 
Now consider (48). If B 3 is a cubic in M 3 it becomes a cubic in m 13 . 
Again, if Aj — MjBj = a cubic Uj in M x 
M 3 2 (B, + A,M,/,//,) = M, 2 (B,{ 1 + M IS /,//,} + U.M, /,//,) 
= M -m ./: (B 1 //,3+U 1 // 3 ), 
VOL. CCXXI.-A. 
I 
