SPHERICAL ABERRATION FOR A SYMMETRICAL OPTICAL SYSTEM. 
47 
The point P x and singular inclination A can therefore be constructed geometrically as 
follows. 
With I 0 as centre and radius n () .GJ. 0 /n 2 describe a circle meeting the refracting 
surface at P 2 ; CJoPj is the angle A required. 
This angle A approaches zero, that is, we get a critical failure of convergency, when 
the two circles approach contact at A x . The limiting case is, therefore, when 
A]I 0 = n 0 . CJ. 0 /n 2 or I 0 divides AjCj externally in the ratio n 0 : n 2 . 
When this happens r 1 + x 0 = n 0 xjn 2 leading to M, = go, a case of true critical 
failure. But clearly, by symmetry, we get a precisely similar result when P1Q1 is 
due to a ray entering the surface at Q 1? and travelling backward through medium 2. 
In this case the limiting position of I 0 divides BjCj externally in the ratio n 0 : n 2 and 
is defined by r x —x 0 — —n 0 xjn 2 or Mi = 
This case would correspond, analytically, to a 2 = i r, and the corresponding equa¬ 
tion (33) would become 
—rjx o = [n 0 /n 2 ) v 7 { 1 — (-A sin X/rj) 2 } - v / { 1 - (n 0 x 0 sin A/w 2 r,) 2 }. 
Now if we examine (34) we find that in the process of clearing roots, r x /x 0 appears 
squared in the final result, which accordingly includes both a 2 — 0 and a 2 = ?r. If we 
write r x fx 0 2 = u, then we should really write equation (33) in the form 
a/u = {n 0 /n 2 ) v / (l — sin 2 \/u) — (1 — n 0 2 sin 2 \jn 2 u), .... (39) 
and the two cases are discriminated by assigning to u one or the other sign. One 
of these cases is necessarily irrelevant since refraction at the posterior surface of the 
sphere is physically excluded. 
Further, if we consider the other two values which make B = 0, viz., 
they correspond to 
and 
M x = ( n 0 + n 2 )/2n 0 and Mi = (n Q + n 2 )/2n 2 , 
n + £c 0 = —n 0 x 0 /n 2 
r x —x 0 = n 0 xjn 2 , 
i.e., to positions of I 0 in which it divides AiC x and BxCx internally in the ratio n 0 : n 2 . 
But these belong geometrically to the limit of cases in which the incident and 
refracted rays lie on opposite sides of the normal, i.e., to a negative refractive index. 
And indeed they are obtained from the two previous points by reversing the sign of 
n,j7i 2 . 
Here again, examination of (34) shows that njn 2 appears squared in it. Therefore 
(34) includes the cases in question. These, however, may be obtained analytically by 
changing \fr Q , or \// 2 , into its supplement, i.e., by reversing the sign of cos \/x 0 or cos A, 
or by changing the determination of the sign of one or other of the square roots on 
the right-hand side of (39). 
h 2 
