44 MR. T. Y. BAKER AND PROF. L. N. G. FILON : LONGITUDINAL 
where 
B = B + Mr 
= i(n 2 —n 0 )~ 2 {— (n 2 2 —n 0 n 2 +n 0 2 ) + (n 2 —n 0 ) 2 M 1 + S(n 2 2 —n 0 n2+n 0 2 )'SIL 1 2 —2n 0 n.M 1 3 }. (27) 
When, however, we come to develop a formula for tangents, similar to (21) for 
sines, it is found that, in the series for tan a 2 /tan a 0 , we do not have all the coefficients 
after the first vanishing together ; for, even at the aplanatic points, the tangent ratio 
is not constant. 
We can, indeed, write 
tan a 2 /tan a„ = M, -1 (l+ BM t -2 tan 2 a 0 )/(l+ CMu 2 tan 2 a 0 ), . . . (28) 
and choose the coefficients B and C so that the developments shall agree as far as terms 
in tan 4 a 0 inclusive. This can, in general, be done in one way only. But we then find 
that B and C are no longer integral functions of the magnification. Their infinities 
have to be taken into account, and generally the method becomes complicated and 
unsatisfactory. 
From other considerations, however, it appears that since the zeroes of a 2 must be the 
same as the poles of Ax 2 , the B in (28) must be the same as the B in (26), and this 
will fix C as follows :— 
tan 2 a 2 = 
sin a a 2 /(l — sin 2 a 2 ) 
Mr 2 sin 2 a u { 1 + B sin 2 a (1 /M, 2 } 2 
{ 1 + C sin 2 ajM^-Mr 2 sin 2 a 0 { I + B sin 2 a./M, 2 } 2 
Mi 2 tan 2 
(1 + tan 2 a 0 ) {1 + (C + M?) M? 2 
a,i{ 1 + (B + M 1 2 )Mr 2 tan 2 « 0 } 2 _ 
tan 2 a 0 } 2 —Mu 2 tan 2 a 0 {1 + (B + M/) Mj -2 tan 2 a 0 } 2 , 
substituting from (21). 
Hence, taking the square root and developing the denominator in powers of tan /3 2 , 
i.e., of Mj -1 tan a„, 
tan a 2 /tan & 2 
where 
_(1 + B tan 2 /3 2 )_ 
1 + tan 2 & (C + |M 1 2 -^) + 1 tan 4 ft {C (1 + M, 2 )+IM, 4 + fM 1 2 -i-2B} 
(l + B tan 2 /3 2 )/(l + C tan 2 /3 2 + D tan 4 /3 2 ),.(29) 
C - C+fM^-i 
D = C(l+M f ') + fM 1 4 + tM 1 2 -i-2B 
= C(l+M 1 2 ) + i(l-M 1 2 )(l + 3M 1 2 )-2B J 
(30) 
(29) is now correct as far as the second order inclusive. 
If we only require the tangent ratio correct up to the first order of aberrations, 
we have the formula 
tan u 2 = tan /3 2 (1 + B tan 2 /3 2 )/( 1 + C tan 2 fi 2 ) 
(31) 
