J. Eur. Opt. Society-Rapid Publ. 22, 30( 2026) 301
tion( see Fig. 1). For each lens m, withm = 1... L, thefirst surface has index k = 2m�1 and the second surface has k = 2m. Outside of the lenses, the refractive indices are n 2m�1 ¼ 1, and inside the lenses we have n 2m ¼ n. Themarginal ray angles before the first and after the last lens of the thin lens group will be denoted by u 1 ¼ a and u 2Lþ1 ¼ b respectively.
2.2 Derivation of the simple spherical aberration expression
Readers primarily interested in the results may skip directly to Section 2.3.
To obtain S expressed entirely in terms of the angles u k, we write A k as
A k ¼ u kþ1 � u k n �1 kþ1
� n�1 k
: ð4Þ
If equation( 3) is used to eliminate u kþ1 in equation( 4), then after simple algebra the refraction invariant A k becomes the one given by the more familiar equation( 2).
To shorten the formulas, consider first only spherical surfaces, i. e. we have G k = 0 for all surfaces( the G k terms will be included later). Because in the thin-lens approximation all distances between the surfaces shown in Figure 1 are considered to be zero, the height of the marginal ray does not change inside this group and we have h k ¼ h ¼ const for all k = 1... 2L. For thin lenses in contact, the contributions S k for odd and even surface numbers result from equations( 1) and( 4) after simple algebra as
S 2m�1 ¼ tu ð 2m�1 � u 2m S 2m ¼ tu ð 2m � u 2mþ1 t ¼
Þ 2 ðnu 2m�1 � u 2m Þ Þ 2 ðu 2m � nu 2mþ1 Þ
hn: ðn�1Þ 2
These two surface contributions can be combined as S k ¼ ð�1
Þ k�1 t ðu~ k � u 2m ð5Þ
Þ 2 ðn u~ k � u 2m Þ; ð6Þ
where for odd-index angles we introduce the notation(
u~ k ¼ u 2m�1 for k ¼ 2m � 1 u 2mþ1 for k ¼ 2m ð7Þ
To derive the new spherical aberration formula for thin lenses in contact, we start by observing that the surface contributions S k in equation( 6) contain an almost perfect cube of the angle difference ~ u k � u 2m, the obstacle being the refractive index appearing in the last bracket. As a first step, we show below that the surface contributions can be written as a perfect cube plus correction terms that give the departure from the cube, such that most of the correction terms cancel each other out during summation over surfaces. To facilitate the construction of these expressions, we also introduce temporary variables l k, that in air are equal to the corresponding angle u k, and inside the lens differ from the angle by a factor q that needs to be determined,
l 2m�1 ¼ u 2m�1; l 2m ¼ qu 2m: ð8Þ
We see from equations( 6) and( 7) that if we expand S k we obtain four terms of total power 3 in the angles u, e. g. for the first surface the result will contain terms corresponding to u 3
1; u2 1 u 2; u 1 u 2 2; u3 2
. In the new variables given by equation( 8) S k will also contain four terms, with coefficients c S1; c S2; c S3; c S4 that need to be determined,
S k ¼ ð�1Þ k�1 c S1 ðl 2m � l~ k Þ 3 þ c S2 l~ 3 k þ c S3l 3
2m þ c S4 l~ 2 k l
2m; ð9Þ
where for odd indices l~ k is defined in the same way as u~ k in equation( 7). Because the goal of this approach is to construct an expression containing a perfect cube, we use the perfect cube in equation( 9) instead of the term l 2
2m l~ k. Consider first the odd surfaces, for which we have ð�1Þ 2m�2 ¼ 1.
By substituting equations( 8) and( 7) into equation( 9) we obtain an expression for S 2m�1 in terms of the angles u that must be equal to that of S 2m�1 in the first of equation( 5). By subtracting the two equivalent expressions for S 2m�1, and by using e. g. m = 1, we obtain after elementary algebra the zero polynomial
0 ¼ tu ð 1 � u 2 Þ 2 ðnu 1 � u 2 Þ�c S1 ðqu 2 � u 1 Þ 3 � c S2 u 3
1
� c S3 q 3 u 3
2 � c S4qu 2 u 2 1
¼ u 3
1 c �
ð S1 � c S2 þ tnÞ�u 3
2 q3 ðc S1 þ c S3 Þþt
� u 2 u 2
1ð
q ð 3c S1 þ c S4 Þþð2n þ 1ÞtÞ þ u 2
2 u �
1 3c S1 q 2 þ ðn þ 2Þt ð10Þ
By annulling the coefficients of the four terms of total power 3 in the angles u we obtain four equations with five unknowns. We can freely choose one of these unknowns, which are the four coefficients c S and q. To simplify the construction of the new expression in equation( 9), wechoose c S4 ¼ 0. After substituting for t the value given in equation( 5) we obtain the system of equations
c S1 � c S2 þ hn2 ¼ 0 ðn�1Þ 2 hn 2nþ1
3c S1 q þ ð Þ ¼ 0 ðn�1Þ 2 ð11Þ 3c S1 q 2 hn nþ2 þ ð Þ ¼ 0 ðn�1Þ 2 q 3 ðc S1 þ c S3 Þþ hn ¼ 0 ðn�1Þ 2
The 2nd and 3rd equation give, after moving one term to the other side, followed by division
q ¼ n þ 2 2n þ 1: ð12Þ
Note that if we set n = 1inequation( 12) we obtain q = 1. Because we have n 2m�1 ¼ 1, and n 2m ¼ n, equation( 8) can also be written as l k ¼ ðn k þ 2Þu k = ð2n k þ 1Þ.
From the 2nd and first equation in equation( 11) we obtain immediately hn 2nþ1 cS1 ¼� ð Þ2 3ðn�1Þ 2 ðnþ2Þ c S2 ¼� hn: ð13Þ 3ðnþ2Þ
The coefficient c S3 results from the last of equation( 11) but, as will be seen below, it is not important.