The distance F′2F′ or x′2 is similarly determined by considering F′1 to be represented by the second system in F′.
We have
| x′2 = | f2f′2 | = − | f2f′2 | = | nr22 | , |
| x2 | Δ | (n − 1)R |
so that
| sF′ = x′2 − f′2 = | r2 (nr2 − R) | , |
| (n − 1)R |
where sF′ denotes the distance of the second principal focus from the vertex S2.
The two focal lengths and the distances of the foci from the vertices being known, the positions of the remaining cardinal points, i.e. the principal points H and H′, are readily determined. Let sH = S1H, i.e. the distance of the object-side principal point from the vertex of the first surface, and sH′ = S2H′, i.e. the distance of the image-side principal point from the vertex of the second surface, then f = FH = FS1 + S1H = −S1F + S1H = −sF + sH; hence sH = sF + f = −dr1/R. Similarly sH′ = sF′ + f′ = −dr2/R. It is readily seen that the distances sH and sH′ are in the ratio of the radii r1 and r2.
The distance between the two principal planes (the interstitium) is deduced very simply. We have S1S2 = S1H + HH′ + H′S2, or HH′ = S1S2 − S1H + S2H′. Substituting, we have
HH′ = d − sH + sH′ = d(n − 1) (r2 − r1 + d)/R.
The interstitium becomes zero, or the two principal planes coincide, if d = r1 − r2.