[ATM] Houghton corrector radii
vladimir sacek
vla at toast.net
Tue Sep 19 00:36:34 JST 2006
Thought it would serve purpose to post equations for
needed lens element radii of curvature for the Houghton
corrector in Newtonian configuration. The primary is
spherical. First comes an aplantic (third-order) form,
and then a plano-lens form that cancels spherical
aberration and significantly reduces coma of the primary mirror.
The aplantic corrector consists from a bi-convex (positive)
front lens followed by bi-concave (negative) rear lens,
the two being in near-contact. The radii of the first/third
and second/fourth surface are nominally identical and,
for the glass refrative index "n", given by:
R1 = 2(n-1)f/(1+q)
R2 = 2(n-1)f/(1-q)
q= [n(n-1)(f/R)^3]/(n+1)
f = -2(1-s)R
where
R1 is the absolute value of the first/third surface radius
(positive and negative, respectively),
R2 is the absolute value for the second/fourth surface radius
(negative and positive, respectively),
"f" is the absolute value of the focal length for the two lens elements, and
"s" is the corrector to mirror separation in units of the mirror radius of curvature R
(which is of negative sign).
Note that "q" is the lens shape factor, also given by (R2+R1)/(R2-R1);
it is normally negative, which in the above relation for "q" results from the (f/R)^3 factor.
The formulae are somewhat simpler than those obtained from
the Rutten/Vennroij book (given on Rick Scott's site).
The plano-lens corrector form consists from the plano-convex
front lens followed by (near-contact) plano-concave rear lens, both facing
incoming light with their curved side. The two radii are of
identical absolute value (front positive, rear negative), given by:
R1,3 = R(n-1)[(n+1)/n(n-1)]^(1/3)
with "R" being the mirror ROC, and "n" the glass refractive. This corrector
form leaves some residual coma, which is in a typical configuration about
1/6 of the mirror coma (about half compared to a Mak-Newt, and about
a third of the Schmidt-Newtonian coma). Astigmatism is negligible in all three.
Vlad
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