Re: [ATM] wedge in lenses

["vladimir sacek" <vla@toast.net>]

Peter Abrahams wrote:

> After the recent discussion on wedge in lenses, and the lack of 
> quantitative
> instructions on tolerances, I looked through many books and also found 
> almost
> nothing.

Mahajan gives it a shot (Optical imaging and aberrations I, p442). It is for 
a
single tilted refracting surface, and after neglecting small term related to 
pupil
magnification and the associated aberration coefficient (Eq. 7-16), both
additional coma and astigmatism caused by surface tilt are approximately
related to the original surface aberration as (t/a), with "t" being the tilt 
angle
and "a" the field angle.

In other words, surface tilt "t" will produce approximately as much of extra
coma and astigmatism, as it would result at a field angle "a" of a correctly
positioned surface.

This is still too general to be practically useful. In order to obtain error 
amounts,
one needs to know specific surface aberration contributions. This is, of 
course
doable, but it's rather cumbersome. Another option is to "standardize" a 
system
in respect to individual surface contributions, and extract appropriate
standardized wavefront aberrations caused by surface tilt.

For the Fraunhofer doublet I have at hand (100mm f/15 w/R1=911mm),
coma contribution for surfaces 1-4 is approximately 1:16:17:2, with the 
surface
1 and 3 having aberration of the same sign, opposite to that at surfaces 2 
and 4.
For the first surface, the p-v wavefront error for the refractive index 1.52 
is
approximated by Wc = (ad^3)/6.5R^2, with "a" being the field angle in 
radians,
"d" the aperture radius and "R" the surface r.o.c.

Coma introduced by this surface's tilt is approximated by replacing the 
field angle
with a tilt angle. For 0.2 degree tilt would result in approx. 1/6.8 wave 
p-v
wavefront error of center field coma (amounts to ~1/38 wave RMS wavefront 
error).

However, this same amount of tilt at the 2nd or 3rd surface, with ~16 times
stronger coma, would result in as much as ~1/2.4 wave RMS wavefront error.
For ~0.9 Strehl degradation factor, coma wavefront error would have to be a 
bit
better than 1/19 wave RMS here - which would require some 8 times smaller 
tilt -
about 0.025 degree.

Sensitivity of the fourth surface is about double of that of the first one, 
but still much
lower compared to the two in the middle.

Astigmatism p-v wavefront error at the first surface is approximated by
Wa = (a^2)(d^2)/1.7R. Again, replacing field angle "a" with surface tilt "t" 
gives the
approximate amount of astigmatism induced by it. For t=0.2 degrees, the p-v
wavefront error of astigmatism induced with the same f/15 lens is only ~1/28 
wave
(amounts to nearly 1/140 wave RMS). With the astigmatism at the remaining 
three
surfaces being approximately 1:4.9:5.3:2, this amount of tilt would cause 
about
1/26 wave RMS wavefront error in the worst case scenario (at the 3rd 
surface).
Note that this astigmatism changes with image height in the focal plane (not 
its square).

According to this, astigmatism is clearly less pronounced than coma, to the 
extent
that can be neglected when determining wedge tolerance (at least in this 
small, slow
system, that is).

In all, for a 100mm f/15 Fraunhofer doublet, tilt tolerance is determined by 
the inner
two surfaces' sensitivity. For 0.9 Strehl degradation factor, it requires 
somewhat less
than 0.025-degree wedge (not more than ~0.04mm, or 0.0017" linear). For an 
8"
f/15, the aberration (coma) doubles, so that angular wedge for the same 
wavefront
error halves, with the linear wedge remaining unchanged.

The tolerance likely lessens exponentially with faster systems (depends on 
both,
specific front surface r.o.c. and individual coma contributions at each 
surface).

This is an outline of what could be the way to get close enough values of 
wavefront errors
caused by a lens wedge. It seems to be coming close enough to those accepted 
general
tolerances, and to agree fairly well with the raytrace. May be good, but it 
sure
needs some check ups. Contrary to what I came up with in my previous post, 
it suggests
that it is wavefront aberrations induced by the wedge error that determine 
the level of
wedge tolerance, not the color separation. I'd go with this one, if for no 
other reason,
because it does consider all four surfaces (previously, I only looked at the 
first and last
one - the two inducing the least amount of wavefront aberrations).

Vlad








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