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Re: FL adjustment for myopics
- From: P3D John Bercovitz <bercov@xxxxxxxxxxx>
- Subject: Re: FL adjustment for myopics
- Date: Thu, 6 Feb 97 11:33:39 PST
Paul Talbot asked about focussing and I said I'd done that
calculation before on the P3D list. Paul found the calculation I
was referring to but it was the wrong calculation: I'd forgotten
what I'd done. So here's the calc Paul wants.
What you should try to do when you focus is to adjust in until the
the thing is focussed and then rack it out about as far as you can
and still have it in focus. This brings the eye to a relaxed
state, approximately, because the eye will relax near its farthest
focus.
If your glasses Rx is -6, then your far focus is at 1/6 meter. So
you want the transparency to appear to be at a distance of 1/6
meter even though it is really at a distance from the lens which
is very roughly equal to the lens' focal length.
We could do the calculation using either the Gaussian or the
Newtonian form of the equation. They are really the same
equation. The Gaussain form gives you distances from the lens and
the Newtonian form gives you distances from the foci of the lens.
So the latter gives you lens shifts directly. However, you're
probably more familiar with the Gaussian form, and it also gives
the operating or effective focal length, so let's go with that:
1 1 1
- = - + -
f s s'
f is the focal length of the lens. It's around 44? mm for the RB
s is the distance from the lens to the transparency, normally
equal to f to make s' infinity.
s' is the distance the transparency appears to be at from a
focussing standpoint.
Since s is the distance to the real object and it is on one side
of the lens, and s' is on the same side of the lens, s' is the
distance to a virtual image and is a negative value.
Rearranging:
1 1
s = [ - - - ] ^-1
f s'
s = [ .0227 - Rx/1000 ] ^-1
Plug & crank time:
Rx s shift
-1 42.1 1.9
-2 40.4 3.6
-3 38.9 5.1
-4 37.4 6.6
-5 36.1 7.9
-6 34.8 9.2
John B
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