Re: 10x Magnifiers for Stereo Viewer

[bercov@xxxxxxxxxxx (John Bercovitz)]

Stephen Kearney writes:

> This is something I still don't understand. How does one determine how
> large the lens diameter must be? 

It's really a ray tracing problem.  First hint is that in tracing oculars, 
one traces backwards as if the light came from the eye and landed on the 
transparency.
                                  .|
                      .            |
              . |\                 |
    _     .     | \                |
 /    \         | |                |
|      |        | /                |
\      /        |/                 |
  ---
  eye           lens         transparency

 ->|13mm|<-20 mm->|<-----50 mm---->|

The dots attempt to illustrate the path of a ray of light from an eye
turned to look at the corner of the transparency.  The ray leaves the
margin (edge) of the pupil of the eye and goes to the margin of the lens.
Then it is refracted slightly and hits the transparency.  That refraction
is calculated by the normal method.  If the ray doesn't reach the corner 
of the transparency (half the diagonal of the transparency away from the 
axis of the optical system), the lens isn't large enough in diameter.
Please make the distance from the anterior of the eye to the transparency
20 mm so we who wear eyeglasses can see too.  Thanks.  8-)  Figure the
pupil of the eye at 6 mm diameter so from the eye's axis to the margin of 
the pupil is 3mm.

> What is a a more realistic field of view figure for typical lenses?

Oh, 35 or 40 degrees for a fairly simple lens such as a cemented doublet.  
After that, things get expensive.   A planoconvex will be somewhat less 
due to chromatic aberration.  If the lens is not viewed through its center, 
then less yet.

John B

PS: Sorry about the crummy ASCII.  Things don't look too well aligned.


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