[MF3D.FORUM:1136] Re: Intro, Deviation and Viewers for a 43mm taking lens

["Michael K. Davis" <zilch0@xxxxxxxxxxxx>]

George,

I appreciate the reply!

At 02:40 AM 8/22/00 -0400, you wrote:
[snip]
>>Lastly, given the FL of a viewer's lenses, how does one calculate the
>>degree of magnification?  
>
>M = 250mm/FL
>
>Take care,
>
>George

Not knowing what to search for previously, I have now searched the archives
for the string 250mm/FL and found several articles.  It seems that this
formula holds true when the image has an apparent distance of Infinity, but
if the image appears to be at a distance of 10 inches, one is supposed to
increase the calculated magnification by a value of 1.  If the apparent
distance to the image is 20 inches, we're supposed to add 0.5, for 40
inches add 0.25, etc.

This progression follows 

   y = 10 * x^-1   

where x is the apparent distance to the image as seen through the loupe.

So, I came up with this:

   Magnification = (10 * x^-1) + (250mm / viewer FL)  

again, where x is the apparent distance to the image.

For x = 10 inches we get Magnification = 1 + 250/FL
For x = 20 inches we get Magnification = 0.5 + 250/FL
For x = Infinity we effectively get Magnification = 0 + 250/FL

OK, I have jumped from the frying pan where I didn't know how to calcuate
Magnification to the frying pan where I don't know how to measure the
"apparent distance to the image."   Joy!

I want to know the real magnification of a viewer as an aide to determining
how  liberal a maximum permissible Circle of Confusion diameter I can
select in calculating my DoF tables, using a spreadsheet I have that takes
image size and viewing distance into account - the goal being to limit my
Near and Far Sharp Circles of Confusion to a diameter equivalent to 4 lpmm
when viewed at a distance of 25cm - that's AFTER MAGNIFICATION.  I can
adjust the target lpmm after determining the ratio of this elusive
"apparent viewing distance" to the 25cm (10-inch) standard. 

So my next question is, how can I measure "the apparent viewing distance"
of the image?  -or- How can I measure the apparent distance to the "window"?

I was thinking that if I already had the true magnification for a viewer, I
could cut out a square of white posterboard that has been scaled to the
viewer magnification, then keep one eye in the viewer and another one on
the white square, moving it fore and aft until the card is in registration
with the 50x50 mounted image.   If I already knew the viewer magnification
was REALLY 4x, for example, I could cut out a 200x200mm square and then
measure the distance from my eye to this square when it is in registration
with the viewer image - that distance being measured from any equidistant
corner of the card to my eye, not from the center along a line
perpendicular to the card.  As it is, I need the apparent distance to
calculate the magnification...

Any ideas?

Thanks!

Mike