Perhaps it would help if I solved this backwards to determine what f is.
Knowing that a 7mm extension results in an object distance (S1) of 37 inches, I
can substitute:
S1 = 37" = 939.8mm
S2 = f + 7mm
which results in a quadratic equation to solve for f.
>
> This is a bit confusing for now as my knowledge of optics is
>insufficient to comprehend the meanings of f, S1, and S2. References
>23 and 24 on that Wikipedia page will help, but I need to read those
>carefully.
>
>
>>
>>Go here <https://en.wikipedia.org/wiki/Lens_%28optics%29> and scroll
>>down about 40% of the page and see the illustration for the "thin lens
>>formula" under "Imaging properties".
>>
>>In the illustration, if we assume that the lens is focused at infinity
>>an image would be formed at the focal length f. If you impose an
>>extension tube the distance to the image is now S2. So you must use the
>>thin lens formula to solve for the image distance S1.
>>
>>1/S1 = 1/f - 1/S2
>>
>>If we plug some numbers: f= 50mm, extension = 10mm, S2 = 60mm we get:
>>1/f = 0.02
>>1/S2 = 0.0167
>>S1 = 1/0.0033
>>S1 = approx 303mm
>>
>>
>>This will be an approximation since your real lens is not a thin lens of
>>negligible thickness and, of course, the object distance S1 is measured
>>from the optical center of the lens which is not precisely known to you.
>>
>
>
>Chris
>
>When the going gets weird, the weird turn pro
> - Hunter S. Thompson
>--
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>
Chris
When the going gets weird, the weird turn pro
- Hunter S. Thompson
--
_________________________________________________________________
Options: http://lists.thomasclausen.net/mailman/listinfo/olympus
Archives: http://lists.thomasclausen.net/mailman/private/olympus/
Themed Olympus Photo Exhibition: http://www.tope.nl/
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