What's the Φ mean?

ScottE wrote:
You mean if I want a 1:1 print of Neil Armstrong's footprint on the moon I just have to use my 28-135 lens and take a snapshot of the full moon then adjust the print magnification? Do you know how big a piece of paper I will need for a 1:1 print of the Rover on Mars? (You did say any original image size.) ;<

Yes. You didn't say anything about resolution, did you?

sorry if I'm wrong here, but the above only applies I think to a simple (single element) lens. For a multi-element lens I think you have to add the distance between the front and rear nodal points to the 4 x focal length.

Mostly, for situations like finding the rotation axis for panoramas, etc, you are only interested in the rear nodal point, but there are two.

Please correct me if I'm wrong.

PhotosGuy wrote:
Apologies for the hijack, Dew!

No problem - glad it stimulated so much discussion!

>>but the above only applies I think to a simple (single element) lens. For a multi-element lens I think you have to add the distance between the front and rear nodal points to the 4 x focal length.<<

Not an expert in optics, but I was under the impression that the 4X rule applied even if compound lens.

These rules of thumb are seldom precise in the application. After all, how many 50mm lense are TRULY 50mm, and not within a couple of mm of actual focal length? So computing extension is a bit of 'sort of' especially when extension tubes not from the original manufacturer of the lens are used. I would guess the 4X may be yet another one of those 'sort of' rules!

For a simple lens, if f=focal length, v=object distance, b=image distance, you get:
1/f=1/v + 1/b

Now put v=b (must be symmetrical for 1:1 magnification) and you get v=2f. So v+b=4f.

Not exact for compound lens, but for a macro, pretty close. I think you'd find, with a 100mm macro, you'd have about 425mm from the film place to the subject. That's the bit I'm not quite sure about.