I know that for extension tubes, the magnification is tube length/lens focal length.
What's the equivalent formula for either:
1) reversing a lens on the body, using a reversing ring. For example, reverse-mounting a 50mm lens on my D200.
2) reversing a lens on the end of another lens? For example, reverse-mounting a 50mm lens on my 70-200.
For (2) it's FL of normal lens divided by FL of reversed lens, plus magnification of normal lens. For instance, reversing a 50mm lens on a 100mm macro lens set to 1:1, you get 100mm/50mm+1 = 3.0
I don't know offhand what the calculation is for (1).
For (2) it's FL of normal lens divided by FL of reversed lens, plus magnification of normal lens. For instance, reversing a 50mm lens on a 100mm macro lens set to 1:1, you get 100mm/50mm+1 = 3.0
I don't know offhand what the calculation is for (1).
Edit... geez get the numbers right kirbic :-P
Ah, thanks! So at 200mm, I'd beat 4:1. Now to find a miniature sun to provide the light. :)
ETA: *gasp* There's something technical about photography that kirbic doesn't know? My world is shaken!
Actually, since my 70-200 has a constant aperture of f/2.8, does that help with the light? How do ya figger out light loss in stops, anyway? I know teleconverters cost you 1 stop of light for every 1.4x magnification.
Yeah there is a shot on my website (the 2 plugs if you care to look in my gallery on my website) was metered at 1/125 at f64 So yeah... I am thinking that Thermonucular lighting may be called for .
So the shot was done at f22 reversed ... so what is that 3.5 stops
Actually, since my 70-200 has a constant aperture of f/2.8, does that help with the light? How do ya figger out light loss in stops, anyway? I know teleconverters cost you 1 stop of light for every 1.4x magnification.
If the clear aperture of the reversed lens is large enough not to artificially "stop down" the normally-mounted lens, there is no light loss (well, only internal reflection due to more glass surfaces, so a small fraction of a stop).
