how to work out object-lens -image distance

[brighamr]

Hi

I would like to project onto a wall to get a very large macro image for paper negs but am having difficulty
with the math
I have held a lens up and got an image just by luck but now trying it again i can't get a big image
how do i work out the object to lens to wall ?

if i was using an actual view camera then i would just go for maximum extension then then
move the whole camera in to the subject with my head under a dark cloth but because of the scale involved
this is not possible

I plan on fixing the lens on a standard and rail for fine focus in a doorway then blacking out around it
then going out the front of my house and around to the back to get to the image
but i need to work out the basic distances to start with
i have a few long apo ronars that should be good for this but there is probably only a millimetre between sharp and totally out of focus ...
i know i will need a lot of light which is not a problem and would need time let my eyes to adjust in the dark side to see the image

robin

[ic-racer]

Are you making a camera obscura? If you can assume the subject is at infinity (like an outside scene) the lens to paper negative distance will equal the focal length marked on the lens. So if you have a 600mm lens, position it about 600mm from your paper negative. You will need some way to fine-tune the focus.

If you want a very large image then I'd recommend a pinhole. Another option is a 0.5 diopter close-up lens. That will have a focal length of 2000mm.

If you are doing a big magnification of say a flower projected onto a wall then you can use a lens with focal length equal to the diagonal of your subject size and approximate the distances with the thin lens equation. So, if your flower is 80mm across, and you want to fill the frame with the flower, you can use an 80mm lens to project it on the wall. A long lens will make a smaller projected image on the wall.

1/f = 1/p + 1/q

f=focal length of lens
p = subject distance
q = film distance

[brighamr]

Thanks ic racer

I plan on doing still life projected onto 4 sheets of 20x24 and if that works out well then i would look to getting a 50" roll

interesting that a longer lens would make a smaller image

I was basically thinking of making a large process camera like the ones that went thru a wall

thanks

robin

[ic-racer]

Here is a calculator for you: http://hyperphysics.phy-astr.gsu.edu...pt/lenseq.html

[polyglot]

A longer lens will give a larger image for a given lens/subject distance, but it requires more extension from the film to do so. You can see a nice diagram here.

What you care about (I think) is magnification and deciding where to put lens and subject to obtain a 20" image of a particular subject, say it's 5" across. 20"/5" means you want a magnification of 4. I'll assume you have a 300mm lens; obviously substitute your actual value into the calculations below...

Using the terms from that diagram (g=lens/subject distance, h=lens/film distance, G=subject size, H=image size):
G = 5" = 125mm, H = 20" = 500mm, f = 300mm

M = H/G = f/(g-f) = 4
300 = (g-300) * 4
g = 375mm

So you put your subject 375mm from the lens' front node.

Likewise,
H/G = (h-f)/f = 4
(h-300)/300 = 4
h = 1500

So you put your lens 1.5m from the paper.

Alternatively, use the 1/f = 1/g + 1/h formula that ic-racer posted here:

1/h = 1/f - 1/g
h = 1/(1/300 - 1/375)
h = 1500

Finally you need to compute the bellows factor to decide on your exposure:
F = (M+1)^2 = 25
So you expose for 25x longer than the metered light level would indicate (or use 25x as much flash power, or open the aperture by 4.7 stops). This extreme correction is required because your lens is 5x it's focal length from the film (huge extension required to obtain huge magnification), which means there is 5^2=25x light-loss due to inverse-square-law spreading of the light. Given the extreme compensation required here, don't forget reciprocity failure if it's relevant to your materials...

[ic-racer]

interesting that a longer lens would make a smaller image

What polyglot posted is, of course, correct but basically you are making a projector which has a fixed subject to film distance with a variable lens/subject distance (your focus wiggle room). I guess I'm one of the few that still shoots and projects B&W movies on the wall. When I want a bigger image on the wall I use the shorter focal length Switar on the projector. Works the same way when enlarging. Longer lenses make smaller images on the baseboard.

[brighamr]

thanks polyglot

will have ago with this on the weekend

robin

[polyglot]

What polyglot posted is, of course, correct but basically you are making a projector which has a fixed subject to film distance with a variable lens/subject distance (your focus wiggle room). I guess I'm one of the few that still shoots and projects B&W movies on the wall. When I want a bigger image on the wall I use the shorter focal length Switar on the projector. Works the same way when enlarging. Longer lenses make smaller images on the baseboard.

Yes, this is absolutely worth noting. Everything scales by similar triangles with the focal length, which means that if you use a shorter lens, everything will be closer together for the same magnification. That means that for a given amount of available room between paper and lens (i.e. your available extension), you can get higher magnifications with shorter lenses. It does mean that your subject will get very close to the lens though, which can make it difficult to light. Do keep in mind the tradeoff on bellows-factor though; using a very short lens to obtain a very large image means that you effectively have a crazy bellows-factor and could end up with an infeasibly dim image.

[ic-racer]

And to further expand on polyglot's points above, your diffracting aperture will not be the one marked on the barrel, it will that number times one plus the magnification. This can get to be a quite large number and ultimately will spoil your results.
For example f8 at 20x mag = aperture of 168. Your Airy disks will be about 8mm on the projected image; it will be a blurry image when viewed up close.

[Emmanuel BIGLER]

The same question was asked to the British LFUK forum.
But sometimes, difficult questions need a combined answer of both the New World and the Old World.