I How to calculate focal length?

AI Thread Summary
To calculate the focal length needed for an Epson 2040 projector to project an image at 8 meters instead of 4.5 meters, a magnification factor of 1.78 is required, which is not available as a direct option. The discussion highlights that while a 1.5x telephoto lens is available, it may not suffice for the desired image size. Users suggest that the lens equation can be adapted by switching object and image definitions, emphasizing the symmetry in lens optics. Ultimately, the consensus is that the best solution may be to reposition the projector at the correct distance, despite logistical challenges. The discussion clarifies that adapting a complete lens is not feasible for this application.
francus
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Calculate the focal length of telephoto lens in front of projector to decrease size projected on the screen
Hello, I am new to this forum and know very little about it. But perhaps somebody can help me.

I have an Epson 2040 projector. The beam it projects fits into the screen at a distance of about 4.5 meters. But logistically I need to put it at 8 meters from the screen, so the projected images are too large for the screen.

This website sells lens for this purpose with 1.5x telephoto lens.
https://navitar.com/products/projection/screenstar-conversion-lenses/standard-screenstar-lenses/

But how much is the magnification I need?
I understand (I may be wrong) that 8/4.5 = 1.78, is what I need, which is not available on that website.

So I imagine I can buy some other telephoto lens, provided it has a format large enough to take the beam of the projector. Am I correct?

Large format lens, such as Hasselblad may be appropriate, such as the following:

https://www.ebay.com/itm/3544447630...IxL5sTpZWOi6Z6OpqObxWGxQhu|tkp:Bk9SR7CZhK6jYQ

But how may I calculate the focal length I need?

I am sorry if I wrote something too much stupid. Be patient, I know very little.
 
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Many thanks for your reply, but there is one thing I do not understand: the above equation considers the standard case of a camera with a small sensor and a large object to be photographed and the lens in the middle. The light goes from the large object to the small screen.

Apparently my case seems different: on one side I have the small projector and on the other side the large screen and the light goes from the small projector to the large screen.

So I am confused how to use the above equation.
 
Light travels the exact same path forward and backward. When modeling lenses of course it matters which way tha light is going but the traces through th glass will be identical.
A parallel incoming beam will converge to the focus, and a point source at the focus will produce a parallel beam.
 
francus said:
Many thanks for your reply, but there is one thing I do not understand: the above equation considers the standard case of a camera with a small sensor and a large object to be photographed and the lens in the middle. The light goes from the large object to the small screen.

Apparently my case seems different: on one side I have the small projector and on the other side the large screen and the light goes from the small projector to the large screen.

So I am confused how to use the above equation.
You can just switch the object and image definitions. As @hutchphd said, everything in the lens equation is symmetric. It sounds to me like you really need to study lenses a bit more. Khan Academy has some really good tutorials on geometrical optics. Check those out.
 
I agree. For instance if you want to focus a point source back down to a point then both source and focal point will be distance 2f from lens (for a "thin lens").
The lens you need to project a 35mm photo slide onto a screen is the same lens you would use take a photograph onto 35 mm film. Of course this assumes the camera lens-film distance is the same as the projector.. But maybe the idea helps
 
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francus said:
So I am confused how to use the above equation.
The 1.5x lens is a "converter" lens to be used in conjunction with the projector. I think your equations are correct that you ideally want 1.78x.
The other lens you reference is a complete lens and not designed to be used in conjunction with the Epson projector lens. Can you call Epson and talk to someone?
 
hutchphd said:
The 1.5x lens is a "converter" lens to be used in conjunction with the projector. I think your equations are correct that you ideally want 1.78x.
Agreed. I think you have three options, in my decreasing order of preference:
  • Put the projector 4.5m from the screen. You will get a better image quality this way.
  • Look for a 1.75x adaptor.
  • Get the 1.5x adaptor and put the projector about 6.75m from the screen, or accept an image about 15% bigger.
 
Many thanks to all, I got a discomforting reply from Navitar: Their telephoto converter 1.5 costs $2700.00. So it is much more than the projector and out of contest.

From your replies, particularly
hutchphd said:
The other lens you reference is a complete lens and not designed to be used in conjunction with the Epson projector lens.

It seems I cannot adapt some used complete lens for this purpose.

So, finally there is no way to solve this problem with lens and the only way is to suspend the projector at the proper distance from the screen. Which is extremely complicated logistically. But since there is no other way I'll have to do just that.

But again many thanks for your help. At least now the situation is clear.
 
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