Lens for a room camera obscura

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Ive been toying with the idea of backing out a bedroom and trying to project an image on the back wall. I have been trying to find a formula that will calculate lens size / strength but have failed. If the distance between the window on which i want to place the lens and the rear wall is 3m. the subject (back garden) has elements that could be the subject anywhere from approx 10m to infinity.
What formula will give the necessary lens diameter and strength?
 
If you use a pinhole, there are at least three different formulae you can use.

D = √(3.6vλ) where D is the diameter, v is the pinhole to image distance and λ is the wavelength of
the light used. Other values are suggested in the literature, both higher and lower. For example, D =
2√(2vλ) or D = √(2vλ). Sidney Ray in Applied Photographic Optics gives a simpler to use empirical
formula, D = k√v where k is a constant of around 0.03-0.04 and D and v are expressed in mm.

For a lens, the normal equation relating focal length to object and subject distance is

1/f = 1/u + 1/v

If you buy a close up lens, the power in dioptres is the reciprocal of the focal length in metres, so a No 1 close up lens has a focal length of 1 metre, a No 2 1/2 metre etc.

And if you Google "abelardo morell camera obscura" you should probably turn up at least three videos on using a room as a camera obscura - but note well the exposure times!
 
Last edited:
Box Brownie said:
Not sure that you even need a lens as such???

Projects here:-
https://petapixel.com/2014/05/12/diy-tutorial-convert-room-camera-obscura/

https://www.photopedagogy.com/camera-obscura.html

HTH :)
Click to expand...

When I was a kid i was able to project an image of what was going on outside onto a wall of my bedroom simply by drawing the heavy velvet curtains together until there was only a small triangular chink left at the top

It was only when I visited the Camera Oscura in Dumfries when I was about ten that that I realised what I had done.
 
Box Brownie said:
Not sure that you even need a lens as such???

Projects here:-
https://petapixel.com/2014/05/12/diy-tutorial-convert-room-camera-obscura/

https://www.photopedagogy.com/camera-obscura.html

HTH :)
Click to expand...

The reason I wanted to find the formulas is to (relatively...) accurately focus on the east wall of the room without a lot of trial and error changing the aperture sizes.

Could it be as simple as mounting a correctly calculated lens on some firm board, taping to the window and then blacking out the remainder of the window and the doorway?
 
The image on the wall will be upside down, thats going to make viewing uncomfortable at best.
 
How does the following calculation look:
Variable
Object distance 18
Focal length 3
Image distance 3.6
Lens (dioptres) 0.3333333333
 
What units are the distances and focal length?
 
Scrub that; the focal length is 3 metres. So, does 1/3 = 1/18 + 1/3.6 becomes the question.

Lacking either calculator or paper, that becomes

6/18 = 1/18 + 1/3.6

5/18 = 1/3.6

1/3.6 = 1/3.6

So, yes, it seems correct.
 
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