What will a pinhole behind a pinhole do?
Can anyone tell me what will happen to the light rays if we setup a pinhole system like the following, will it simply crop the image?
Also what about a thick pinhole? Is the following correct, and will only let a small field of view in and project only a small area limited to the thickness of the pinhole?
What I wanted to do is project a very small spot from something like an LED light in a dark chamber into a very small spot, say ~2 microns, without resorting to an expensive or complicated optical setup, not the whole light, even just a portion of it is fine.
So I thought one of these two approaches, might be able to 'crop' the projected image down to ~2 microns.
It doesn't take a physicist to know you just invented a square changing bag.
In case the new aperture would be smaller than the original one, it then would act as the actual pinhole. But then just for that tiny angle of view offered by original pinhole, which now would be the vignette.
The exact description is you will get a grey or white spot in the centre of your field of view...
If you erase the red lines in your drawings, you'll see exactly what you'll get: a very small blue line representing a very dimm ray of light and no image at all.
Imaging yourself looking through a tube with a diameter of 1/2 inch and 20 feet long. No angle of view what so ever remains ....
Maybe a faint light at the end of the tunnel, if the light source is strong enough (and no!!! don't ever look into the sun or a laser!!)
"Have fun and catch that light beam!"
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No way. In a regular pinhole camera, diffraction sets a minimum spot size on film of order a few 100s microns; same here. Plus two holes in succession will severely limit the light throughput.
What I wanted to do is project a very small spot from something like an LED light in a dark chamber into a very small spot, say ~2 microns
To produce a 2 micron spot on film one needs a lens f/4 or faster, free from aberrations at the level of the diffraction limit; a far cry from a pair of pinholes.
For a thin aperture and imaging, well an entire image with off-axis rays, I agree with you.
For on-axis rays, with the elimination of oblique rays in a thick aperture, and forming of a non-image spot, I am questioning it.
I understand what diffraction is, the most common example is a rippple simulation, waves, water for example, hitting an aperture and expanding outwards.
Make that a long cylinder, like a sewer perhaps, and they would bounce off the wall, interefere with itself and propagate forwards, and expand outwards (diffract) as it leaves.
The problem here is that, this analogy does not account for absorbance. The material itself isn't reflective, refractive or diffractive for light on a significant level, it is absorptive. Assume the material is black, and not polished aluminium.
In the case of light, what I imagine, is that as the wave initially expands due to diffraction as it enters a thick pinhole is that once it strikes the material of the pinhole, it will be absorbed, and not reflected back (unlike water) to interfere with itself. As the wave propagates forward, this would eliminate more and more of the outer portions of the wave, leaving a narrower and narrower angle, leaving a diverging wave at the angle of survival exiting the pinhole.
What I'm saying is there is a fundamental difference in the way the material works that is normally unimportant in a ripple simulation idea of diffraction.
photon exhibit wave–particle duality
so .. see particle behaviour.
with waterwaves model a pinhole will produce a white spot, bright in the center having brightness diminished towards the edges.
illumination decreases with the square of the distance and pinhole circle seen at an angle appears as an ellipse with area less than the circle and hence additional vignetting.
The easy answer - Try it and report the results.
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Curious: what do you need a 2u LED spot for?