I have a sensor and I want to make really small FOV, like 0.1 degree.
So, I use lens, which focus light into pinhole. So FOV is Tan(D/f)^-1, where D is pinhole size, f is focal length. It means, If I want small FOV - 0.1 degree I should use really small pinhole size, around 10 um for focal length -4.5 mm.
But the problem is that, even parallel beam make spot size in the focus around 100 um. So all my energy doesn go through the pinhole..
I use aspheric lens.
How can I make really small FOV collimator?
Nikolay, Since angular field of view (AFOV) of a collimator, of which entrance pupil diameter is h and focal lengh is f, FOV=2xtan^-1(h/2f), to make h reasonable size, say a few mm, you should use a collimator with very long focal length. In case FOV=0.1 deg=0.001745 rad,
h=tan(0.001745/2)*2*f=1.745mm for f=1000mm.
=3.49mm for f=2000mm.
https://www.edmundoptics.com/resources/application-notes/imaging/understanding-focal-length-and-field-of-view/
Shigeo
Nikolay h should be read as image sensor size, which is smaller than EPD of collimator. Thanks, Shigeo
Nikolay: 0.1 mm spot at the focus of a 4.5 mm lens is very large if the incident beam is accurately collimated.
What is the diameter of the beam at the lens? A very small beam waist will increase diffraction, but would need to be less than 0.1 mm for a 0.1 mm spot with a 4.5 mm lens.
How is the lens oriented? Aspheric lenses are usually strongly curved on one side. This should face the collimated beam. The flatter surface should face the pinhole.
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