# Why does this illumination pattern of an LED light going through a small pinhole not emulate the point spread function?

I have an optical setup where a component LED passes through a 50 um pinhole, and creates an illumination pattern on an image sensor approximately 5 cm away that looks like the following:

Now, I'm curious why it is the case that the central pattern is not a circle and instead is rectangular with two holes. Is this because there is no optic element that collimates the incoming LED light? Thanks!

• You have built a pinhole camera/microscope and are imaging the die in the LED. The two round things with tails are the bond wires. Cool! :-) If you want to see the point spread function, then you have to illuminate with a "point", i.e. collimate the light of the LED on a small second aperture far enough away. – CuriousOne Jan 16 '16 at 6:52
• Thanks for the answer! What exactly do you mean by collimating the light on a small second aperture? – ArKi Jan 16 '16 at 7:02
• Right now your pinhole is imaging an object much larger than itself. Use a lens to collimate the light on a small aperture to get a point source of max. intensity some distance (one or even multiple meters) away from the pinhole that you are trying to measure. You may have to expose for quite some time since the total amount of light falling on your pinhole will be small. If you have a second 50um pinhole then you will have to place it 10m away to get 1 second or arc resolution, I believe. – CuriousOne Jan 16 '16 at 7:11

You've selected a pinhole that is roughly 100 wavelengths in diameter. That's the problem: even if you focussed the LED onto the pinhole, many secondary maxima (diffraction rings) pass thru the pinhole. To get a "clean" output, ideally you want the pinhole to be about the size of the primary maximum. So if you can calculate the focal length of a lens that will produce a spot on the order of $50 \mu m$ , you'll get a cleaner output. Keep in mind that this calculation requires a decent estimate of the lens' aperture so that you can use the diffraction pattern formulas.