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So in optics, the Point Spread Function (PSF) describes how an optical system responds to a point source of light. My understanding is that this is due to diffraction and the wave-like nature of Light.

This would lead me to believe that this should apply to any light entering the optical system (indeed, we know there is a diffraction limit for the resolving power of an optical system). After all, if I drew a ray from my camera to any object reflecting/emitting light into my camera, that light should be diffracted the same as if it were a point source, no?

Where I'm getting tripped up is that point sources seem to be spread far more than I'd expect from looking at an image. A star for example, can be spread across dozens of pixels despite being as close to a point source as is practically possible. And this effect occurs even on diffraction limited systems where the resolution of the sensor is at or lower than the diffraction limit of the optics and so, I would assume that diffraction could not be observed. If the same kind of blurring observed in stars was applied to the rest of the image, all fine detail would be lost.

So what is different about a star/true point source of light, versus everything else? Does the PSF apply to all light, but is extremely narrow and so is only noticeable for extremely intense light sources like a star? Or is there something "special" about a truly point source of light that causes it to be "blurred" more than other "broader" features in an image?

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    $\begingroup$ A ray is not a point source. A ray is the gradient vector of the constant phase surface of the wave. Do you know about Fraunhofer diffraction? See, en.wikipedia.org/wiki/Fraunhofer_diffraction $\endgroup$
    – hyportnex
    Jan 31 at 17:05
  • $\begingroup$ I recognize a ray isn't necessarily a point source, but what I was envisioning is looking at 2 things extremely far away (think a star and a nebula for example, where they're effectively 'at infinity' and the light from those targets is then a plane wave). So in that sense, it would seem that if a star is blurred over many pixels, then so should any other incoming plane wave. But what we see are stars blurred over an area much larger than much of the fine detail captured in the image, and I'm confused as to why that is $\endgroup$
    – Chris Gnam
    Jan 31 at 17:55
  • $\begingroup$ the amount of "blurring" depends on the direction of the point source, it is not the same for all and this is the case even for geometric optics ignoring diffraction. The off-axis aberrations such as coma depend strongly on the off-axis angle of the object point. The pixel itself size is unrelated to the optical blur we are talking about. The pixel size is related to the amount of money you wish to spend on the detector array, the "diffraction" and "other" blur is related to the physical size and amount of money you wish to spend on the "glass". $\endgroup$
    – hyportnex
    Jan 31 at 18:32
  • $\begingroup$ The PSF applies to a single wavelength for a point source at infinity. Are you saying that the size of the diffraction limited Airy disk is smaller than the size of a pixel, yet the disk appears to cover a dozen pixels? Make sure that your detector is not saturated. If the intensity is high enough to saturate the pixels, light at the fringe of the Airy disc could be intense enough to light up a pixel more than you expect making the appearance of the disk to be larger than it actually is. $\endgroup$
    – garyp
    Jan 31 at 18:42
  • $\begingroup$ @garyp moving my comment as you did: I'm asking why the PSF applies to a point source at infinity, but doesn't seem to apply to non-point sources. For example when you look at a hubble image where everything in the image is "at infinity" and so all of the light entering the optical system is a plane wave, the stars are smeared out over a large distance (even if not fully saturated) yet there is fine detail in the background nebulae or galaxy or whatever, that is clearly not as smeared/blurred as the stars. $\endgroup$
    – Chris Gnam
    Jan 31 at 19:25

3 Answers 3

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Given an imaging system that takes incoming light and focuses it on a detector, the point spread function is a measure of the systems image quality.

Ideally a point image should produce a point at the detector. There are many reasons why it might not. For example, pixels in the detector have a finite size. Lenses can have aberrations, or simply be out of focus. Light is a wave and diffracts.

An image can be broken up into many point sources of light. Each of these produces a spot of light in the detector whose shape is given by the point spread function. The total image is the sum of the point spread function.

This can be useful. You can compare optical systems by comparing their point spread functions.

Given an image at the detector and a known point spread function, it is possible to figure out what the original input image was like. This has been used to sharpen the focus of out-of-focus photographs.

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Point spread function characterizes optical system without questioning the underlying physical phenomenon. If I am imaging the infinitesimal dot, how it will spread on the detector?

It could be due to the diffraction as well as reflections in the optical elements, currents in the detector or whatever else. But you are not questioning that, you just assume your incoming image is I(x,y) and you will get it convolved with your point spread function when reading the data $$I(x,y)*PSF(x,y).$$

This way you can apply PSF to any input image, not just the dot like patterns. If you are imaging star, there might be some diffraction so that your incoming $I(x,y)$ is not the dot, but that is unrelated to your PSF. $I(x,y)$ is how it looked like before your system not when emitted. And of course PSF is simplification as the response to different intensity might be different, there might be systematic biases of your optical system and other effects.

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For a point source the spread varies as inverse square of the distance. However for a linear source it varies as receprocal of the distance near the source. So the variation of intensity depends on the geometry of the source specially at close distance.

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  • $\begingroup$ @ChrisGnam whoops. I'll move mine to where it belongs. Perhaps you should, too. I'll address your question after you move yours $\endgroup$
    – garyp
    Jan 31 at 18:41

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