6
$\begingroup$

Consider hypothetical camera with pinhole (in mathematical sense) lens and is sensor, like in below image. Is it the case that the central region of the sensor will receive more light (pixels will be brighter) than the pixels closer to the border?

Will the darkening be proportional to the cosine of the sensor normal and the direction which the light comes from?

Is this phenomenon occurring in digital cameras, is it corrected digitally?

enter image description here

EDIT:

I mostly interested in case of idealized film. Digital cameras exhibit this effect due to light attenuation caused be some participating media. I'm asking about pure geometrical effect like in Lambertian law.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ You will see circles, due to diffraction $\endgroup$
    – Milo Lu
    Commented Mar 24, 2017 at 1:46
  • $\begingroup$ What if we assume infinitesimal pinhole and classic optics (no diffraction). $\endgroup$
    – ciechowoj
    Commented Mar 24, 2017 at 20:41

1 Answer 1

Reset to default
5
$\begingroup$

This effect is called vignetting. The cases you talk about are specifically natural and pixel vignetting as explained on the wiki page. Digital cameras compensate these effects, but optical vignetting has to be done in image software for DSLR's with interchangeable lenses. For example if you import an image to Lightroom, there's "lens correction" option, which removes vignetting, distortion and chromatic aberration caused by the lens based on the lens profile.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yeah, I've read that article. But does this phenomenon occur if we consider idealized film? Where there is no shadowing by the pixel sensor edges and no attenuation due to nonzero color filers thickness. I'm asking about phenomenon caused just be geometry. $\endgroup$
    – ciechowoj
    Commented Mar 24, 2017 at 20:08
  • 1
    $\begingroup$ Yes, that's the "natural vignetting" on the wiki page $\endgroup$
    – JarkkoL
    Commented Mar 27, 2017 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.