13
$\begingroup$

Small aperture in a photo camera leads to well-known starburst effect:

sample photo with a starburst (source)

It's generally believed to be caused by diffraction.

But since diffraction is a wavelength-dependent phenomenon, it should have rainbow-like fringes — or, for monochromatic light, simply fringes instead of continuous spikes — like the diffraction in reflecting telescopes:

sample photo from a reflecting telescope(source)

But all my attempts at reproducing the fringes have failed. I've tried making photos of laser pointer spot on the wall with a DSLR camera (Canon EOS 1100D Kit) at the smallest aperture it supports, here are results for a red 640 nm and green 532 nm pointers:

enter image description here

green laser pointer photo

The green image was shot with focal length $49\,\mathrm{mm}$ and aperture $f/22$, and the red one with focal length $18\,\mathrm{mm}$ and same aperture setting of $f/22$. Here I can't notice any fringes — the spikes still look continuous.

So, my question is: if the starburst effect is due to diffraction, then why are the spikes continuous, lack any fringes?

$\endgroup$
  • $\begingroup$ Could it be that you have overexposed the image? The images in you first link all have a lot of background as well as the very bright source and they used starburst filters. $\endgroup$ – Farcher May 18 '17 at 12:37
  • 1
    $\begingroup$ @Farcher no, the filters were not used. After the "Starburst filters" paragraph all discussion of filters on that page ended. All the following paragraphs speak about diffraction on aperture. As for overexposing, I did try lowering exposure time, nothing changed qualitatively, only dimmed. $\endgroup$ – Ruslan May 18 '17 at 12:50
  • $\begingroup$ If you found my answer useful and think that it correctly and completely answers your questions you can mark it as the correct answer. If you feel like my answer is not complete and is missing something, leave a comment, I will do my best to improve it. $\endgroup$ – Ilya Lapan May 25 '17 at 11:15
  • $\begingroup$ @IlyaLapan your answer is fine, but I'd rather have the question in the bounty queue until the bounty expires. Then in the grace period I'll both accept and bountify the answer. This way both the question and the answer get more views, so someone might add their own answers, and yours might get some more upvotes ;) $\endgroup$ – Ruslan May 25 '17 at 11:24
9
+50
$\begingroup$

I think your aperture is too big. To get the idea of the scale, for the red light, you say your aperture size is $f/22$, i.e. $\frac{49\,\mathrm{mm}}{22} = 2.227\,\mathrm{mm}$, then the angular diameter of the first peak is: $$ \sin(\theta) = \frac{\lambda}{d} = \frac{640\,\mathrm{nm}}{2.227\,\mathrm{mm}} = 2.8738\times10^{-4}$$ $$ \theta = 2.8738\times10^{-4}\,\mathrm{rad} $$ That is too small to notice on such a picture. Your peaks and black spots just blend in together on this photo, you do not notice them. You are limited by the resolution of the camera matrix. Make the aperture smaller and the angular size of the elements on your photos will become bigger and you will be able to notice them. Or, increase your focal length. Or increase the resolution of your camera. Telescopes have far larger focal lengths, so that is why the diffraction pattern is so noticeable on the picture you've posted. Here is a nice visual example. Look at this picture of an Airy pattern:enter image description here

Now, if I shrink this image into 60 by 40 pixels:

enter image description here

You do not see the pattern anymore, just see the bright spot with some blur around it. That is my guess for what is happening in your case. So your camera has a $22.2\,\mathrm{mm}\times14.7\,\mathrm{mm}$ image sensor and a matrix of $12.2$ Megapixels. That gives about a $4.81 \times 10^{-3}\,\mathrm{mm}$ per pixel in one of the directions. The angular size of the interference pattern maximum I've calculated was $ \theta = 2.8738\times10^{-4}\,\mathrm{rad}$. With focal length of $49\,\mathrm{mm}$, the size of the first order maximum as it appears on the camera matrix is $ l = f\sin(\theta) = 2.4\times10^{-3}\,\mathrm{mm}$. Those calculations are very rough and are completely ignoring a lot of effects, assuming the image is perfectly in the focal plane of the lens but the point is: the size of the first order peak as it appears on the camera matrix is about the same size as the camera matrix pixel. And that is the largest element on your image (the bright spot in the middle of you diffraction pattern).

$\endgroup$
0
$\begingroup$

A few weeks ago I asked this question about Newton's rings. My question was "how is it possible that a diffraction phenomenon occurs with white, non-coherent light." The respond given by WetSavannaAnimal aka Rod Vance, aside from answering my question in full, described a blurring effect due to the fact that the light was white.

Maybe it is possible that there are fringes in the spikes, but because the incident light is a superposition of all the visible spectrum, there is a blurring effect that masks them.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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