See for example this photo:

Example photo

From my experience, the longer the exposure the stronger can this effect be observed. Is this correct? Are there any other factors that influence the creation of these stars (is there a better word for it, by the way?) And what exactly does technically happen?

  • \$\begingroup\$ Is there a way to avoid this effect with filters or something else? \$\endgroup\$ Commented Aug 12, 2012 at 18:41
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    \$\begingroup\$ @Luis: According to the answers, I suppose you can just open up your aperture. \$\endgroup\$
    – eflorico
    Commented Aug 12, 2012 at 20:30

8 Answers 8


This appears to be a beautiful example of Fraunhofer diffraction. It is due to the wave nature of light. The effect depends on the wavelength (that is, the color). It is most pronounced when bright light from a practically infinite distance passes through narrow slits, causing the light to spread perpendicular to the slits. This spreads a point-like beam of light into a pair of streaks.

Using a small aperture creates slit-like situations at the corners formed by adjacent blades. Thus, when you have a combination of relatively intense, pointlike, monochromatic light sources in the image and a narrow aperture, you should see a streak (of the same color) emanating from the points in two directions perpendicular to the blades. When your diaphragm is formed by straight blades, this will cause there to be twice as many streaks as blades. However, the streaks for parallel blades will coincide. Thus, for a diaphragm with an odd number of blades (where no two blades are parallel) there will be twice as many radial streaks as blades but for a diaphragm with an even number of blades (where opposite blades are parallel) the streaks overlap in pairs, giving the same number of streaks as blades (but each streak is twice as bright).

A classic example is shown in the first image in the Wikipedia article on diffraction, for Fraunhofer diffraction through a square aperture. You see four well-defined streaks.

The theory is further explained here. This explanation was published in 1967 by CA Padgham. Ken Rockwell mentions it in his discussion of bokeh.

We should expect a certain amount of diffraction always to be present. It is usually slight and averaged out in most pictures: it just contributes a tiny amount to the blurriness that is present in any image when looked at closely enough. Only in images that bring together several factors--points of intense monochromatic light, small apertures, straight diaphragm blades--will it become prominent. This information shows how you can make the stars more prominent or how you can suppress them, by altering these factors for your exposure (to the extent you can).

Finally, length of exposure is related to the occurrence of this effect, as you have observed, but only because exposures with bright points of light are almost always made much longer than needed to record the lights: you're trying to see the rest of the scene, which is much darker. The brightness of the diffraction streaks decreases so rapidly away from their sources that if you used a sufficiently short exposure to properly expose the lights themselves, the streaks would be practically invisible. For instance, there are dimmer but still prominent light sources in your background: they look like windows in the far distance. They, too, must have their own streaks, but those streaks are too dim to see. (Appropriate software filtering might be able to bring them out.)

  • \$\begingroup\$ This is clearly the most detailed answer. Thank you! \$\endgroup\$
    – eflorico
    Commented Jan 9, 2011 at 1:34
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    \$\begingroup\$ This isn't due to fraunhofer diffraction, but merely diffraction. The diffraction integrals are very difficult to solve, so there are two cases that make them simpler; the fresnel diffraction integral for moderate distances, and the fraunhofer diffraction integral for far distances. In this case, the fraunhofer diffraction integral will give an incorrect solution, because the detector is very near to the source of diffraction (the aperture). The fresnel calculations would have to be done, or possibly the complete calculation depending on the elements behind the aperture stop. \$\endgroup\$ Commented Apr 10, 2016 at 17:29
  • \$\begingroup\$ @Brandon Your clarification is most welcome, but I'm having trouble squaring it with popular descriptions of Fraunhofer diffraction, such as the one on Wikipedia: "In optics, the Fraunhofer diffraction equation is used to model the diffraction of waves when the diffraction pattern is viewed at a long distance from the diffracting object, and also when it is viewed at the focal plane of an imaging lens." That latter qualification would seem to apply explicitly to an image captured by a camera. \$\endgroup\$
    – whuber
    Commented Apr 10, 2016 at 18:30
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    \$\begingroup\$ @BrandonDube sorry, you are incorrect. This effect is much more accurately modeled by Fraunhofer diffraction. The confusion here is that although the observation distance is techincally short (i.e., apparently in the Fresnel regime), because the lens is focusing the plane waves to a common point (the focal plane), the viewing distance is effectively the same as viewing at infinity. See the ["Focal plane of a positive lens"]en.wikipedia.org/wiki/… section of the WP article on Fraunhofer diffraction. \$\endgroup\$
    – scottbb
    Commented Apr 12, 2016 at 14:42
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    \$\begingroup\$ @BrandonDube You're right, I chose my words poorly. Saying the effect is "much more accurately modeled by Fraunhofer diffraction" was wrong: the Fresnel integral is always more precise than the Fraunhofer integral. I should have said that in photography, diffraction spikes such as in the OP (or in Matt Grum's answer) are always in the far-field (i.e., Fraunhofer) regime, for the reason that the observation (image) plane is effectively the same as viewing at infinity (i.e., having a small Fresnel number). \$\endgroup\$
    – scottbb
    Commented Apr 13, 2016 at 2:27

It's due to diffraction where the aperture blades meet as stated by John and Pearsonartphoto. It's a neat way to test how many aperture blades you have!

To answer your second question, the length of the exposure doesn't directly affect the effect. There are two main factors, the first is the size of the aperture (it needs to be small), and long exposures tend to go with a small aperture. The second factor is you need to be shooting into the lightsource. This tends to only happen at night with artificial light, so again people tend to wind up using long exposures.

Here's an example (not mine!) of the effect with a very short exposure to demonstrate the point:

(c) photogeek133

Ok I lied it was a long exposure with moving flashes set to strobe, but each light was on for a very brief period. The other two ingredients - shooting into the strobes, and small aperture (f/14) are what is producing the star patterns.

  • \$\begingroup\$ That certainly is an impressive photo! I've heard the term diffraction before. It was mentioned as a problem - does (and how does) diffraction appear anywhere besides from shots like these (directly into the light source)? From my understanding, it shouldn't be a problem usually. \$\endgroup\$
    – eflorico
    Commented Jan 8, 2011 at 16:45
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    \$\begingroup\$ In simple terms (see whuber's answer for a detailed analysis!) diffraction causes light to spread out. This is obviously a problem if all points of light spread out, as this will give a blurry image. Diffraction happens all the time, it's just the spreading is not noticeable for big gaps or dim lights. What we have here is a very small gap, and a bright source, so the star pattern which would usually be too dim is clearly visible. \$\endgroup\$
    – Matt Grum
    Commented Jan 8, 2011 at 17:06

What you are seeing is the result of the shape of the aperture in your camera. If you put, say, a heart shape or other "filter" on the front of your camera, you'd see a different shape in place of those lights.

You're almost correct in your guess that the longer the exposure, the stronger this affect can be observed. What's actually happening is the smaller your aperture, the more this affect will appear.


There are filters, starlight, that are designed to do this, but without a filter, the effect is usually seen with tighter apertures on lenses that have straighter aperture blades. The straighter the blades, the more pronounced the effect.

So, what is happening is these bright and stationary light sources are having their light bent by the aperture of your lens and the star pattern is being created by the sharp points being defined by the hexagon from the six blades of your aperture. You'll notice that the star rays are all going in the same direction for the lights, that's because of the aperture blades.

By the way, I like the shot.

  • \$\begingroup\$ LOL, I answered at almost the same time, so... \$\endgroup\$ Commented Jan 8, 2011 at 16:02

Why do light sources appear as stars sometimes? Well, I changed my opinion and share now the prevalent one that the stars come from diffraction effects. The strongest argument for favoring diffraction over reflection comes from the symmetry properties of the star pattern, namely, if N is odd, then N iris blades generate 2*N spikes.


I believe that you will find the answer to your questions under http://www.stfmc.de/misc/diffcontrarefl/tlf.html

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    \$\begingroup\$ Stephan, we'd like answers to the questions rather than pointers to other places people can go research. It looks to be a helpful link, but could you summarise what it says here, as it applies to the question? \$\endgroup\$
    – MikeW
    Commented Aug 10, 2014 at 18:52
  • \$\begingroup\$ I read your excellent computational exploration of the diffraction. You said in the linked article that you are exploring further. Have you any new analysis to follow? \$\endgroup\$
    – Hans
    Commented Oct 30, 2023 at 0:36

This is not a true answer, but an expansion on calculating diffraction patterns from @whuber's answer.

First, we have the diffraction integral. The function \$U_\text{p}\$ describes the complex amplitude in the observation plane at a distance \$(x_\text{p},\,y_\text{p})\$ from the optical axis, and a distance \$L_z\$ from the source (some kind of diffractive object, e.g. pinhole, camera aperture, etc.)

\$U_\text{s}\$ is a function that describes the complex amplitude in the source plane; for an extremely small pinhole you could use a dirac delta function. The third variable in \$U_\text{s}\$ is 0 because for convenience we say the diffractive object is the origin of the coordinate system. The variables \$x_\text{s}\$ and \$y_\text{s}\$ in its arguments bookkeep for that fact that the object may have some size in the \$xy\$ plane.

$$ U_\text{p}\!\left(x_\text{p},y_\text{p},L_z\right) = \!\iint{\mathrm{d}x_\text{s}\,\mathrm{d}y_\text{s}}\,\frac{e^{-ikr_\text{sp}}}{r_\text{sp}} \,U_\text{s}\!\left(x_\text{s},y_\text{s},0\right) $$

This may not look like such an awful integral, but \$k\$ and \$r_\text{sp}\$ are both just notation for something bigger:

$$ \begin{align} k &= \frac{2\pi}{\lambda} \\ r_\text{sp} &= \sqrt{(x_\text{p}-x_\text{s})^2 + (y_\text{p}-y_\text{s})^2 + (L_z)^2} \end{align} $$

Integrating a function with a radical with square terms in it both in the argument of \$e\$ in the numerator and in the denominator is a very nasty integral indeed.

One simplifies the integral by removing the square roots by using the binomial series representation and truncating higher order terms. The Fraunhofer integral holds when one needs 2 terms; the Fresnel integral is for when one needs 3 terms. There is some nuanuce to the proof of that, but it is outside the scope of this.

When we start manipulating these things to get the Fresnel and Fraunhofer diffraction integrals, we get three quantities.

$$ \begin{align} N_\text{F}^\text{d} &= \frac{\Delta d^2}{\lambda L_z} && \text{Detector Fresnel #} \\ N_\text{F}^\text{s} &= \frac{\Delta s^2}{\lambda L_z} && \text{Source Fresnel #} \\ \theta_\text{d} &= \frac{\Delta d}{\lambda L_z} \end{align} $$

If \$N_\text{F}^\text{d} \cdot \left(\theta_\text{d}\right)^2 \ll 1\$, the Fresnel integral is valid. If that is true and \$N_\text{F}^\text{s} \ll 1\$, the Fraunhofer integral holds.

The two integrals are:


$$ U_\text{p}\!\left(x_\text{p},y_\text{p},L_z\right) = C\!\left(x_\text{p},y_\text{p},L_z\right)\!\iint{\mathrm{d}x_\text{s}\,\mathrm{d}y_\text{s}}\, {U_\text{s}(x_\text{s},y_\text{s},0)\, e^{-i\frac{k}{2L_z}\left(x_\text{s}^2 + y_\text{s}^2\right)}\, e^{+i\frac{2\pi}{\lambda L_z}\left(x_\text{s}x_\text{p} + y_\text{s}y_\text{p}\right)}} $$


$$ U_\text{p}\!\left(x_\text{p},y_\text{p},L_z\right) = C\!\left(x_\text{p},y_\text{p},L_z\right)\!\iint{\mathrm{d}x_\text{s}\,\mathrm{d}y_\text{s}}\, {U_\text{s}(x_\text{s},y_\text{s},0)\, e^{i2\pi\left(\nu_xx_\text{s}+\nu_yy_\text{s}\right)}} $$


$$ C\!\left(x_\text{p},y_\text{p},L_z\right) = \frac{e^{i\frac{k}{2L_z}\left(x_\text{p}^2 + y_\text{p}^2\right)}e^{-ikL_z}\cdot i}{\lambda L_z} $$

and \$\nu_x\$ and \$\nu_y\$ are the size of the source in a given dimension divided by the wavelength of light times the distance to the source. Normally it would be written \$\nu_\text{s} = d\,/\left(\lambda x_\text{s}\right).\$

To answer @whuber's question as to why you may need one or the other, despite what Wikipedia states, requires a bit of thought.

The "at the focal plane of an imaging lens..." comment is probably lifted from a textbook, and the implication is that the source of the diffraction (i.e. the pinhole, slit, whatever — these equations are agnostic as to the geometry of the source) is very far away. Unfortunately, not only could the lens be at any distance and closer than the Fraunhofer integral allows, but the diffraction also originates inside the lens system for a camera.

The correct model for diffraction from a camera's aperture is an n-sided aperture (n is the # of aperture blades in the lens) illuminated by a point source at the location of the thing in the image that produces the starburst pattern.

When the objects are really far away (a few meters would be fine), the point sources behave as if they are plane waves and the derivations carried out on Wikipedia are fine.

For example, the aperture for a double gauss 50 mm lens is on the order of 40 ~ 60 mm from the image plane. It is imaged by a couple of lenses behind the physical stop to a distance greater than that (this is the location of the exit pupil), but the exit pupil is not where the \$U_\text{s}\!\left(x_\text{s},y_\text{s},0\right)\$ function is centered!

For 500 nm and a 1 mm radius aperture light, we can check if the Fraunhofer integral is valid. It is equal to (0.001)2 / (500×10-9 × 50×10-3), or 40, which is ≫ 1 and the Fraunhofer integral is invalid. For visible light, as long as the aperture stop is on the order of millimeters from the detector, \$N_\text{F}^\text{s}\$ will never be anywhere near 1, let alone much smaller.

These equations may differ somewhat from the ones on Wikipedia; I would reference OPT 261, Interference & Diffraction at the University of Rochester Institute of Optics taught by professor Vamivakas. The equations in Optics by Hecht should be fairly similar. The equations are for the complex amplitude, to get the Irradiance (a.k.a. intensity or brightness), you would take the magnitude squared of the result.

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    \$\begingroup\$ For the reasons cited in my comments in the accepted answer, your statement "Nfs will never be anywhere near 1, let alone much smaller", is never correct for real-world photography. These diffraction spikes are Fraunhofer diffraction precisely because the Fresnel number is < 1 (technically 0), because the lens focusing effect means the observation distance is as if it were at infinity. \$\endgroup\$
    – scottbb
    Commented Apr 13, 2016 at 2:32

Here is an example and personally, I love the effect. It can add a bit of artistry to the photo as in the one I will link to.

The cause is due to the aperture blades on my nifty 50mm.

Exposure is a secondary to the stars because I have to close the aperture in order to not overexpose the photos with all the bright lights I am shooting into. If I expose just for the lights I will see nothing but black in the photo where I want to expose the building.

Thus to compensate for the small aperture setting (f/20 in this shot) I must increase my time exposure (20sec) in order to get the proper exposure. Thus, the diffraction occurs or is greatly amplified as I increase the number on my aperture or close it down to prevent overexposure.

Note the exif info:

  • Canon EOS-1Ds Mark III
  • Canon EF50mm f/1.8 II
  • ƒ/20.0
  • 25 sec
  • ISO-100


  • 1
    \$\begingroup\$ How is your answer helpful to the OP question ? I think that the previous answers have already said it all... \$\endgroup\$
    – Olivier
    Commented Apr 9, 2016 at 13:35

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