Matt in his answer on What does a hexagonal sun tell us about the camera lens/sensor? mentions

Incidentally the number of (distinct) points to the star is equal to double the total number of unique orientations* in the sides of the aperture shape i.e. three blades would be six points, six blades would also be six points as a hexagon has only three unique orientations in its sides.

* a hexagonal aperture has six sides but only three unique orientations as there are three pairs of parallel sides.

I think It makes sense to have three pointed star in case of three blades.

But how does a six blade aperture also result in a three pointed star?


2 Answers 2



As has been stated in the answer you linked to, the propagation of light from the pupil of an imaging system to the image plane can be modeled by a very close approximation called Fraunhofer Diffraction.

$$ U(x,y) \propto \int\int u(\xi, \eta) e^{-i\frac{2\pi}{\lambda z}(x\xi+y\eta)} d\xi d\eta $$

where $\xi$ and $\eta$ are the x- and y- coordinates in the pupil plane, $u(\xi, \eta)$ is the optical field in the pupil plane, and $U$ is the optical field in the image plane.

Under the substitutions

$$ \begin{eqnarray} \mathcal{f}_x &= &\frac{x}{\lambda z} \\ \mathcal{f}_y &= &\frac{y}{\lambda z} \\ \end{eqnarray} $$

Fraunhoffer diffraction is simply a Fourier transform (FT). Thus it is easier to write:

$$ U(x,y) \propto \mathcal{F}\left[ u(\xi,\eta) \right] $$

where $\mathcal{F}$ will represent the FT, including the necessary scale factors and substitutions mentioned.

A property of the FT important for this discussion is called Convolution Theorem. In terms relevant to this answer, this theorem states that the multiplication of two patterns in the pupil of the lens is equivalent to the convolution of the FT of those patterns in the image plane. We can write this as:

$$ \mathcal{F}\left[ u(\xi,\eta) \right] \otimes \mathcal{F}\left[ p(\xi,\eta) \right] \propto \mathcal{F}\left[ u(\xi,\eta) \times p(\xi,\eta) \right] $$


$$ U(x, y) \otimes P(x,y) \propto \mathcal{F}\left[ u(\xi,\eta) \times p(\xi,\eta) \right]$$

where $P(x,y)$ is the FT of the pupil function $p(\xi,\eta)$, and $\otimes$ is the convolution operator.


So how does this apply in the case of a camera lens? Well, in the pupil of a camera lens, there is some optical field -- i.e. some pattern of light on its way to the image plane. Keep in mind that it is not yet a clearly focused image, but an optical field which will become an image by the time it arrives at the image sensor. Also in this plane is the aperture stop, which blocks some of the light. This aperture stop can be considered as a function which is multiplied by the light field on its way through the pupil. It multiplies the incoming filed by 1 where there aperture is clear, and by 0 where light is blocked. In other words:

$$ p = \left\{ \begin{array}{rl} 1 & \text{where the aperture is clear} \\ 0 & \text{where the aperture blocks light} \end{array} \right. $$

So, based on the background given above, we know that the image we see will be the convolution of the ideal image $U$ with the FT of the pupil function, $P$. This will make bright light sources in the image take on the shape of the FT of the pupil function, which is why $P$ is typically called the point spread function (PSF) of the camera lens*.

So what does $P$ look like? The exact computation of the necessary FT is messy and complicated, but there is a basic rule of thumb which provides a good qualitative description of the FT of some basic shapes:

Sharp edges in the input to the FT are represented by bright streaks in the output, with the orientation of the streak being perpendicular to the edge which produced it.

So, a triangular aperture has 3 sharp edges, and therefore the resulting PSF has 3 streaks, which produces a 6-pointed star. In the images below you can see an example in which I have produced a triangular shape to represent the lens aperture, and the resulting PSF:

aperture shape PSF

In the (more realistic) case of a hexagonal aperture, there will technically be 6 streaks, but because the opposite sides of a hexagon are parallel to each other, the streaks in the PSF will be in parallel pairs, so they will overlap and you will only see 3 distinct streaks in the image.

*This is only true in the case where the lens is optically perfect. In reality, aberrations in the optics also contribute to the PSF, but a description of that is beyond the scope of this answer.


Fraunhofer diffraction is virtually a two-dimensional Fourier transform. So we just have to attempt to make a Fourier transform of an appropriate function and see how many rays the resulting "star" look like.

  1. Hexagonal aperture. I take the following function: $$ \begin{split} h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot \left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\ \left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right] \end{split}$$ With $\theta(x)$ -- is a Heaviside step function.
    While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this: $$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2} -\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}} {\pi \omega_x(\omega_x^2-3\omega_y^2)}$$ ...plus some singular terms that I've dropped.
    Plotting square of the result (from -100 to 100):
    enter image description here

  2. Triangular aperture. I take the following function: $$ t(x,y) =\theta(2y+\sqrt{3})\theta(\sqrt{3}(x+1)-y)\theta(-\sqrt{3}(x-1)-y)$$ After Fourier transform, taking square and plotting:
    enter image description here

  • $\begingroup$ I beat you to it by a few seconds! Wow! Good answer though. Our answers compliment each others. $\endgroup$
    – Colin K
    May 16, 2011 at 15:00
  • $\begingroup$ Nah! Yours is better. $\endgroup$
    – Kostya
    May 17, 2011 at 8:29
  • $\begingroup$ Sorry to bring a thread back from the dead, but Kostya, I have been trying to come up with a Fraunhoffer simulation of what the diffraction from a hexagonal aperture looks like. I started with sinc functions but they are not quite the same; however your solution (and the graphical FT of a hexagon) gives me exactly what I want. I was wondering, what are the units, though? You plotted your solution to 100. Is that degrees? Or how can I scale it? $\endgroup$ Apr 4, 2013 at 21:28
  • $\begingroup$ @ChristianR Lets just link your follow-up here. physics.stackexchange.com/questions/60070/… $\endgroup$
    – Kostya
    Apr 5, 2013 at 9:28
  • $\begingroup$ +1 I've referenced and used your expression here. $\endgroup$
    – uhoh
    Apr 18, 2020 at 2:01

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