I have to do a short calculation, but, quite frankly, I have no idea how to even start...

Suppose you have a 10-m telescope (f/10) with an infrared camera that observes at 10 micrometer. The seeing is 2 arcsec (FWHM of the PSF, 2”= 1/100000 rad) and you would like to implement an adaptive optics system.

The question is: How many lenslets do you need in the Shack-Hartmann wavefront sensor array? Below is a picture of the working principle of such an array. The telescope would also include a coronagraph, but I think (might be wrong) you don't need that information for this calculation.

Wavefront sensor

  • $\begingroup$ @B--rian, PSF means "point spread function". It's the image of a point source, e.g. an airy disk that shows the diffraction rings $\endgroup$ Jan 3 '21 at 18:31
  • $\begingroup$ I haven't really crunched any numbers, but I would logically assume it would be easiest to the same number of lenslets as you do segments in your adaptive optics adjustments. I.e., if you have a square-pattern of 50 mirrors (arbitrary #), then have one lenslet per mirror. If you have a hex-patterned mirror, it may be easiest to have hex-patterned lenslets. This way, each point of actuation on your mirror is controlled by one point on the Shack-Hartmann image array. It would also work easily to have any integer number of lenslets per actuation point if the pattern still matches $\endgroup$ Jan 4 '21 at 1:39

Generally speaking as many as possible because the number of lenslets determines the lateral wavefront resolution. But in reality there are a few factors to be considered.

Wavefront sensor

Say your lenslet array is of focal length $f$ and subaperture size $D$, given a maximum desired detection angle $\alpha$ (in your case is $2'' \approx 1\times10^{-5}\, \text{rad}$), there is the following relationship:

$$ 2'' = \alpha \approx \tan \alpha = \frac{\Delta x}{f} \leq \frac{D}{2f}. \tag{1} $$

  • To ensure a good resolution and reduce cross-talk effect, $D$ is chosen to be at least a few image sensor pixel size. In other word, for a given image sensor to measure the seeing angle, the number of lenslets are limited to a maximum value.

  • Usually you do not design the lenslet arrays; those are pre-designed, e.g. https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2861. So Eq. (1) serves as a metric to select between different microlens array designs.

  • For AO applications, the primary goal of wavefront sensors is the measurement speed rather than the lateral resolution. The sensors need to be fast. The smaller the number of pixels, the smaller the sensor latency is, resulting in a low number of pixels, and hence low number of lenslets.

Adaptive optics (AO) system

$w \in \mathbb{R}^m$: Measurement wavefront of a lenslet array of number $m$.

$c \in \mathbb{R}^n$: Correction wavefront of the deformable mirror that has $n$ number of actuators.

An adaptive optics (AO) system works by solving a linear system to output $c$ given $w$:

$$ w = A c $$ where interaction matrix $A \in \mathbb{R}^{m \times n}$ is pre-calibrated before usage.

  • To compute $c$ in a least squares sense, it requires $m \geq n$, meaning an over-determined system, mentioned also by @nflemming2004.

  • As for the aperture sizes of the two, it does not really matter that much because the conjugate lenses/mirrors in your optical system can scale down/up the apertures. That said, the sub-aperture of the wavefront sensor (i.e. $D$) and the actuator cover area $D_{\text{deformable_mirror}}$ do not have to be equal.

  • If $m \gg n$ (as in most cases), as you can imagine, the extra number of lenslets only serves to make a robust estimate, and is wasted from AO's perspective because of insufficient correction degree-of-freedom.


Using WFS30-7AR as our wavefront sensor, that $f = 5.6 \,\text{mm}$ and $D = 150\,\text{um}$, given Eq. (1) we know:

$$ 2'' \leq \frac{D}{2 f} = 1.34 \times 10^{-2} \, \text{rad} \tag{satisfied} $$

The number of lenslet is $m = 80 \times 80$, far larger than $n = 140$ the actuator number of deformable mirrors (say https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3208).

So for this design, we conclude it satisfies the requirement of this question.

  • $\begingroup$ supposing there was infinite precision on the A.O. system (continuously deformable mirror), would there be an upper limit on the amount of lenslets or a lower limit on their size, due to the infrared wavelength? $\endgroup$ Jan 4 '21 at 8:12
  • 1
    $\begingroup$ @Sarah: (1) For AO it means $n \rightarrow \infty$, there is no upper limit on $m$ the amount of lenslets (though, this AO system is under-determined since $n > m$); (2) But for wavefront sensing, Eq. (1) tells us there is a upper limit on $m$ because of the 2'' seeing requirement. $\endgroup$
    – WDC
    Jan 4 '21 at 8:22

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