<p>Not sure what you're asking here.  Pretty much any telescope will be unable to simultaneously produce infinity (magnified) output <strong>and</strong> an aberration-free intermediate focus.  Further, an aspherical wavefront can't be produced purely by a radiating point source (all objects are collections of point sources), so no optical system can refocus an ugly wavefront.  </p>

<p>The way a Shack-Hartmann WFS works is basically not to care about aberrated images. Each subaperture's image is unresolved to begin with, and all the sensor needs is to determine the approximate centroid location. </p>

CC BY-SA 4.0
2018-05-25T15:00:25.723Z
Thanks for the answer. To articulate, my question is can a Kepler telescope conjugate a large aspheric wavefront, e.g. scale down the incoming wavefront to another conjugate plane. The paper above said no, and I am wondering the reason why.
WDC
CC BY-SA 4.0
2018-05-25T14:52:27.523Z
2018-05-25T14:52:27.523Z
Carl Witthoft
<p>To my understanding, a Kepler telescope is designed to conjugate one plane to another. For example scale down/up the incoming wavefront onto a wavefront sensor for measurement. It works well for small wavefronts, i.e. the wavefront is relatively flat (e.g. $&lt;10 \lambda$ wavefront RMS).</p>

<p>What would happen if a large aspherical wavefront is impinging onto the same Kepler telescope system? Paper <a href="https://www.spiedigitallibrary.org/conference-proceedings-of-spie/5252/0000/Calibration-issues-with-Shack-Hartmann-sensors-for-metrology-applications/10.1117/12.513462.full" rel="nofollow noreferrer">Calibration issues with Shack-Hartmann sensors for metrology applications</a> suggests to place the aperture at the shared focal plane for performance improvement, as below:</p>

<p><a href="https://i.sstatic.net/Cid9R.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Cid9R.jpg" alt="enter image description here"></a></p>

<p>How to understand the intuition behind? Why would Kepler telescope fail for large aspherical wavefronts?</p>

<telescope><optics>
How to understand Kepler telescope under large aspherical wavefront
WDC
# How to understand Kepler telescope under large aspherical wavefront

(Asked by: username_0 on 2018-05-24)

To my understanding, a Kepler telescope is designed to conjugate one plane to another. For example scale down/up the incoming wavefront onto a wavefront sensor for measurement. It works well for small wavefronts, i.e. the wavefront is relatively flat (e.g. $<10 \lambda$ wavefront RMS).

What would happen if a large aspherical wavefront is impinging onto the same Kepler telescope system? Paper [Calibration issues with Shack-Hartmann sensors for metrology applications](https://www.spiedigitallibrary.org/conference-proceedings-of-spie/5252/0000/Calibration-issues-with-Shack-Hartmann-sensors-for-metrology-applications/10.1117/12.513462.full) suggests to place the aperture at the shared focal plane for performance improvement, as below:

[![enter image description here](https://i.sstatic.net/Cid9R.jpg)](https://i.sstatic.net/Cid9R.jpg)

How to understand the intuition behind? Why would Kepler telescope fail for large aspherical wavefronts?

**Tags:** telescope, optics

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## 1 answers

### Answer by username_1 on 2018-05-25. Score: 1

Not sure what you're asking here. Pretty much any telescope will be unable to simultaneously produce infinity (magnified) output **and** an aberration-free intermediate focus. Further, an aspherical wavefront can't be produced purely by a radiating point source (all objects are collections of point sources), so no optical system can refocus an ugly wavefront.

The way a Shack-Hartmann WFS works is basically not to care about aberrated images. Each subaperture's image is unresolved to begin with, and all the sensor needs is to determine the approximate centroid location.

> **Comments:**
> **username_0**: Thanks for the answer. To articulate, my question is can a Kepler telescope conjugate a large aspheric wavefront, e.g. scale down the incoming wavefront to another conjugate plane. The paper above said no, and I am wondering the reason why.  

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