# Telescope Resolving Power Limits

1. ### kmartin

10
Is it possible for a telescope to resolve images beyond the diffraction limit? In other words, is the information in the light entering a telescope insufficient to resolve beyond the diffraction limit, or is information lost when the light is focused onto a screen?

Attached to this post is a diagram of my thought experiment telescope design. The collimated output from a telescope is split into two beams. Each beam is then passed through a birefringent crystal, and then the two beams are focused down onto a single screen.

If, for example, the telescope is looking at two far away point sources (Point A and Point B), then the collimated beam will consist of two overlapping plane waves, traveling in slightly different directions. Because of the small difference in propagation direction, they will experience slightly different refractive indices when going through the birefringent crystals. This causes a difference in path lengths for light from point A compared to light from point B. If the path lengths are arranged so that light from point A constructively interferes at the focus point, and the correct length of birefringent crystals is chosen, then light from point B will interfere destructively. This allows light from point A to be separated from light from point B, even if this would be impossible conventionally due to the resolving power limit. By my calculations, if the collimated beam is 5mm in diameter, the birefringent crystals are made of calcite and 1.5cm long, and the light wavelength is around 500nm, then the resolving power of the telescope would be doubled. Using thicker crystals, or a more exotic crystal (e.g. calomel) could produce a much higher increase. Also the beam could be split multiple times to improve the resolving power in both dimensions, and to reduce the effects of interference patterns.

Note that the light experiences a 'walk off' as it enters and exits the crystal, as shown in the diagram, however this shouldn't affect the operation of the telescope. Also the phase difference from the change in refractive index is proportional to the angle between the propagation directions of the light from the two sources. This is much more significant than the geometric path difference, which is proportional to the square of the angle.

My telescope design can be thought of as similar to a system that combines the output from several telescopes into a single image. However in my design the alignment is much simpler, and there would be no spatial coherence issues, allowing for better nulling.

So would my telescope design work?

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### Staff: Mentor

The resolution power is usually given by the size of the first, big mirror*. Even with perfect analysis of the light reflected by that mirror, you won't get beyond the resolution limit.

*without atmosphere or with adaptive optics

3. ### Andy Resnick

5,780
What you are describing is very similar to an imaging technique in microscopy- differential interference contrast. There are telescopes that use interference:

http://en.wikipedia.org/wiki/Astronomical_interferometer

I saw the Palomar testbed during construction- very impressive technological feat. One thing you are forgetting is that the incoming wave is not a plane wave- clear air turbulence means the incoming wavefront has a stochastic component. Keeping the two 'arms' of the interferometer a constant length is another trick.

4. ### kmartin

10
Thanks for the replies, but I don't think they really answer my main question.

To put it another way, if I could measure the phase of the light at every point of the focused image of a telescope, as well as the intensity, wouldn't that give me more information about the light source than a conventional telescope that just measures intensity?

Also I would be grateful if someone could give the fundamental reason why my thought experiment telescope design in the opening post doesn't work.

5. ### sophiecentaur

13,384
The 'resolution limit' is not a hard and fast figure. The intensity / angle figures of two images which is just at the so-called Rayleigh criterion shows a dip to half power between the two maxima. This is a good rule of thumb as a limit for human eyballing (I guess??) but if you move the images much closer and if the system noise level is low enough (and if you know the detailed spatial impulse response of the optics) you can still resolve the two. The limit is, theoretically, only imposed by the system noise (and your processing power). Whilst it is clearly better if you know the phase profile of the waves, the intensity at the detector can still give you more information than is suggested by the criteria that have been used historically. This is the same process they use for 'image enhancing' photos of car number plates and faces which, at first sight, apprear illegible.

6. ### Andy Resnick

5,780
Why do you think it would not work? As I mentioned, there are existing systems that use interferometry. You may not appreciate the need to account for atmospheric turbulence, because your incoming wavefront is not a plane wave.

Edit: you may appreciate David Saint-Jacques dissertation "astronomical seeing in space and time"

7. ### kmartin

10
If it worked then it could potentially allow the construction of a small relatively cheap space telescope that has a resolving power much greater than any current telescope*. Also it could make conventional astronomical interferometry obsolete, as only one big telescope is required. So its probably not going to work, but its worth asking for other peoples opinions in case there is something in the idea.

*Admittedly there might be a bit of a brightness problem.

8. ### Andy Resnick

5,780
Again, this has already been proposed and developed:

http://en.wikipedia.org/wiki/Space_Interferometry_Mission

10
10. ### sophiecentaur

13,384
I must say, I couldn't see how the design in the OP gave any increase in actual Aperture - which is the essence of the improvement in resolution for an interferometer. It has that great lens and then messes about with it and loses most of its energy gathering power (another factor in telescopes) - not to mention the energy loss in the polarisers.

### Staff: Mentor

I think it boils down to the following.
Can you split the light after it enters the aperture and then recombine it later to give you a better resolution?

12. ### sophiecentaur

13,384
Would it give you a bigger effective aperture? I don't think so. Interferometers are based on increasing the baseline to extend over a much bigger distance than the aperture of just one telescope. This system doesn't do that.

13. ### kmartin

10
I think the more important question is:

Can you use a birefringent crystal to adjust the phase in order to mimic the effect of translating a telescope laterally?

14. ### kmartin

10
Here's a (hopefully) better explanation.

Consider two identical telescopes a few metres apart, both pointing at the same far away light source. The images formed by the two telescopes are indistinguishable because the source is far way. However if the images are combined they can create a higher resolution image. How is this possible if the images are identical? The answer is that there is a difference in the phase of the light. For example, consider the attached diagram, where two telescopes are looking at two far away point sources (coloured blue and green for illustrative purposes). The light entering telescope 1 is momentarily in phase, where as the light entering telescope 2 is out of phase. The only difference between the light entering telescope 1 and the light entering telescope 2, is that in telescope 2 the light from the 'green' source has been delayed by half a wavelength. This difference is what allows the light from the two telescopes to be combined together to form a high resolution image. Is it possible to modify the light from telescope 1 to make it identical to the light from telescope 2? Possibly yes, if the light is passed through a birefringent crystal where light travelling in different directions experiences different refractive indices. The light from the green source would experience a higher refractive index, giving it a longer optical path length, causing it to lag behind the light from the blue source. So now we can take light from telescope 1, split it with a beam splitter, pass half of it through a birefringent crystal, then recombine the light to create the same image we would have got from combining the light from telescopes 1 and 2.

This technique, if it works, would allow the construction of a small relatively cheap single mirror space telescope that has a resolving power much greater than any current telescope.

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15. ### sophiecentaur

13,384
How do you think a single aperture (defined by the dimensions of the single telescope) can be transformed by a wider aperture, merely by jiggling about with two different polarities?

What you are saying, implies that a single slit can be made to have the same interference pattern as two similar slits, separated by a significant distance. Can you demonstrate this to be true, with some calculations and geometry? Where do you get your angle dependent path differences?

16. ### kmartin

10
The angular dependent path difference comes from the angular dependence of the birefringent crystal. Small changes in propagation direction cause small changes in the effective refractive index, which cause small changes in the optical path length.

17. ### sophiecentaur

13,384
The way an interferometer works is that the wide baseline gives a useful phase change as the angle of arrival changes by a small amount. I understand, now, what you are saying but can you explain how the path length through a crystal can change by what must be at least one wavelength for a change in angle of a few arc seconds? I looked for some information but couldn't find more than general stuff and nice pictures.

18. ### sophiecentaur

13,384
I guess one practical problem with this would be to get that thickness of calcite crystals and to ensure good optical quality.

19. ### kmartin

10
The effective refractive index is given by $n(\vartheta)$ where

$\frac{1}{n(\vartheta)^2} = (\frac{cos(\vartheta)}{n_o})^2+(\frac{sin( \vartheta)}{n_e})^2$

For calcite
$n_o=1.658$
$n_e=1.486$
http://en.wikipedia.org/wiki/Birefringence

We want to align the crystal at an angle $\vartheta_m$, where the rate of change of refractive index is maximum, and the response is most linear. This gives a maximum rate of change of refractive index as

$\frac{dn(\vartheta_m)}{d\vartheta}≈ 0.17$

So a 1.5cm length of calcite would give a path length change of 2.5μm per milliradian.

However the act of focusing the light and collimating it amplifies the angle. So for a telescope with an aperature of say 50cm, which collimates the incoming light to a diameter of 0.5cm, this gives 250μm per milliradian, or 0.25μm per microradian.

However I am far from being an expert on birefringence, so if someone could confirm my thinking it would be good.

Last edited: Nov 17, 2012
20. ### sophiecentaur

13,384
Sounds more convincing now. Thanks for the figures.