Telescope Resolving Power Limits


by kmartin
Tags: limits, power, resolving, telescope
kmartin
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#19
Nov17-12, 03:36 PM
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Quote Quote by sophiecentaur View Post
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.
The effective refractive index is given by [itex]n(\vartheta) [/itex] where

[itex]\frac{1}{n(\vartheta)^2} = (\frac{cos(\vartheta)}{n_o})^2+(\frac{sin( \vartheta)}{n_e})^2[/itex]
http://www.radiantzemax.com/kb-en/Kn...icle50260.aspx

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

We want to align the crystal at an angle [itex]\vartheta_m[/itex], 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

[itex]\frac{dn(\vartheta_m)}{d\vartheta}≈ 0.17[/itex]

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.
sophiecentaur
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#20
Nov17-12, 05:53 PM
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Sounds more convincing now. Thanks for the figures.
kmartin
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#21
Nov18-12, 09:02 AM
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Quote Quote by sophiecentaur View Post
Sounds more convincing now. Thanks for the figures.
Its nice to know I'm not going completely mad. Thanks
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Nov18-12, 10:31 AM
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This has set me thinking and I think I have found snags. If your source is unpolarised then what will be the relationship between the two linearly polarised signals that emerge from your polarisers?
The only time that the interference between the two components will produce the required pattern will be for the contributions that are exactly in the appropriate plane. I mean that the two phasors will only cancel properly when the two waves which are polarised at 45degrees to the polariser planes so that the two vectors have equal magnitudes. You would need to pre-polarise the incoming light to make it linearly polarised in this plane.
Also, the resultant interference pattern on the screen will be modified by geometry of the 'two slit' arrangement where the two beams recombine. I'm not sure how relevant this is as it may be producing a very fine set of fringes as the 'throw' is so short.
kmartin
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#23
Nov18-12, 12:31 PM
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Quote Quote by sophiecentaur View Post
This has set me thinking and I think I have found snags. If your source is unpolarised then what will be the relationship between the two linearly polarised signals that emerge from your polarisers?
The light has to be polarised as the angular dependent refractive index only works when the polarisation is in the same plane as the crystal's optic axis and the propagation direction. However it would be possible to split the light into its two polarisations and process them separately to prevent any significant brightness loss.
sophiecentaur
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#24
Nov18-12, 01:37 PM
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Yes, I understand about the function of the two different paths for the two components. Your original diagram doesn't have a polariser at the input and I assumed that the your system would just deal with light straight from an object. How would you propose to split and recombine the light so that the two input polarisations could be treated completely separately? I think there would be other interference problems if the two images were to be allowed to combine on a single screen.

I have no experience of optics but I do know about RF antenna arrays. Their beamwidths are normally 'aperture limited' but it is possible to make 'super gain' (narrow beam) arrays by suitable weighting of array elements. This gives a narrower beam than normal but it is very difficult to go very far without producing large sidelobes, for instance. Now, with RF, one has the chance of controlling phase and amplitude much easier than with optics and it is still pretty well impossible to obtain a big improvement. It worries me that you are, in effect, trying to do the same sort of thing (a phase slope on half of an RF array would possibly achieve a similar thing) but with a necessarily more crude arrangement. I don't have a specific objection because I can't exactly put my finger on it but, somewhere, I feel there has to be a fundamental problem. (Maverick Cop: "There's something wrong, sergeant - it just doesn't feel right") Could it be to do with the assumed performance of the extraordinary ray's path through the birefractor? Diffraction will also be at work there and be limited by the aperture. It may cast doubt on your simple assertion that the path length is only affected by ray angle. We are not actually dealing with 'rays' in a diffraction limited system.
This just might be spoiling your day. If so, sorry.


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