Takahashi FSQ 85 ED 光學討論

檢測望遠鏡光學部份,赤道儀機械組件,目鏡配件和拍攝器材 CCD等...
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昇仔
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文章 昇仔 » 週四 17 6月, 2010 15:16

Skyobs 寫:
Skyobs 寫:By the way, how does FFT pattern (is that a plot of magnitude, phase, or combined?) relate to diffraction pattern? :roll:
With some search on the web, now I understand why use FFT to show diffraction pattern. (It's because far field diffraction pattern has the same form as Fourier transfomer of the aperture. Here's a lecture on the topic http://academicearth.org/lectures/appli ... iffraction )

However, FFT diagram with fringes as shown has some drawbacks:
1) It only represents a single wavelength.
Star light is broadband. Fringes and dips (of multiple wavelengths) beyond the first few would mix together and become a smooth flare rather than distinct fringes.
2) It contains artifact caused by discrete rectangular processing (during transform and display, and maybe after JPEG compression).
You can expect diffraction from circular aperture is the same at any angle. But the diagrams show particular "spikes" in 0,45,90,...deg.

So I would suggest apply maximum filter of width approximately 2-3 fringes on the FFT result to smooth out those fringes. This could hopefully get a better visualization of the diffraction pattern.
I wrote something about the Fourier transform in the case of Fraunhofer diffraction here before. But suddenly I realized that there is more straightforward way to handle this problem.

:idea: :idea: :idea: :idea: :idea:

I may be able to solve it and will return to u soon. :P
最後由 昇仔 於 週四 17 6月, 2010 20:47 編輯,總共編輯了 1 次。

anguslau
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文章 anguslau » 週四 17 6月, 2010 18:47

I did some test with my FSQ106+5D. Following are the results.

A) FSQ106 + 0.73x + 5D

First image is the full-size image show the two test lamps at at the lower-left corner.

Second image is a crop from three images for the same setup as above:
Left-> in-focus
Mid-> at-focus
Right-> out-focus
I tried to keep the amount of in/out travel to be the same using a vernier caliper. Of course there can be errors. In particular, not sure I got the at-focus image very accurate.

Some observations. Firstly there are some observable dark bands due to non-circular aperture. But seems not as severe as FSQ85 since the obstruction seems to be less and there are less sharp corners in the aperture. The in/out images are very similar in shape, turned 180 degrees as expected.

Third image tries to see how identical are the in/out images. I rotated the in image by 180 degree and stack it on top of the out image, setting blending mode to difference. You can see that the two images are not exactly the same size, probably due to slight differences in in/out travel. Shape is pretty close.

In the fourth image, I racked out the focuser a bit more to make the aperture larger so as to reduce the effect of the rectangular shape of the test lamps. The three crops show:
Left-> test lamps placed near lower-left corner
Mid-> test lamps placed near middle of image
Right-> test lamps placed near upper-right corner
You can see that the middle image shows circular aperture, indicating no vignetting. The other two images mirror each other.

The fifth image are at-focus images when the test lamps are placed near the corresponding places of the image. You can see that there are no dark band diffraction pattern in the middle while dark band patterns rotate for the corners.
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anguslau
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文章 anguslau » 週四 17 6月, 2010 18:58

B) FSQ106 + 5D

First image are at-focus images for test lamps placed near corresponding places within the image. You can see that there are no dark bands in the middle image as expected. Dark bands can be seen for the corner images, but much more subtle compared to the with reducer config.

Second image are the in-focus images for test lamps placed near corresponding places within the image. Again, circular aperture in the middle, indicating no vignetting. Some vignetting for the corners, but less compared to the with reducer config. And very little sharp corners.

Third image are the out-focus images for test lamps placed near corresponding places within the image. Similar to the in-focus images except rotated 180 degrees.
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昇仔
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文章 昇仔 » 週四 17 6月, 2010 19:01

Good work angus :D :D :D

It seems that 106 has less dark bands than 85. But your test seems to indicate that the in-out focus pattern are symmetical w.r.t to the focal plane as expected. Have I done anything wrong? But in my case the asymmetry is quite apparent and is unlikely to be a mistake. :?:

anguslau
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文章 anguslau » 週四 17 6月, 2010 19:08

昇仔 寫:It seems that 106 has less dark bands than 85. But your test seems to indicate that the in-out focus pattern are symmetical w.r.t to the focal plane as expected. Have I done anything wrong? But in my case the asymmetry is quite apparent and is unlikely to be a mistake. :?:
I suspect your in/out travel distance (from at focus point) are not the same, hence their sizes are different. I found that under such circumstances, the shapes can be perceived to be quite different. :roll:

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昇仔
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文章 昇仔 » 週四 17 6月, 2010 19:15

anguslau 寫:
昇仔 寫:It seems that 106 has less dark bands than 85. But your test seems to indicate that the in-out focus pattern are symmetical w.r.t to the focal plane as expected. Have I done anything wrong? But in my case the asymmetry is quite apparent and is unlikely to be a mistake. :?:
I suspect your in/out travel distance (from at focus point) are not the same, hence their sizes are different. I found that under such circumstances, the shapes can be perceived to be quite different. :roll:
This is quite unlikely :? . When I moved more in-focus and out-focus, the shape of the image essentially remained the same, only shrinking or enlarging in size.

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昇仔
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文章 昇仔 » 週五 18 6月, 2010 01:24

昇仔 寫:
Skyobs 寫:
Skyobs 寫:By the way, how does FFT pattern (is that a plot of magnitude, phase, or combined?) relate to diffraction pattern? :roll:
With some search on the web, now I understand why use FFT to show diffraction pattern. (It's because far field diffraction pattern has the same form as Fourier transfomer of the aperture. Here's a lecture on the topic http://academicearth.org/lectures/appli ... iffraction )

However, FFT diagram with fringes as shown has some drawbacks:
1) It only represents a single wavelength.
Star light is broadband. Fringes and dips (of multiple wavelengths) beyond the first few would mix together and become a smooth flare rather than distinct fringes.
2) It contains artifact caused by discrete rectangular processing (during transform and display, and maybe after JPEG compression).
You can expect diffraction from circular aperture is the same at any angle. But the diagrams show particular "spikes" in 0,45,90,...deg.

So I would suggest apply maximum filter of width approximately 2-3 fringes on the FFT result to smooth out those fringes. This could hopefully get a better visualization of the diffraction pattern.
I wrote something about the Fourier transform in the case of Fraunhofer diffraction here before. But suddenly I realized that there is more straightforward way to handle this problem.

:idea: :idea: :idea: :idea: :idea:

I may be able to solve it and will return to u soon. :P
I originally thought that the Fraunhofer approximation is not applicable since the aperture of the telescope a is not small, so a^2/(lamda f) << 1 is not satisfied and the diffraction integral cannot be reduced to the Fourier transform of the aperture function. But I soon realize that as light rays converge to the focus, the wave fronts are spherical and the phase of light wave on each of these wave front is a constant. Thus the diffraction integral can be simplified and Fraunhofer diffraction applies. But the validity of the approximation hinges on the conditon rho^2/(lamda f) << 1, where rho is the typical length scale of the diffraction spikes on the focal plane.

Write a short note of it, hope it is useful. Revised on 22/6. See the following two messages.
最後由 昇仔 於 週二 22 6月, 2010 00:20 編輯,總共編輯了 1 次。

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Wah!!
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文章 Wah!! » 週五 18 6月, 2010 09:25

見到呢堆式子就頭痛, 睇都睇唔明..... (讀U時D數學好唔掂, 而家更係乜都唔記得晒 >.<")
昇爺有時間不如搞個光學理論講座呀!

anguslau
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文章 anguslau » 週五 18 6月, 2010 11:21

Wow! This is heavy stuff and reminds me of my school days. Great job and good to see some theory behind what we are discussing here. :wink:

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昇仔
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文章 昇仔 » 週五 18 6月, 2010 11:32

其實冇乜用,因為轉了一個大圈,只是了解到 Wah 師兄方法背後的道理。 :wink:

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Skyobs
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文章 Skyobs » 週六 19 6月, 2010 12:02

Thanks a lot to 昇仔 for the sharing.
[Good Job]
(There seem to be a typo on 3rd line below Fig 1. in sentense "The distance r between ??? ...")

I'm extremely weak in equation derivation skills, and not familiar with detailed maths & optics. There are two points I'm a bit puzzled.
1) Although diffraction spike is small, shouldn't wavefronts from the whole aperture (which subtend a significant angle) be integrated for each point on the spike? Higher order terms in expanding the phase term seem cannot be totally ignored base on a small rho.
2) The final result looks like the form of Fresnel diffraction, the Fourier transform is modulated by an R-dependent phase term. This seems to show that Franuhofer approximation doesn't apply here.

But from practical point of view, can we simply ignore the phase term (including possibly higher order phase terms) in final result because difference in R is so small within the extent of recordable diffraction spike?
By the way, can we simply do FT in spherical coordinates (centre on each image point) on the converging wavefront to get the result without going through those approximations?

(Just throw out some thoughts to those interested to think. No need to try reply in much mathematical detail for the sake of these questions. :lol: )

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文章 昇仔 » 週二 22 6月, 2010 00:51

Skyobs,

Thanks for the comments. The derivation I presented earlier was essentially correct though I admit that the intermediate steps were not "clean" enough because I did it with "bare hand" without making reference to textbooks or online reources. I have revised the notations and some of the details, and used vectors in the derivation instead of moving to spherical coordinates at once. An analytical solution of circular aperture is also included as an example. See the next message. :D
Skyobs 寫: 1) Although diffraction spike is small, shouldn't wavefronts from the whole aperture (which subtend a significant angle) be integrated for each point on the spike? Higher order terms in expanding the phase term seem cannot be totally ignored base on a small rho.
We have indeed integrated da' over the whole wavefront for each point on spikes. Note that (x,y) is fixed on the focal plane but (x',y') are the integration variables which run over the aperture.

No, the higher order terms would not add up to a significant value. The reason is that when the higher order terms in the phase << 1, Exp(i higher order term) ~ 1,

Integrate[ Exp[ i k (useful terms) + i k (higher order term)]]

= Integrate[ Exp[ i k (useful terms)] x Exp [i k (higher order term)] ]

Thus the integrand is only modulated by a function very close to 1.
Skyobs 寫: 2) The final result looks like the form of Fresnel diffraction, the Fourier transform is modulated by an R-dependent phase term. This seems to show that Franuhofer approximation doesn't apply here.
No. it is not. The phase Exp(i kR) is an arbitrary constant phase which is independent of the position on the focal plane, it is not a modulation. In fact R ~ f, the focal length of the objective.
Skyobs 寫: But from practical point of view, can we simply ignore the phase term (including possibly higher order phase terms) in final result because difference in R is so small within the extent of recordable diffraction spike?
In handling problems of wave optics, it is always the variation of the phase factor that determines the result. The reason is that k r = 2 Pi r / lamda is extremely senitive to very small changes in r because lamda ~ 10^-7 m is very small, i.e. delta r / lamda could be very large even if delta r is small in the ordinary sense. Thus the phase varies very rapidly over the domain of integration, leading to many cancellation and that is why wave optics problems are quite complicated.
Skyobs 寫: By the way, can we simply do FT in spherical coordinates (centre on each image point) on the converging wavefront to get the result without going through those approximations?
The phase has to vary linearly with x', y' in order to identify the result with the FT. In comparsion with the rapidly changing phase, other modulation in the integrand is not important at all so many approximations can be made. In fact wave optics problem always involve these kinds of approximation.
Skyobs 寫: (Just throw out some thoughts to those interested to think. No need to try reply in much mathematical detail for the sake of these questions. :lol: )
Revised the derivation as follows anyway. Perhaps there are "cleaner" treatment in Fourier optics textbooks, but I just do it in that way that I can think of :lol:
最後由 昇仔 於 週二 22 6月, 2010 10:37 編輯,總共編輯了 1 次。

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昇仔
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文章 昇仔 » 週二 22 6月, 2010 00:54

Revised derivation:
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Wah!!
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文章 Wah!! » 週三 23 7月, 2014 10:41

無意中找到一篇提到繞射圖案與 FFT 關係的文章:
http://www.erbion.com/index_files/Moder ... s/Ch11.pdf

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