Jump to content


Photo

Diffraction spikes and secondary vanes

  • Please log in to reply
82 replies to this topic

#51 Starman1

Starman1

    Vendor (EyepiecesEtc.com)

  • *****
  • Vendors
  • Posts: 22830
  • Joined: 23 Jun 2003
  • Loc: Los Angeles

Posted 13 August 2012 - 06:28 PM

Does that mean that when I see a spike from a bright star extending over a degree away from the star that the entire length of that spike is within the first maximum away from the star?



Does that imply a 2-degree visual spike? It is easy to figure out, since the entire central maximum is approximately D/W times the Airy disc diameter, where D is the aperture diameter and W the vane width. It is never seen in its entirety, though, since the portion close to the minimum is of extremely low intensity. Of course, a sufficiently bright point-like object would ultimately show second and/or third maxima of the vane pattern, but such objects are not a part of the usual sky inventory.


Oy vey!
That means that as the spider vanes get thinner, the length of the spike gets longer, though the minimum after the first maximum near the star should be invisible.

On a 317.5mm primary with 0.5mm blades, that would make the spike length 635X the width of the Airy disc.

That would be substantially less than the visible length of the spike I see.

The Airy disc on the 12.5" f/5 scope is .00665mm (2.43932 x lambda in mm x focal ratio) , so the length of the spike should be 4.23mm.

At the image scale of the 12.5" f/5 scope, .0361 degrees per millimeter, a 2 degree spike (and it's longer than that, but I'm not sure by how much) would cover 55.4mm, or over 13X the width of the first maximum.

Ergo, I am seeing well out into further maxima well beyond the width of the first maximum if your formula D/W x Airy Disc holds up.

The spike length (as you would expect) shortens as the magnitude of the star goes down. By the time I get to 3rd magnitude, the spikes fit entirely inside a field 22mm wide.
And, by 5th magnitude, inside a 10mm field.

So tell me, is there something wrong with my math, or does that mean that many more maxima are visible in the star spikes than just the first one near the star?

#52 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 13 August 2012 - 07:36 PM

Don,

Your math is correct, and the relation is what the theory says (if the sinc function is the right one, the spike length would be somewhat less, D/1.22W times the Airy disc). But if you see that far from the star (assuming zero magnitude), that is at least the 12th subsidiary maxima, which is according to Suiter's graph - which should apply in that respect regardless of the degree of light coherence - some 1,600 times fainter than the central spike maxima. Since the maxima itself is, for this vane area (~160sq.mm) about 500 times fainter than mirror maxima (about 1/10 of the first bright ring in aberration-free aperture, or about 3rd bright ring level, assuming intensity proportional to the area), that makes it some 800,000 times - or 5.9 logarithmic units - fainter than the mirror maxima. For zero magnitude star, this translates into 14.8 magnitude down for that last maxima.

That would be comparable to the brightness of the 38th bright ring in monochromatic light. Since the starlight is not monochromatic, the ring structure is suppressed quite a bit, and one probably can't see more than 10 rings, or so, under any circumstances. But if you really se that far down, you should be able to see nearly ten rings with zero magnitude stars (with the end ring still significantly brighter than the spike end).

Also, the spike should be much (several magnitudes) fainter in its outer portion. You should clearly see the maximas separated (in about a dozen sections), since the minimas are of much lower intensity. Does this fit what you see?

Vla

#53 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 13 August 2012 - 10:19 PM

Vla,

I agree with the thrust of your posting on 8/10 at 8:49 PM. I want to comment and elaborate on the following snip from that post:

Actual vane is much more likely acts as an elongated (inverse) aperture emitting incoherent light, with the first minima lengtwise given by 1.22Lambda/W and width-wise by 1.22Lambda/H. Coheremce factor is important in determining the spike intensity: with coherent light, the complex amplitude is summed up in the pupil and squared at the focus, so doubling the flux increases peak intensity fourfold. With incoherent light, the complex amplitude is squared (individually) in the pupil and summed up at the focus, so doubling the flux doubles the peak intensity. Hence, a four times wider vane will produce as many times brighter spike, not 16 times brighter as on Suiter's graph. Also its full half-length in this case is closer to 3 arc minutes.


Fraunhofer diffraction of a spider vane is indeed modeled by a rectangular aperture (Babinet's principle) illuminated by temporally incoherent but *spatially* coherent light. The image plane receives an accordant spike of widely diffracted light in the form of a spike having a sinc-squared intensity profile (if filtered to narrow the wavelength band enough to throw out the patterns from other wavelengths that fill in the dark fringes.)*

In such a profile, 90% of the light diffracted by the vane falls between the first dark fringes to either side of the peak. If a star of four times the brightness relative to another is observed, the image of the diffraction spike (same vane width) will indeed be four times as bright at the peak of the spike relative to that from the first star. However, if the same brightness star is observed but the vanes are quadrupled in width, then the *total* light contained in the diffracted spike will be (almost exactly) four times that from the narrow vanes. However, the *peak intensity* of the pattern will indeed be 16 times greater. This is because four times the total power (due to the greater obscuration or "inverse aperture" as you so aptly and descriptively stated) will be concentrated into one fourth the area (due to the first dark fringes spanning one fourth the spacing, as demanded by diffraction theory). The two factors of 4 create the combined increase at the peak of 16.

This effect works in a similar way as the impact on Airy disk peak intensity from a star when masking an existing aperture to 1/2 the original diameter. The peak intensity goes down by a factor of 16 in that case as well. The first factor of 4 is due to the area obscuration (half the diameter gives ¼ the area) and another factor of 4 is due to the fact that the transmitted light is spread over a 4x greater area (twice as wide = 4x the area).

Similarly, placing a mask of one quarter the diameter over a telescope aperture drops the peak intensity at the focal place by a factor of 16 squared, or 256 times. The equation for the “intensity” at the peak of the Airy disk (e.g., see Born & Wolf, 6th Ed., eq. 8-5 (14),) is proportional to the square of the aperture area for a constant illumination. This is the same result as that of a rectangular aperture, though in the case of vane obscuration, the length is constant so area changes linearly with the changing parameter, not by the square.

I am confident that Dick Suiter's plot is not a plot based on "narrow" vane width in the sense that the width is close to the wavelength. In our article on spider vanes, we were careful to sample the vane sufficiently (which was a bit of a challenge for a PC in those days.) I also believe that the plot in Dick’s website is completely correct for a constant uniform illumination (for monochromatic light) on common telescope spider vanes of varying width. My recollection from lab measurements of wire diffraction from decades ago is consistent with this, but I will repeat the experiment in the next couple of months and report back. (I’m working on a related investigation and can easily take some pertinent data.)

Cheers,
Bill

*This means that the Fraunhofer diffraction pattern between the temporally-and-spatially coherent point source (e.g., laser) and the spatially-coherent-only point source (star) are identical, but that the treatment of *extended* sources must be done differently between the two, in precisely the way you describe.

#54 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 13 August 2012 - 10:32 PM

Vla,

Thanks for the image in your 5:13 PM image. Similar 'real' images can be found by browsing through the HST image archives. Often in such images, when the Airy disk is saturated, features of the spider vane diffraction can be seen in wonderful detail, including the first diffracted 'line' to either side of the main spike. (The spike appears as three parallel lines with sinc-squared modulation along their length. There are of course an infinite number of such parallel lines arrayed to either side of the main spike, but they have no significant number of photons in them and therefore so not show.) Often these are in color as well, with red outside and blue inside as one looks further from the star image.

Best,
Bill

#55 Starman1

Starman1

    Vendor (EyepiecesEtc.com)

  • *****
  • Vendors
  • Posts: 22830
  • Joined: 23 Jun 2003
  • Loc: Los Angeles

Posted 14 August 2012 - 12:04 AM

Don,

Your math is correct, and the relation is what the theory says (if the sinc function is the right one, the spike length would be somewhat less, D/1.22W times the Airy disc). But if you see that far from the star (assuming zero magnitude), that is at least the 12th subsidiary maxima, which is according to Suiter's graph - which should apply in that respect regardless of the degree of light coherence - some 1,600 times fainter than the central spike maxima. Since the maxima itself is, for this vane area (~160sq.mm) about 500 times fainter than mirror maxima (about 1/10 of the first bright ring in aberration-free aperture, or about 3rd bright ring level, assuming intensity proportional to the area), that makes it some 800,000 times - or 5.9 logarithmic units - fainter than the mirror maxima. For zero magnitude star, this translates into 14.8 magnitude down for that last maxima.

That would be comparable to the brightness of the 38th bright ring in monochromatic light. Since the starlight is not monochromatic, the ring structure is suppressed quite a bit, and one probably can't see more than 10 rings, or so, under any circumstances. But if you really se that far down, you should be able to see nearly ten rings with zero magnitude stars (with the end ring still significantly brighter than the spike end).

Also, the spike should be much (several magnitudes) fainter in its outer portion. You should clearly see the maximas separated (in about a dozen sections), since the minimas are of much lower intensity. Does this fit what you see?

Vla

Yes, at least a dozen minima in the spike (I was calling it waves). At the minimum, the spike intensity is nearly invisible.
For stars like Vega and Deneb and Altair, the spike only dies when the star passes out of the field of view of the scope, and then it abruptly and instantly disappears entirely. If my UTA had a larger opening, my secondary were larger, and the baffles in the mirror box had a larger opening, chances are likely I could still see the spike much farther out. I have little trouble picking up stars in the 16.5-17.0 range, so I would expect to see the spike fade into oblivion at about that magnitude with averted vision.

Oddly, I never really noticed this phenomenon until I owned a primary mirror from Carl Zambuto. The contrast between the sky and faint features in extended objects is truly phenomenal.
Bright star images have substantially less flare and are noticeably smaller than I have seen in other scopes of similar aperture. And the spikes are quite intense. I recently noticed some spikes on a star of eighth magnitude (though they were short) and now you have me interested in counting the minima in the spikes for Vega to see how far out I can see them.

This has been a true learning thread. Thanks. :bow:

#56 Starman1

Starman1

    Vendor (EyepiecesEtc.com)

  • *****
  • Vendors
  • Posts: 22830
  • Joined: 23 Jun 2003
  • Loc: Los Angeles

Posted 14 August 2012 - 12:08 AM

Vla,

Thanks for the image in your 5:13 PM image. Similar 'real' images can be found by browsing through the HST image archives. Often in such images, when the Airy disk is saturated, features of the spider vane diffraction can be seen in wonderful detail, including the first diffracted 'line' to either side of the main spike. (The spike appears as three parallel lines with sinc-squared modulation along their length. There are of course an infinite number of such parallel lines arrayed to either side of the main spike, but they have no significant number of photons in them and therefore so not show.) Often these are in color as well, with red outside and blue inside as one looks further from the star image.

Best,
Bill

Really?
Is that why the spikes on bright stars often appear prismatic?
Incredible.

#57 careysub

careysub

    Surveyor 1

  • *****
  • Posts: 1900
  • Joined: 18 Feb 2011
  • Loc: Rancho Cucamonga, CA

Posted 14 August 2012 - 04:19 AM

Don't forget that eye brightness response is not linear; doubling the intensity will result in a significantly lesser increase in the apparent brightness. Also, a thinner vane will produce somewhat fainter spike, but it will be longer, which may in part offset the brightness decrease, as perceived by the eye.


Thanks, this is an important point to remember if we are talking about visual - not imaging - scopes. Regardless of the underlying physics, we are really interested in what the eye can detect.

Similarly I have seen it argued that curved vanes are pointless since they "only" distribute the same amount of scattered light (everything else being equal) over the image instead of being concentrated in relatively intense spikes. While physically true, if the eye cannot detect the light distributed this way (or its ability is greatly diminished) this is a win.

#58 Jon Isaacs

Jon Isaacs

    ISS

  • *****
  • Posts: 43848
  • Joined: 16 Jun 2004
  • Loc: San Diego and Boulevard, CA

Posted 14 August 2012 - 09:15 AM

While physically true, if the eye cannot detect the light distributed this way (or its ability is greatly diminished) this is a win.



It is there and it decreases the overall image contrast the same way any diffraction effect does, the energy is less concentrated, it is spread out, smeared. It is a small effect and only matters when viewing the planets.

Jon

#59 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 14 August 2012 - 10:45 AM

The design owes something to native american petroglyphs that I've seen ("she who watches").



Fascinating; talking about modern art :) Wonder if "she" is God, or mother - or it comes to the same?

Vla

#60 mark cowan

mark cowan

    Vendor (Veritas Optics)

  • *****
  • Vendors
  • Posts: 4084
  • Joined: 03 Jun 2005
  • Loc: salem, OR

Posted 14 August 2012 - 01:37 PM

Similarly I have seen it argued that curved vanes are pointless since they "only" distribute the same amount of scattered light (everything else being equal) over the image instead of being concentrated in relatively intense spikes. While physically true, if the eye cannot detect the light distributed this way (or its ability is greatly diminished) this is a win.


I don't think its a win. Being necessarily thicker they distribute more diffracted light closer to the airy disc. True, if done right they distribute it in a circular pattern - but nevertheless the impact is greater overall, if less obvious visually. Contrast is affected more strongly.

Best,
Mark

#61 Jarad

Jarad

    Fly Me to the Moon

  • *****
  • Posts: 6384
  • Joined: 28 Apr 2003
  • Loc: Atlanta, GA

Posted 14 August 2012 - 02:47 PM

My current scopes include one with 3 curved vanes, one with 2 curved vanes in the configuration that Don likes, and one with 3 vanes in a T configuration (the focuser is opposite the top of the T). I have previously owned another with a standard 4-vane configuration.

Here are my observations:
1 - Total diffraction increases with more vanes and thicker vanes (as you would expect).
2 - Curved vanes produce a diffuse glow around bright objects (again, as you would expect).
3 - The main difference between curved vanes and straight vanes is aesthetic - the straight ones have obvious spikes, the curved ones have the diffuse glow. Actual observable detail doesn't seem to change much. On fainter objects, the diffuse glow becomes invisible while the spikes are still quite obvious.

For small to medium scopes where you can use curved vanes that are reasonably thin and still be acceptably stable, I think a curved spider gives a more aesthetically pleasing image (personal opinion). Yes, it will have slightly more total diffraction than a 3-vane straight spider, but the actual impact on planetary detail is very minor.

For larger scopes, it becomes impossible to adequately support large secondaries without going to really thick vanes. At that point, I think options like a 3-vane spider (Y or T configurations), wire spiders, or a standard 4-vane are better, primarily due to the need to maintain good collimation and prevent vibration.

The big difference is aesthetic, so I view it as a personal preference issue more than a right or wrong issue. The actual amount of difference in total diffraction bewteen a good curved 3-vane spider and straight 3- or 4-vane spider is very small. You can calculate it as non-0, but I doubt you will be able to observe additional detail on a planet with one vs. the other. This is from direct side-by-side comparison of my 14.5" with curved vanes next to my 14.7" with the 3 straight vanes, both with good mirrors.

The only exception I can think of is double stars with a faint companion where you can rotate the tube to move the faint companion into the gap between the spikes. If that is what you want to do, go with a straight spider.

Jarad

#62 killdabuddha

killdabuddha

    Apollo

  • *****
  • Posts: 1137
  • Joined: 26 Aug 2011

Posted 14 August 2012 - 03:11 PM

Similarly I have seen it argued that curved vanes are pointless since they "only" distribute the same amount of scattered light (everything else being equal) over the image instead of being concentrated in relatively intense spikes. While physically true, if the eye cannot detect the light distributed this way (or its ability is greatly diminished) this is a win.


I don't think its a win. Being necessarily thicker they distribute more diffracted light closer to the airy disc. True, if done right they distribute it in a circular pattern - but nevertheless the impact is greater overall, if less obvious visually. Contrast is affected more strongly.

Best,
Mark


Been waitin for the curved vanes to enter the fray, to say that we're really surprised by the amount of haloed light comin off some of the stars, and to the degree that this diffraction pattern actually washes out nearby stars (with the 11mms and stronger). And yeah, not surprisingly, it also infringes from outside the FOV. Almost wonder whether we wouldn't have been better off with spikes to "carry away" this diffracted light. Just sumthin that we never encountered when researchin curved vanes.

#63 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 15 August 2012 - 12:15 PM

Hi Bill,

Good to talk to you again - and, about diffraction, what else :question:

Fraunhofer diffraction of a spider vane is indeed modeled by a rectangular aperture (Babinet's principle) illuminated by temporally incoherent but *spatially* coherent light.



Any point source emits spatially coherent light, but it will be nearly coherent only if it is nearly monochromatic. Temporal coherence time for white light with even effective intensity over the wavelength range is in the small fractions of a trillionth of a second, which makes it highly incoherent. For the effective intensity peaking in a relatively narrow range, and dropping to near zero toward the ends, coherence time is much longer, but still in the millionths or billionths of a second. So the star light is effectively incoherent.

Suiter's graph is, obviously, for monochromatic light. Polychromatic would have much shallower minimas, with the intensity plot resembling continuous, decaying wavy line flattening farther out. The difference in pattern shape may not have much importance, but monochromatic (coherent) vs. incoherent should.

The image plane receives an accordant spike of widely diffracted light in the form of a spike having a sinc-squared intensity profile...



That is correct, and my assumption that the intensity relation for circular aperture applies, in a general form, to rectangular aperture, is not. Hecht gives relation for rectangular aperture as a product of the squared sinc functions for both, height and width. The first minima is Lambdaf/W for the spike length, and Lambdaf/H, for spike width, f, W, H being the focal length, vane width and height (length), respectively.

However, if the same brightness star is observed but the vanes are quadrupled in width, then the *total* light contained in the diffracted spike will be (almost exactly) four times that from the narrow vanes. However, the *peak intensity* of the pattern will indeed be 16 times greater. This is because four times the total power (due to the greater obscuration or "inverse aperture" as you so aptly and descriptively stated) will be concentrated into one fourth the area (due to the first dark fringes spanning one fourth the spacing, as demanded by diffraction theory). The two factors of 4 create the combined increase at the peak of 16.



Yes, and this is where the controversy comes from: the general rule for how combining waves produce intensity is that coherent waves, due to their sustained overlapping, have their fields (amplitude) added first, and then squared for the irradiance. On the other hand, incoherent waves have their individual amplitudes squared first, and then added up.

In other words, for any given flux of x waves of amplitude A, irradiance produced by coherent light is (xA)^2 vs, xA^2 for incoherent light, i.e. larger by a factor of x. Doubling x increases irradiance by a factor x^2=4 in coherent light, and by a factor of x=2 in incoherent light.

If we assume that the central intensity for incoherent light is a sum of individual wave amplitudes squared, then it only - and directly - depends on the size of flux, i.e. aperture (vane) area, and is not affected by the changes in pattern size.

So, what's the catch 22?

The equation for the “intensity” at the peak of the Airy disk (e.g., see Born & Wolf, 6th Ed., eq. 8-5 (14),) is proportional to the square of the aperture area for a constant illumination.



Don't have Born and Wolf, but Schroeder cites them giving quite simple relation for the actual central intensity of PSF as:

I=PiE/(2LambdaF)^2

where E is the flux (proportional to the aperture area) and F the focal ratio.

Vla

#64 Starman1

Starman1

    Vendor (EyepiecesEtc.com)

  • *****
  • Vendors
  • Posts: 22830
  • Joined: 23 Jun 2003
  • Loc: Los Angeles

Posted 15 August 2012 - 12:26 PM

Suiter's graph is, obviously, for monochromatic light. Polychromatic would have much shallower minimas, with the intensity plot resembling continuous, decaying wavy line flattening farther out. The difference in pattern shape may not have much importance, but monochromatic (coherent) vs. incoherent should.
Vla


The light bulb in the brain turns on--slowly, like a fluorescent, but it does go on.
That is exactly what I see. A prismatic, wavy line with minima that never quite disappear.
Also, the very outer end of the spike, way out from the bright star, appears monochromatic.
May I presume that is because the intensity of the maxima has dropped to the point where my vision is now scotopic, or is it due to the nature of the spike?

#65 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 15 August 2012 - 12:31 PM

Bill,

Thanks for the image in your 5:13 PM image. Similar 'real' images can be found by browsing through the HST image archives. Often in such images, when the Airy disk is saturated, features of the spider vane diffraction can be seen in wonderful detail, including the first diffracted 'line' to either side of the main spike.



My pleasure :) OSLO is quite inefficient in simulating diffraction effects of such small obstructions. One has to use high number of aperture divisions, which slows calculation down too much to allow for really good image (i.e. for high enough number of aperture divisions). It also bases the image on the nominal intensity, which is much different than the actual visual image based on nearly logarithmic eye intensity response. So to bring out faint portions, I had to normalize pattern to 1/33 of its peak intensity (which makes any intensity 1/33 or higher appear white), plus bring out some more of it using photo-enhancing software. So, like you say, central maxima is *very* saturated.

There may be a better way to do it in OSLO, but I haven't figured it out yet.

Vla

#66 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 15 August 2012 - 12:46 PM

Don,

This has been a true learning thread. Thanks.



Likewise. That happens all the time, even when I don't participate directly. I learn and realize things, and then go back to add it or correct on my site. I had no idea that thin vanes can produce such long visual spikes. Out of curiosity, I tried to illustrate it on a star field (Pleiades, 1-degree field, the faintest stars with spikes are about 7-th magnitude; actual view probably wouldn't show that much, due to the presence of nebulosity). Didn't know it can be that bad at that aperture size (you call that "good mirror" ;)).

Vla

Attached Files



#67 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 15 August 2012 - 12:55 PM

That is exactly what I see. A prismatic, wavy line with minima that never quite disappear.
Also, the very outer end of the spike, way out from the bright star, appears monochromatic.
May I presume that is because the intensity of the maxima has dropped to the point where my vision is now scotopic, or is it due to the nature of the spike?



If you see colors closer to the star, then the eye is not scotopic. Most likely, the intensity is too low to allow detection of any but wavelengths to which the eye is most sensitive.

Vla

#68 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 15 August 2012 - 08:03 PM

Hi, Don,

RE your 1:08AM, 8/14 response to my earlier post concerning colors in the diffraction spikes:

Colors in the diffraction spikes occur because light at every wavelength in the incident beam will create its own sinc-squared intensity pattern, scaled by that wavelength. The angular spread in radians between the center of the pattern and the first dark fringe, at a given wavelength, is just the wavelength divided by the vane width. Keeping with Vla's reminder that a stellar source is spatially -but not temporally- coherent means that each wave adds as intensity, or incoherently. So the first diffracted lobe, which is an overlap of all the wavelengths in the beam, will be white in the center (for a white star) with the red red pattern - larger because of its larger wavelength - leaking into the first dark fringes to either side. The first bright fringes on either side of the center lobe will be blue toward the inside of the pattern and red toward the outside. Looking out past a few fringes, the colors mix up again (in a predictable way), because of the wavelength-dependent scaling of the pattern width. (See Vla's post at 8:49PM on 8/10 where he cites the wavelength dependence as well.)

Attached is an image of a star taken by the Hubble showing the colors and other structure. I've put others showing many of the phenomena discussed in this thread.

By the way, thanks for posting your insights on the telescope FOV impacts on the spikes. I'd not thought of that or seen it anywhere else. Good stuff.

Best,
Bill

[image]http://www.cloudynights.com/photopost/showphoto.php?photo=26021&password=&sort=7&thecat=500[/image]

#69 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 15 August 2012 - 08:28 PM

Dang, I just can't get the hang of imbedding images in these posts! Anyone, please educate me! Anyway, my gallery has several snips from Hubble images showing spikes around stars. Type in "wzmek" in the Member's Gallery search box to get to the images. The images labeled "Hubble Vane Diffraction 7", "Hubble spider diffraction", and the image "Butterfly2 ..." show fine features of the patterns, and the images "Hubble diffraction 3" and "Hubble diffraction 4" show the colors that bleed into the dark fringes.

Bill

#70 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 15 August 2012 - 09:05 PM

Hi Bill,

Good to talk to you again - and, about diffraction, what else



Same here! Hope all is well with you.

RE Schroeder's equation, B & W have the same equation too, and next to it in their text, a variation using power (energy per unit time) instead of source strength per unit area. This form of the relation has the area dependence as follows: I(0) = PA/(wavelength F)^2, where I(0) is the intensity at the image peak, P is the optical power through the aperture area A, and F is the distance to the image plane. The area can be written as WH using your convention in your earlier posts. Therefore, if the vane area is quadrupled by making the width equal 4W, then the power through the new vane area is 4P, so the peak intensity goes up by a factor of 16. Cool, no?

I hope everyone can see the vane diffraction images. I wanted to look up the HST relative vane width but was unable to find it in the STSci.org website. I may be able to dig it up at work. Much of the old build paperwork still exists in storage one floor above me, but I dread having to look through it. Some of the old-timers here may have something squirreled away. My very faint recollection is that the relative width was 0.04 but I'll let you know.

Best,
Bill

#71 Jon Isaacs

Jon Isaacs

    ISS

  • *****
  • Posts: 43848
  • Joined: 16 Jun 2004
  • Loc: San Diego and Boulevard, CA

Posted 15 August 2012 - 10:28 PM

Dang, I just can't get the hang of imbedding images in these posts! Anyone, please educate me! Anyway, my gallery has several snips from Hubble images showing spikes around stars. Type in "wzmek" in the Member's Gallery search box to get to the images. The images labeled "Hubble Vane Diffraction 7", "Hubble spider diffraction", and the image "Butterfly2 ..." show fine features of the patterns, and the images "Hubble diffraction 3" and "Hubble diffraction 4" show the colors that bleed into the dark fringes.

Bill


Bill:

I don't know what the H you are talking about, I think you do.. However I can help with the images. :)

If you copy the webpage of the image, then you need to use the URL function.

Bill's Photo

If you want to "hot link" the image, then you need to "copy the image location", it will be a jpg file.

Posted Image

Jon

#72 mark cowan

mark cowan

    Vendor (Veritas Optics)

  • *****
  • Vendors
  • Posts: 4084
  • Joined: 03 Jun 2005
  • Loc: salem, OR

Posted 16 August 2012 - 02:26 AM

OK, it's not Hubble but if you want to see firsthand all the nice chromatic spider spike effects and the lobes in action just lay a yardstick (or ruler as applies) across the front of your OTA while viewing Vega or Sirius. Then turn it sideways (minimum obstruction) and back again while watching through the EP. It's quite a light show, and will show you exactly what's going on with changing vane thickness!

Best,
Mark

#73 wh48gs

wh48gs

    Surveyor 1

  • *****
  • Posts: 1613
  • Joined: 02 Mar 2007

Posted 16 August 2012 - 08:33 AM

Bill,

RE Schroeder's equation, B & W have the same equation too, and next to it in their text, a variation using power (energy per unit time) instead of source strength per unit area. This form of the relation has the area dependence as follows: I(0) = PA/(wavelength F)^2, where I(0) is the intensity at the image peak, P is the optical power through the aperture area A, and F is the distance to the image plane.



The equation should be I(0)=PA/(Lambda*f)^2, where "f' is the focal length. With A=(Pi)D^2/4, it can be written as (Pi)P/(2LambdaF)^2, which is same as Schroeder's form, with the power in the pupil P equaling electromagnetic flux. Doubling the aperture quadruples the flux, but those two will increase the intensity 16x only if the focal length remains unchanged, i.e. if the focal ratio halves. Thus, that portion of intensity change is due to the pattern size, not the flux/aperture (which is logical, since aperture area and flux are the two sides of the same coin, and do not give separate contributions to image intensity).

But that is exactly where the apparent controversy is. The relation indicates that for any given flux, the resulting central intensity will scale inversely to F^2, i.e. inversely to the base area of the 3-D PSF, which makes logical sense. We can only fit the same amount of energy in a given volume with the base reduced in half, linearly, if we quadruple its height. But if we look at the flux itself, represented by an "x" number of wave emitters in the pupil, their individual contributions to the center of diffraction pattern does not change with the change in F-number and change in pattern size caused by it. Since, by definition, in aberration-free aperture all these individual contributions meet in phase with the maximum constructive interference, this intensity value should be constant. How does it get to change inversely to the square of focal ratio, at a given constant flux?

Vla

#74 Mike Spooner

Mike Spooner

    Vendor (mirrors)

  • -----
  • Vendors
  • Posts: 147
  • Joined: 06 Aug 2010

Posted 16 August 2012 - 11:02 AM

RE Schroeder's equation,


Oh, sigh,star,sigh - the things I used to know. :(

Anyway, back on topic, I use curved spiders and I'm not sure I really notice an increase in diffused light but I do marvel at how often the diffraction spikes of a regular spider line up with axis of a double star orbit or planetary moon observation - I mean if only my chances in Vegas were that good!
Next time I get a chance to observe, I'll do a little more critical looking for the effects on stars close to and just outside the FOV. Interesting topic- thanks!

Best,
--Mike

#75 wzmek

wzmek

    Vostok 1

  • -----
  • Posts: 145
  • Joined: 01 Feb 2007
  • Loc: Bethlehem, CT

Posted 16 August 2012 - 04:44 PM

Hi, Vla,

...if we look at the flux itself, represented by an "x" number of wave emitters in the pupil, their individual contributions to the center of diffraction pattern does not change with the change in F-number and change in pattern size caused by it. Since, by definition, in aberration-free aperture all these individual contributions meet in phase with the maximum constructive interference, this intensity value should be constant. How does it get to change inversely to the square of focal ratio, at a given constant flux?



Here's what I think answers your question, assuming I'm understanding it correctly. I think that you are apparently making reference to Huygens point sources filling the aperture as an interpretation of the incident field. I think that the resolution to the dilemma can be found in the Fresnel-Kirchhoff diffraction equation, which shows that the electric field strength at an observation point (say at the focal plane of a telescope), from a wave incident on a diffracting aperture (say a telescope objective), is equivalent to an infinite number of point sources of spherically expanding electric fields all in phase. The factor in the Kirchhoff equation representing these wavelets is U*exp(ikr)/r, where U is the field strength of the wave, k is 2pi/lambda, and r is the distance from the source point in the aperture to the observation point. The key is the 1/r factor. The electric field strength during propagation is therefore inversely proportional to the distance to the observation point. To get from field strength to intensity, the field strength must be squared (up to a constant). This says that each wavelet contributes a fixed energy to the total pattern (U is independent of distance to the observation point) but the field as the wavelet propagates dies down due to being spread out, exactly at the rate that produces the 1/r^2 dependence for flux. This will cause the peak intensity to "...change inversely to the square of focal ratio, at a given constant flux." What do you think?

Best,
Bill






Cloudy Nights LLC
Cloudy Nights Sponsor: Astronomics