What defines the size of the etalon "sweet spot"??

[Merlin66]

I'm continuing my education (following Valery's recent input) on the how's and why's of the etalon's we all use....
A couple of things so far:
The parallel gap we know about between the etalon plates...the incoming beam must always be at a slight (how slight???) angle to get it to work!! This is independent of any subsequent Pressure/ tilt tuning.
(May be the angular size of the solar disk is sufficient?) Which leads to the next issue:
What factors define the size of the sweet spot?
Don't know yet.....if anyone has the answer - please share.

[Merlin66]

http://www.astro.umd.edu/~veilleux/mmtf/basic.html
Amazing what you can find on the net nowadays!!
Also:
http://arxiv.org/ftp/arxiv/papers/1209/1209.1027.pdf

Very interesting paper on various external solar etalons....I find it difficult to accept the "base" tuning to be almost exactly at Ha centre band and not the red edge of the red wing....
Need to follow up.

[swisswalter]

Hi Ken

most interesting stuff. Just right for a rainy day. Thanks for sharing

[Valery]

What factors define the size of the sweet spot?

1. Objective diameter to etalon diameter ratio. As larger the ratio as smaller and more distinct the sweet spot is.
2. band wide. As narrower the band wide, as smaller and more distinct the sweet spot is.
3. Etalon inclination. In the ideal case it must be dead square to the optical axis. Pressure tuned and solid heated etalons are preferable.


Valery.

[marktownley]

I reckon valery has got all bases covered there Ken.

Some interesting reading too, shall dive into that this weekend...

[Merlin66]

Valery,
Thanks for the input.
But talking about external etalons....
The angles of incidence (from the middle/ edges or the solar disk) are constant.
The bandwidth is nominally fixed by the etalon design (finesse/ bandwidth - say a 0.7A system) - say also a constant.

The variation, as far as I can see, then depends on the reflectivity and accuracy of the optical plates.
Yet not all external etalons have the same sized "sweet spot" (or should that now be "JACQUINOT Spot")

[marktownley]

Okey doke, I couldn't wait till the weekend to read these, so read them tonight instead. They're interesting articles Ken, thanks for finding them

[Merlin66]

I got a reply from Cyril, re the "red wing" limits of the etalons he tested....he's 100% sure that they did not extend into the "red wing".
He's sending me further data from his researches.......

[Valery]

Valery,
Thanks for the input.
But talking about external etalons....
The angles of incidence (from the middle/ edges or the solar disk) are constant.
The bandwidth is nominally fixed by the etalon design (finesse/ bandwidth - say a 0.7A system) - say also a constant.

The variation, as far as I can see, then depends on the reflectivity and accuracy of the optical plates.
Yet not all external etalons have the same sized "sweet spot" (or should that now be "JACQUINOT Spot")

Also the temperature and barometric pressure. Normally all etalons work worst during hot wheather.

[Bob Yoesle]

1. Objective diameter to etalon diameter ratio. As larger the ratio as smaller and more distinct the sweet spot is.

Hi Valery,

I just want to note that the objective & etalon diameters might be somewhat proportional for the sweet spot size due to geometry, but the actual sweet spot optical size is defined via field angle magnification, which is solely defined via the focal lengths, and the acceptance angle of the etalon.

The field angle at the Sun's limb is ~ 0.25 degree. This is what a front mounted etalon sees, and why it will generally offer the largest "sweet spot" of on-band performance.

For an internal etalon, if one uses an objective with a 1000 mm FL, and the collimator lens (such as the PSTs) has a FL of 200 mm, the field angle magnification will be 1000 / 200 = 5, and the limb of the Sun will have a field angle of 5 x 0.25 = 1.25 degrees. If a 0.7 angstrom etalon has an acceptance angle of 0.5 degree to be on band, we can see that the extent of on band performance will be relatively small. A narrower bandpass etalon will have a smaller acceptance angle, and the sweet spot will therefore be even smaller.

If we use a collimator lens with a longer focal length (which also requires a proportionally larger etalon to prevent vignetting) we get better performance: Using a collimator with a 500 mm FL results in a magnification of 1000 / 500 = 2, and a field angle of the Sun's limb is now 2 x 0.25 = 0.5 degree, and most if not all of the disk will be within the sweet spot of the etalons acceptance angle.

Bob

[Valery]

1. Objective diameter to etalon diameter ratio. As larger the ratio as smaller and more distinct the sweet spot is.

Hi Valery,

I just want to note that the objective & etalon diameters might be somewhat proportional for the sweet spot size due to geometry, but the actual sweet spot optical size is defined via field angle magnification, which is solely defined via the focal lengths, and the acceptance angle of the etalon.

The field angle at the Sun's limb is ~ 0.25 degree. This is what a front mounted etalon sees, and why it will generally offer the largest "sweet spot" of on-band performance.

For an internal etalon, if one uses an objective with a 1000 mm FL, and the collimator lens (such as the PSTs) has a FL of 200 mm, the field angle magnification will be 1000 / 200 = 5, and the limb of the Sun will have a field angle of 5 x 0.25 = 1.25 degrees. If a 0.7 angstrom etalon has an acceptance angle of 0.5 degree to be on band, we can see that the extent of on band performance will be relatively small. A narrower bandpass etalon will have a smaller acceptance angle, and the sweet spot will therefore be even smaller.

If we use a collimator lens with a longer focal length (which also requires a proportionally larger etalon to prevent vignetting) we get better performance: Using a collimator with a 500 mm FL results in a magnification of 1000 / 500 = 2, and a field angle of the Sun's limb is now 2 x 0.25 = 0.5 degree, and most if not all of the disk will be within the sweet spot of the etalons acceptance angle.

Bob

Hi Bob,

Most if not all front mounted air-spaced etalons are 0.7A or a bit less. So, their acceptance angle is 0.5deg. If we have internal etalon and do not wasting it's diameter and the collimated beam is the same diameter as the etalon, then the sweet spot diameter will be (De/D)degree. De- etalon diameter, D - objective diameter.

[Merlin66]

Valery,
Sorry I disagree....
Bob,
I understand your logic....
The question I'm trying to answer is what's the size of the sweet spot, say in pixels, for a nominated etalon (external/ internal). That way we could possibly judge the "optimum" camera/ etalon combo.
According to discussions with Sylvain Veilleux (MMTF), the area of the sweet spot is constrained by a 1/4 wave variation across the spot....in his opinion the sweet spot for an external etalon and focal lengths we use will exceed the size of the chips we generally use...
I'm still working with him to better define the pixel size of an internal etalon arrangement....

[swisswalter]

Hi Ken

very interesting thread, I'm following it with great pleasure, but I'm not able to contribute

[Sunwatcher]

very interesting thread, I'm following it with great pleasure, but I'm not able to contribute

Well said Walter, "very interesting"!

[swisswalter]

Thank you Tommy

you can't dance on every marriage (our saying)

[Valery]

Valery,
Sorry I disagree....
Bob,
I understand your logic....
The question I'm trying to answer is what's the size of the sweet spot, say in pixels, for a nominated etalon (external/ internal). That way we could possibly judge the "optimum" camera/ etalon combo.
According to discussions with Sylvain Veilleux (MMTF), the area of the sweet spot is constrained by a 1/4 wave variation across the spot....in his opinion the sweet spot for an external etalon and focal lengths we use will exceed the size of the chips we generally use...
I'm still working with him to better define the pixel size of an internal etalon arrangement....

1/4 wave variation - what does it mean?

OK, let's assume that the lens has a deviation of the wave front in one third wavelength.. Does this mean that there will be no area in the focal plane that we can define as a "sweet spot" according to Sylvain Veilleux???

[Merlin66]

Valery,
The 1/4 wave limit is the bandwidth "limits" of the etalon in which it would still appear "on band" and generally gives the physical limits of the sweet spot "disk" (Jacquinot Spot) - the ITF/blocking filters if they are doing their job should suppress the other spectral orders on either side of the "on band" tuned setting.
(Each resonance spectral order is approx. 0.7A FWHM bandwidth and sit 10A apart)

[Valery]

Valery,
The 1/4 wave limit is the bandwidth "limits" of the etalon in which it would still appear "on band" and generally gives the physical limits of the sweet spot "disk" (Jacquinot Spot) - the ITF/blocking filters if they are doing their job should suppress the other spectral orders on either side of the "on band" tuned setting.
(Each resonance spectral order is approx. 0.7A FWHM bandwidth and sit 10A apart)

OK, 6562,8A is the wave length. The 1/4 of it is =1640,7A Where to use this number in the calculation of the sweet spot size (according to you and Sylvain Veilleux (MMTF)???


Valery.

[Merlin66]

Valery,
Sorry for giving mis-information....
I was not reading my own notes!!! The MMTF website gives all the necessary details:
To quote:
For Fabry-Perot etalons to operate as tunable filters, the plate spacing must be small. This ensures that the monochromatic spot at the center of the field of view is large, or equivalently the spectral resolution is low. The monochromatic, or Jacquinot spot, is defined to be the region over which the change in wavelength does not exceed √2 times the etalon bandpass. For a given wavelength, small plate spacing translates into low interference orders.

More specifically, the diameter of the Jacquinot spot is given by
DJ = 2√(2√2) F / √(nN)
~ 3.3636 F / √(nN)

where F = camera focal length (measured in camera pixels)
n = order of interference
N = etalon finesse
Unquote

For the standard Lunt/ Coronado etalons, with a bandwidth of 0.7A, and spacing of 10A, the n value is 656, (for Ha) and the finesse is around 15 (10/0.7).
The "camera focal length" in our case is the focal length (expressed in camera pixels) of the re-imaging lens behind the etalon.

An example:
PST - refocusing lens =200mm
DMK51 camera = 4.4micron pixel
the camera lens therefore = 200/0.0044 = 45450 pixel
The spot diameter
= (3.3636 x 45450)/(sqrt(656 x 15)) = 1540 pixel
This represents about 100% of the chip size of the DMK51 (1600 x 1200 pixel)
I'm still trying to get this formula ( which Sylvain says DOES work) to apply to all our applications.

[swisswalter]

Hi Ken

most interesting, thanks for working it out

[Valery]

Valery,
Sorry for giving mis-information....
I was not reading my own notes!!! The MMTF website gives all the necessary details:
To quote:
For Fabry-Perot etalons to operate as tunable filters, the plate spacing must be small. This ensures that the monochromatic spot at the center of the field of view is large, or equivalently the spectral resolution is low. The monochromatic, or Jacquinot spot, is defined to be the region over which the change in wavelength does not exceed √2 times the etalon bandpass. For a given wavelength, small plate spacing translates into low interference orders.

More specifically, the diameter of the Jacquinot spot is given by
DJ = 2√(2√2) F / √(nN)
~ 3.3636 F / √(nN)

where F = camera focal length (measured in camera pixels)
n = order of interference
N = etalon finesse
Unquote

For the standard Lunt/ Coronado etalons, with a bandwidth of 0.7A, and spacing of 10A, the n value is 656, (for Ha) and the finesse is around 15 (10/0.7).
The "camera focal length" in our case is the focal length (expressed in camera pixels) of the re-imaging lens behind the etalon.

An example:
PST - refocusing lens =200mm
DMK51 camera = 4.4micron pixel
the camera lens therefore = 200/0.0044 = 45450 pixel
The spot diameter
= (3.3636 x 45450)/(sqrt(656 x 15)) = 1540 pixel
This represents about 100% of the chip size of the DMK51 (1600 x 1200 pixel)
I'm still trying to get this formula ( which Sylvain says DOES work) to apply to all our applications.

Normally, 0.7 angstrom etalon has an acceptance angle of 0.5 degree to be on band (DJ = 1degree).
OK, remove the front objective and collimation lens from the PST (your example). Refocusing lens will be the objective while the etalon will be as a front monted etalon. The sun disk size (1/2 degree) will be 400pixels. This mean that DJ (sweet spot) will be almost 4 solar diameters or almost 2 degree?


And the next example: http://solarchat.natca.net/viewtopic.php?f=4&t=11780 The first photo will prove you wrong too.

[Merlin66]

Valery,
I can't strongly disagree!
Even Sylvain has "problems" with our "amateur" set-ups...
We are still trying to work out why there are so significant differences between our results and his formula....
One of his last comments......." I don't have a good explanation for the discrepancy."

IMHO there seems to be a difference of a factor of ten between the size of the sweet spot we record and the MMTF formula....this would give a "sweet spot" based on your example set up of about 1/3 the solar diameter.
(154/420)
(In an actual PST where the final image is at 400mm the "sweet spot" based on the 1/10 correction would be 154/834 = >1/5 the solar disk, 20%.....)

The quest continues......

[Valery]

(In an actual PST where the final image is at 400mm the "sweet spot" based on the 1/10 correction would be 154/834 = >1/5 the solar disk, 20%.....)

The quest continues......

OK, I tell you where is the error.

1. Bandpass difference between actual bandpass and what were taken for calculation.

2. In that example the system was constructed literally at the knee and hence the variations.

If we assume that etalon is 0.7A and has acceptance angle 0.5degree, that internal etalon works at its full aperture, then the sweet spot
size will be (De/D)degree.
In a PST as is - 0.5degree (full sun disk), in 100mm F/10 telescope (at magic -200mm distance before focal plane of the objective) - 0.2 degree or 0.4 of sun disk.

If the De>= Dcl (where Dcl - diameter of the collimation lens, Dcl - collimator lens) then sweet spot diameter Dss = Fc/Fo (degree) or Dss = 2Fc/Fo (sun disk)
where Fc - collimator lens focal length, Fo - objective focal length.

Very good practical rule of thumb.

[marktownley]

A very interesting thread - work is so busy at the moment I don't have time to dive into this in the detail I would like, but I shall be diving in when things quieten down shortly!