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Focal length of a Barlow

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Sputnik

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Posted 27 October 2020 - 03:46 PM

I am trying to determine the “true” magnifying factor of a Celestron X-cel LX 2X Barlow for use with a LogitechPro 4000 webcam (the Logitech CCD's pixel size = 5.6 micron) in planetary imaging. I am using the Barlow without an eyepiece inserted at the prime focus of my Celestron C8 f10 CAT, then directly inserting the nose piece of the webcam into the Barlow - with or without an appropriate extension between it and the webcam. The amount of distance between the Barlow optics and the webcam sensor will determine the resolution of the image. That distance will depend on, among other factors, the focal length of the Barlow.

For planetary imaging a resolution of around 0.3 arcsec/pixel may be close to optimum for my setup. The resolution ("sampling") of a webcam sensor is (206 x Pixel size)/FL, or 1154/FL for the Logitech, where FL is the amplified (or increased) focal length of the C8 produced by the Barlow. For my setup, FL = 2000 x A, where A is the amplification of the Barlow. (Note: A is not the same thing as the magnifying power of the Barlow, "2X" as indicated on the side of the Barlow.) More precisely A is a function of both the Barlow focal length and the distance between it and the sensor. The equation (from Thierry Legault, Astrophotography) is : A= (d/f) + 1, where d is the distance to the sensor and f is the Barlow focal length. I can somewhat control the distance so that if I know the Barlow focal length I can determine the resolution for any given distance. There will be an optimum distance (or, amplification) for my setup to achieve a reasonable resolution at the sensor.

So, what is the Barlow focal length? I have read that most Barlow's have a focal length between 50 and 100 mm, but that is a large range resulting in a large range for the distance needed to achieve an optimum resolution. For example, if the Barlow focal length is 50 mm then to get 0.3 arcsec/px I would need to separate the Barlow and sensor by about 46 mm. But for a Barlow focal length of 100 mm I would need a separation of twice that (i.e., about 92 mm). That is approximately the difference between using the Barlow alone versus the Barlow plus extension.

Even though a Barlow is a negative lens it nevertheless does have a focal length. So, my question is: how can I determine its focal length? This specification is nowhere to be found, either in the manufacturer's literature or markings on the body of the Barlow. Does anyone know what is the focal length of this Celestron Barlow, or how it can be directly measured?

Celetron AVX

Celestron Orange Tube C8 and f6.3 focal reducer

Williams Optics Z61 with WO field flattener

Pentax K7 DSLR

LogitechPro 4000 webcam

Edited by seckroaad, 27 October 2020 - 03:48 PM.

#2 sg6

sg6

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Posted 27 October 2020 - 05:23 PM

If "A= (d/f) + 1" and you have a 50mm negative lens then with a d of 46mm you get:

A = -0.92+1 = 0.08.

Having said that the focal length is negative you seem to have then just ignored the -ve signage.

Your focal length f is -50mm not +50mm.

Very unsure of "d", I would half suspect it is not the distance from the barlow to the sensor but the distance of the barlow from the original focal plane. In effect it is using the position of the original image formed as the object distance for the negative lens. That could make more sense - well to me. That way both d and f would be negative so d/f becomes positive.

That would however then not be where the sensor has to go. That would appear to be saying that if you inserted a -50mm lend a distance of 46mm back from the original focal plane then you get an image increase of 1.98. What is is not specifing is where that image is created.

That would have to come from 1/F = 1/f1+1/f2 -(D(f1*f2))

Putting the values from above in it gives a focal plane (image) at 31meters, which is probably about right for a -50mm lens 46mm from the object, about infinity is reasonable.

If the barlow was intended as approximately 2x for a refractor of (in general) say around 600mm then putting one in the optics of a 2032 mm focal length system just throws the idea of 2x out the window. To get a guaranteed 2x increase in image you need a different thing to a "barlow". You need a small optical system where the second lens (L2) is 2x the focal length of the first lens (L1). Then the image from the SCT is used as the object for the small system and is placed at the focal length of L1 thus collimating the image and that goes into L2 collimated and forms an image at its focal length that is always 2x the one from the SCT (or any other scope). Guess that is called a Power mate or Tele extender.

Easy enough to make, 2 achro lens of say 20mm and 40mm better. As the L1 image plane is outside or at the end of any tube the overall length would be around 60mm.

#3 Tulloch

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Posted 27 October 2020 - 06:31 PM

Hi there, you don't need to worry about the focal length of the Barlow, for planetary imaging the best focal ratio for your setup is 5x the pixel size in microns (see proof below). So for your camera with 5.6 micron pixels, the focal ratio to chase is around f/28 (=5x5.6), which might be difficult. A 2x barlow with an extension tube might get you close to this (as the magnification from a barlow is not generally constant).

To determine the actual focal length of your setup, use this formula once you have the size of the image of the planet in pixels and arcsec.

Hope this helps, Andrew

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#4 RedLionNJ

RedLionNJ

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Posted 27 October 2020 - 10:54 PM

I'm not QUITE asking the same question as Andrew, but it's along similar lines....

Since you achieve focus by changing the separation of the two mirrors AND, as you do that, potentially moving the separation between the secondary and the sensor (or field lens of the barlow) to achieve focus, the only way to really determine your image scale is to take an image and measure it with something easy like Winjupos, or painfully with a lot of manual effort and math,

In practice, though, I think you'll find the C8 can do better than at 0.30 arcsec/pixel - I would be trying for nearer 0.20.

Sputnik

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Posted 27 October 2020 - 11:54 PM

If "A= (d/f) + 1" and you have a 50mm negative lens then with a d of 46mm you get:

A = -0.92+1 = 0.08.

The formula uses the absolute value of the focal length, so the result is as I have posted. See Legault's book cited in my post. His discussion of this is on page 66.

Sputnik

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Posted 27 October 2020 - 11:58 PM

Very unsure of "d", I would half suspect it is not the distance from the barlow to the sensor but the distance of the barlow from the original focal plane. In effect it is using the position of the original image formed as the object distance for the negative lens. That could make more sense - well to me. That way both d and f would be negative so d/f becomes positive.

"d" is the distance from the focal plane of the Barlow to the webcam sensor. See Legault's book cited in my post, page 66.

Sputnik

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Posted 28 October 2020 - 12:20 AM

Hi there, you don't need to worry about the focal length of the Barlow, for planetary imaging the best focal ratio for your setup is 5x the pixel size in microns (see proof below). So for your camera with 5.6 micron pixels, the focal ratio to chase is around f/28 (=5x5.6), which might be difficult. A 2x barlow with an extension tube might get you close to this (as the magnification from a barlow is not generally constant).

To determine the actual focal length of your setup, use this formula once you have the size of the image of the planet in pixels and arcsec.

Hope this helps, Andrew

Andrew, great help. It's late so I'll absorb this tomorrow. I think that the equation given in the reference is consistent with Legault's recommendations.

Can you point me to the title, author, etc. for the pages you included?

#8 Tulloch

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Posted 28 October 2020 - 12:53 AM

Andrew, great help. It's late so I'll absorb this tomorrow. I think that the equation given in the reference is consistent with Legault's recommendations.

Can you point me to the title, author, etc. for the pages you included?

No worries, I realise it wasn't the actual question you asked, but sometimes it's just easier to measure it directly instead of making assumptions and trying to get it exact. Once you take an image, compare it to the actual size of the planet in arcsec, it's much easier.

The equation came from here, I meant to include a link to the website as well, sorry about that.

http://www.wilmslowa...re/formulae.htm

Andrew

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