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I took an image of Jupiter through my 8" Dobsonian Telescope, attaching a DSLR and a 1.25" Barlow Lens where the eyepiece goes, as shown in this video: https://www.youtube.com/watch?v=reFxoF3XoaU

Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.

http://www.rocketmime.com/astronomy/Telescope/Magnification.html:

The above website answers the following question, the calculations of which I attempted to mimic.

My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?

Here are the calculations I did in hopes of getting the above result:

$$\textrm{Diameter} = 8'' = 203.2\ \textrm{mm}$$

$$f_{\textrm{ratio}} = \frac{\textrm{focal length of objective}}{203.2\ \textrm{mm}}$$

$$\therefore \textrm{focal length of objective} = 203.2 \cdot 5.9 = 1200\ \textrm{mm}$$

$$\textrm{Magnification} = \frac{\textrm{focal length of objective}}{\textrm{focal length of eyepiece}} = \frac{1200}{x}$$

As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the "focal length of the eyepiece." How can I proceed to calculate the magnification?

James K
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StopReadingThisUsername
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1 Answers1

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The idea of magnification is not relevant to astro-photography, what is relevant is the image scale.

The image scale depends on the focal length of the objective and the size of the sensor. The Barlow effectively modifies the focal length, so if you have a $B\times$ Barlow and an objective of focal length $f_O$ the effective focal length is $f_e=B\times f_O$.

If your sensor has side of length $l$ then the field of view that corresponds to that side is (in radian) $\theta_{rad}=l/f_e$ or in degrees $\theta_{deg}=(l/f_e)\times(180/\pi)$

So if your sensor has a short side length of $\approx 0.015 \mbox{m}$ and a the telescope has a focal length of $1.2 \mbox{m}$ and you are using a $\times2$ Barlow you have the short side of the image corresponds to a field of view of $\approx 21$ minutes of arc.

Conrad Turner
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  • Nice! Are Barlows ever labeled something like "2X" by themselves (in the context of a camera body adapter), or does it depend (watching the mechanical details of the video) both the focal length of the telescope and the distance from the Barlow lens itself to the camera focal plane? How can the questioner obtain this? Or am I missing the point and the Barlow comes together with the telescope and camera adapter? – uhoh May 02 '16 at 10:29
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    Barlow's are generally sold as x2 or x3, without pulling mine out of the telescope kit box I can't tell you if it is so labelled, but find it difficult to believe it is not. This one is clearly labelled – Conrad Turner May 02 '16 at 15:54
  • Great! But now I'm puzzled. The X2 label suggests the magnification of a given eyepiece fully inserted to the stopping point will be twice, but doesn't that actually depend a lot on the focal length of the telescope? This is a geometrical optics question that I'm putting in a comment which I shouldn't. But I think in this context it's necessary to know both focal lengths and the distance explicitly. – uhoh May 02 '16 at 16:30
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    @uhoh the x2 refers to the focal length multiplier not magnification directly. Thus a 1m focal length objective with a 2x Barlow will have an effective focal length of 2m, and with a 3x Barlow an effective focal length of 3m. The magnification is the effective focal length divided by the eye-piece focal length. With a given objective and eye-piece the effextive magnification with a x2 Barlow is as a result twice that without. Note the typos: x2 and x3 rather than 2x ans 3x – Conrad Turner May 02 '16 at 16:56
  • Thanks! This tutorial saved me from doing the math myself. It gives M = 1 + x/f, where f is the Barlow lens focal length, and x is the distance between the Barlow lens element(s) and the focal plane of the eyepiece (or camera sensor). If an eyepiece is mechanically seated properly, then the actual change in magnification will be given by the value for M stamped on the unit. (continued...) – uhoh May 02 '16 at 18:42
  • ...(continued) And if the camera adapter unit then puts the camera sensor at the same location as an eyepiece focal plane would fall, then M will give the correct change in magnification compared to using the camera sans Barlo. So as long as these distances are fixed by the hardware, the X2 will in fact give a 2X boost to the magnification!! – uhoh May 02 '16 at 18:43
  • I'm sorry, but I'm still not exactly sure I have an answer to my question. Because I am using a DSLR in place of any eyepiece with a Barlow, I don't know the focal length of the eyepiece. I know that the Barlow is a 2x Barlow, and that this would be multiplied to the focal length of the eyepiece, but cannot process because I don't know the focal length of my "eyepiece". What should I do/use in place for its value? – StopReadingThisUsername May 03 '16 at 13:52
  • Also, I'm not exactly sure what you mean by the side length of my sensor...could you kindly explain that term? – StopReadingThisUsername May 03 '16 at 13:53
  • @Arjun You camera has a CCD/CMOS chip, it has a size, probably something like 23x15mm for a DSLR, look at the manual/specification – Conrad Turner May 03 '16 at 17:23
  • @ConradTurner Thanks so much. I have one final thing to ask - is there any published source I could use to learn more about this in detail? Like why it works, how we came to this conclusion? I tried looking online but was not very successful.... – StopReadingThisUsername May 03 '16 at 23:37
  • @Arjun Get a book, "Star Ware" by Philip Harrington is generally a good place to start – Conrad Turner May 04 '16 at 03:34