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

<p>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 $f_e=B\times f_O$.</p>

<p>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 $\theta_{deg}=(l/f_e)\times(180/\pi)$</p>

<p>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.</p>

CC BY-SA 3.0
2016-05-02T10:29:53.083Z
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
CC BY-SA 3.0
2016-05-02T15:54:29.507Z
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](http://www.rothervalleyoptics.co.uk/skywatcher-x2-economy-barlow-lens-125.html)
Conrad Turner
CC BY-SA 3.0
2016-05-02T16:30:15.540Z
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
CC BY-SA 3.0
2016-05-02T16:56:12.100Z
@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
CC BY-SA 3.0
2016-05-02T18:42:33.683Z
Thanks! [This tutorial](http://astunit.com/astunit_tutorial.php?topic=barlow) 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
CC BY-SA 3.0
2016-05-02T18:43:36.640Z
...(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
CC BY-SA 3.0
2016-05-03T13:52:39.177Z
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
CC BY-SA 3.0
2016-05-03T13:53:34.480Z
Also, I'm not exactly sure what you mean by the side length of my sensor...could you kindly explain that term?
StopReadingThisUsername
CC BY-SA 3.0
2016-05-03T17:23:35.310Z
@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
CC BY-SA 3.0
2016-05-03T23:37:11.680Z
@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
CC BY-SA 3.0
2016-05-04T03:34:50.543Z
@Arjun Get a book, "Star Ware" by Philip Harrington is generally a good place to start
Conrad Turner
CC BY-SA 3.0
2016-04-30T03:59:41.633Z
2016-05-02T16:54:19.143Z
2016-05-02T16:54:19.143Z
Conrad Turner
<p><a href="https://i.sstatic.net/BfYBZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BfYBZ.jpg" alt="enter image description here"></a></p>

<p>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: <a href="https://www.youtube.com/watch?v=reFxoF3XoaU" rel="nofollow noreferrer">https://www.youtube.com/watch?v=reFxoF3XoaU</a></p>

<p>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.</p>

<p><a href="http://www.rocketmime.com/astronomy/Telescope/Magnification.html" rel="nofollow noreferrer">http://www.rocketmime.com/astronomy/Telescope/Magnification.html</a>:</p>

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

<blockquote>
  <p>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?</p>
</blockquote>

<p>Here are the calculations I did in hopes of getting the above result:</p>

<p>$$\textrm{Diameter} = 8'' = 203.2\ \textrm{mm}$$</p>

<p>$$f_{\textrm{ratio}} = \frac{\textrm{focal length of objective}}{203.2\ \textrm{mm}}$$</p>

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

<p>$$\textrm{Magnification} = \frac{\textrm{focal length of objective}}{\textrm{focal length of eyepiece}} = \frac{1200}{x}$$</p>

<p>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?</p>

<optics>
Magnification of a telescope
StopReadingThisUsername
# Magnification of a telescope

(Asked by: username_0 on 2016-04-29)

[![enter image description here](https://i.sstatic.net/BfYBZ.jpg)](https://i.sstatic.net/BfYBZ.jpg)

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?

**Tags:** optics

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## 1 answers

### Answer by username_1 on 2016-04-30. Score: 6

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.

> **Comments:**
> **username_2**: 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?  
> **username_1**: 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](http://www.rothervalleyoptics.co.uk/skywatcher-x2-economy-barlow-lens-125.html)  
> **username_2**: 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.  
> **username_1**: @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  
> **username_2**: Thanks! [This tutorial](http://astunit.com/astunit_tutorial.php?topic=barlow) 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...)  
> **username_2**: ...(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!!  
> **username_0**: 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?  
> **username_0**: Also, I'm not exactly sure what you mean by the side length of my sensor...could you kindly explain that term?  
> **username_1**: @Arjun You camera has a CCD/CMOS chip, it has a size, probably something like 23x15mm for a DSLR, look at the manual/specification  
> **username_0**: @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....  
> **username_1**: @Arjun Get a book, "Star Ware" by Philip Harrington is generally a good place to start  

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