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Swithin StCleeve

A 'working out the magnification' question

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I've bought a dinky little 4" mirror scope, (the Skywatcher Heritage 100), which I've been having lots of fun with. I looked at Saturn the other day with the widefield eyepiece, and it was tiny. So I started to work out the magnification.

At 400mm focal length, with a 25mm eyepiece, I've got a 16 times magnification. Is that right? (400/25 = 16). Which would explain the apparent smallness of the planet. The 10mm eyepiece gives me 40X, and yes, I could see more detail, (and nice and sharp too).
Then I attached my camera, and using a T-ring on my SLR, I got a decent enough shot, in that it was a good representation of what my eyes saw - but I had to use the barlow where the eyepiece should be. So my question to the group is... how do I work out the magnification of my image when there's no eyepiece, just a barlow?

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PeterCPC    1,720

If you want to see what the image scale is, you could use the FOV calculator - look in the top under resources and then FOV calculator. Put in your equipment and it will show a representation of what you will see.

Peter

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Cornelius Varley    1,725

The term "magnification" can't be applied to imaging. The correct term is FOV or image scale is measured in arcsec/pixel. By using the FOV calculator in the resources section you will be able to compare different focal lengths and sensor sizes.

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ollypenrice    17,056
22 minutes ago, Swithin StCleeve said:

So my question to the group is... how do I work out the magnification of my image when there's no eyepiece, just a barlow?

You don't, because the term 'magnification' has no meaning without an eyepiece. When we use the term 'magnify' we have to mean 'make something larger.' In visual observing the 'something' is the size of the image on the retina and so the observer's perception of its size. This has no equivalent in photography. We can talk about the size of the image projected onto the chip and this is dependent on the focal length of the system, so a 2x Barlow will make your image twice as big on the chip (more or less, because the spacing has a significant effect as well.) But twice as big as what? As it was before, but how big was that?  There is no 'natural' size for a photographic image as there is for an observed one.

In the days of film the professionals' term was 'plate scale,' expressed as arcseconds (of sky) per mm (on the plate.)

In digital imaging it is more useful to define resolution in terms of arcsecnds (of sky) per pixel.

Olly

PS Crossed with Peter, above, but we agree. :icon_biggrin:

Edited by ollypenrice
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I went on the FOV bit and I have to give a value for the eyepiece, but I'm not using one.

If I attach my DSLR to my 8" Skywatcher (which has with the barlow (and no eyepiece), the moon easily fills the field of view. So there is magnification there, but you're saying you don't call it magnification?

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furrysocks2    472

I loved my little 76mm f/4.6 table-top reflector for it's relaively low magnification - my most memorable views of the Double Cluster have been through that scope, balanced on top of a bin in a park at my first star party. Your 100mm f/4 looks like a nice little scope!

If you're mathematically inclined, take a spreadsheet and it's not too difficult to do the trig to calculate apparent field of view and arc-sec/pixel for a given camera/barlow combination. I'd like a larger sensor than the 1/4" webcam I'm using now for a wider field of view, but mind that the size of the secondary etc will put a limitation on how large a sensor it can fully illuminate. All these things are connected if you "do the math".

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Looks like it's more complicated than I thought. When I look through my telescope eyepiece, I'm looking at an object 30X, or 60X what I'd normally see. But when I look through my camera eyepiece, and I see a similar 'enlarged' image - I can't put a number that one, because I'm looking through my camera, and not my eyepiece?

And yet, it's still magnified. I mean, I can see it's bigger.

Can't get my head around that.

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PeterCPC    1,720
15 minutes ago, Swithin StCleeve said:

I went on the FOV bit and I have to give a value for the eyepiece, but I'm not using one.

If I attach my DSLR to my 8" Skywatcher (which has with the barlow (and no eyepiece), the moon easily fills the field of view. So there is magnification there, but you're saying you don't call it magnification?

If you want to use FOV calculator with a camera then swap to imaging in the FOV calculator.

Peter

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furrysocks2    472

Take one of your commonly used eyepieces... the magnification is given by telescope focal length / eyepiece focal length, so the "true field of view" (how large or small a portion of the sky is visible) changes but the "apparent field of view" (how wide that portion of sky looks in the eyepiece) should remain the same. It depends on the eyepiece, 60 or 80 degrees or whatever. With that, you can draw a comparison against eyepiece magnification, eg that sensor in that scope gives the same true field of view as this eyepiece would, or as it would if it was Nx magnification.

For example, an APS-C sensor is 22.3 x 14.9mm and in your 100/400 scope should give a field of view of 1.60° x 1.07°. A 60 degree 25mm eyepiece would give a magnification of 16 and therefore a true field of view of 60/16 = 3.75°. A 10mm eyepiece would give a magnification of 400/10 = 40x and so a true field of view of 60/40 = 1.5°. Of course, you'll see more field of view through an 80° eyepiece as opposed to a 60°, but you can work out approximately what magnification you get on a camera, with the caveats that you have to know what eyepiece you're comparing it to, and that the pixel size will only give you so much resolution or arc-sec/pixel.

Apologies in advance if I've made any mistakes in the above.

Edited by furrysocks2
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ollypenrice    17,056
59 minutes ago, Swithin StCleeve said:

Looks like it's more complicated than I thought. When I look through my telescope eyepiece, I'm looking at an object 30X, or 60X what I'd normally see. But when I look through my camera eyepiece, and I see a similar 'enlarged' image - I can't put a number that one, because I'm looking through my camera, and not my eyepiece?

And yet, it's still magnified. I mean, I can see it's bigger.

Can't get my head around that.

When you look through a telescope or bins, the apparent size of the moon is bigger, by a definable amount, than than the image as you perceived it unaided. When you look through the viewfinder of the camera in a telescope then, yes, you could still calculate the magnification that the scope and 'viewfinder-as-eyepiece' were giving you. It would make the image you perceive in real time larger by a factor of 'x.' But this is the image on your eye, it is not the photograph. Consider the two identical camera bodies in identical scopes with the same sized chips. One of those chips has pixels half the size of the other, though, so twice as many of them. You will see no difference whatever through the viewfinder but when the digital images from each system are shown at full size (1 camera pixel = 1 screen pixel) on a PC screen the small pixel camera will produce an image four times larger by area than the large pixel camera. 

The key point is that both the chip and the optics influence the final size of your target in a full size image. By changing either you can, say, make an image of the moon at full size 2x larger than it was before. This may be useful to you but how does anyone else know how big it was before? However, when we say that binoculars magnify 8 times they magnify something we can all recognize by 8x, namely the size we will see the target.

If I tell you, 'I'm imaging at 1 arcsecnd per pixel,' then you know exactly what I mean.

If I put a 2x Barlow in my system and tell you that I'm imaging at 0.5 arcseconds per pixel that is also meaningful to you.

But if I say nothing about my system and post an image via the Barlow which I describe as being at 2x magnification it means nothing to you (or anyone else) because they just ask '2 x what?'

Olly

 

Edited by ollypenrice
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IThanks guys. The FOV calculator has given me quite a bit of info, none of which I understand in terms of how much bigger the image is compared to your normal field of view. I though it was a simple question, looks like it isn't.

IMG_3049.jpg

Here's an un-cropped picture of the moon on Tuesday, taken from the little telescope. If I showed someone that photo, there's a chance they'd say "what magnification is that?". I'd feel a bit silly saying "you can't say magnification because it's a photo, but it's so many arc-seconds and so many pixels".

I'd probably say "about 60 times", because that's what it looks like.

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Thanks for the last two replies Olly & Furry. There's quite a bit there to get my head around. I suppose I'm coming at it from the simple layman's terms of 'magnification', and there's a lot more to it than that. I just really wanted a number to tell people if I showed them my photo.

Edited by Swithin StCleeve
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ollypenrice    17,056
19 minutes ago, Swithin StCleeve said:

IThanks guys. The FOV calculator has given me quite a bit of info, none of which I understand in terms of how much bigger the image is compared to your normal field of view. I though it was a simple question, looks like it isn't.

IMG_3049.jpg

Here's an un-cropped picture of the moon on Tuesday, taken from the little telescope. If I showed someone that photo, there's a chance they'd say "what magnification is that?". I'd feel a bit silly saying "you can't say magnification because it's a photo, but it's so many arc-seconds and so many pixels".

I'd probably say "about 60 times", because that's what it looks like.

I'm asked this all the time as an astrophotographer and imaging tutor/provider. I answer correctly by saying the question cannot be answered - and I say why.

If I look at your moon picture above (it's a nice one, by the way) I see it occupying about a quarter of my screen. You tell me it's an image which magnifies by 60x. Hmmm: I hit 'Control+' twice and your image is suddenly almost full screen. So has it suddenly become an image at 120x magnification? But then I slide my chair back from the screen so I'm further away. Has it now gone back to being a 60x image?  You see the problem? Magnification is a meaningless term in the context of photographs.

Olly

PS I should stress that there is absolutely nothing wrong with your question and, as I say, many people ask it. It often comes up on here, too. Furrysocks2 has propsed a way of making a comparison between visual and photographic observing by making field of view a reference point. While this can be done, personally I wouldn't choose to come at the discussion that way.

Edited by ollypenrice
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furrysocks2    472
20 minutes ago, ollypenrice said:

Furrysocks2 has propsed a way of making a comparison between visual and photographic observing by making field of view a reference point.

Agreed, it's contrived. The best reference point for imaging is arc-sec/pixel as you say, which does not compare visual to photographic at all.

It is at all meaningful, therefore, to compare "magnification" against a given scope/EP and the limit of resolution at an observer's eye? Conditions and patience would be of course be factors too and the reference point now a function of the individual, rather than the field of view. Perhaps this is overly contrived, too.

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Dave S    109

I've come to hate the term magnification as it is synonymous with cheap terribly nasty scopes. They get advertised with a real push on what magnification they give. I saw one the other day and it was just called techlon (or something) 160x magnification scope. Not even a hint of the physical dimensions of the thing. 

On a practical note. FOV for me is infinitely more useful not least because software like stellarium lists objects with their size in Deg,min,sec of arc which means I know straight away which piece of kit I need to see/image it as I know what FOV my setups have.

Dave...

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I can understand why it's impossible to technically give a magnification for what I'm asking. But seeing as I can look in my camera's viewfinder, and see an image of the 'bigger moon', just the same as when I look in my telescopes eyepiece, it's not un-natural for me to compare apparent magnification, (or whatever word you want to substitute for the fact that the image I'm looking at is X times bigger because I have a telescope).

If I held my phone up to the telescope's 25mm lens, with 1 400mm focal length, and took a photo, would I then be taking a photo of the moon at a 16X magnification? Would it be okay to use the word in that situation, I mean.

Edited by Swithin StCleeve
typo

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ollypenrice    17,056
15 minutes ago, Swithin StCleeve said:

 

If I held my phone up to the telescope's 25mm lens, with 1 400mm focal length, and took a photo, would I then be taking a photo of the moon at a 16X magnification? Would it be okay to use the word in that situation, I mean.

16x what?

Olly

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27 minutes ago, Dave S said:

I've come to hate the term magnification as it is synonymous with cheap terribly nasty scopes. They get advertised with a real push on what magnification they give. I saw one the other day and it was just called techlon (or something) 160x magnification scope. Not even a hint of the physical dimensions of the thing. 

Dave...

 

Yea. My first telescope was a cheap nasty thing advertising massive magnification.

5 minutes ago, ollypenrice said:

16x what?

Olly

16 times bigger than what I'd see if I hadn't got a telescope?

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Alien 13    3,460

I think that the question is quite reasonable in reverse, for example you see a lunar image on your PC screen and want to know what sort of magnification you would need to see something similar through an eyepiece..

Alan

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ollypenrice    17,056
1 minute ago, Swithin StCleeve said:

 

Yea. My first telescope was a cheap nasty thing advertising massive magnification.

16 times bigger than what I'd see if I hadn't got a telescope?

You take an image of the moon through your powerful scope. I take one of the moon through my granny-style flip phone, no scope. We both blow up the images to give a print of the moon that is 20 cm across. They have the same magnification. By definition. Will my image be as good as yours? Mine will show an array of blobby pixels, yours will show craters. But the moon is the same size in both. So what do you mean by magnification? You mean resolution of detail. How do you quantify resolution of detail? In arcseconds per pixel. And that's it.

Olly

 

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steppenwolf    4,095

I have to say that I agree completely that 'magnification' is meaningless in imaging, I can magnify your image to any scale that I choose in relation to the original data, using image processing but I still won't be able to give you a magnification factor over the original!

This won't answer your question - there is no answer - but in the good old days of film cameras, the standard lens that came with a SLR camera was a 50mm lens. If you looked through the viewfinder of the camera with one eye and with the naked eye with your other eye then the two 'images' would pretty much overlap. Lenses longer than 50mm became 'telephoto' lenses and lenses shorter than 50mm became 'wide angle' lenses. However, that 1 to 1 view purely applied TO THE VIEW THROUGH THE VIEWFINDER.

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Stub Mandrel    5,423

At the risk of being mocked, there is the concept of 35mm equivalent focal length.

A 50mm lens on a 35mm camera or full frame DSLR gives a 'natural looking' perspective to images that we perceive as showing things as we see them with the naked eye if taken at the same spot.

A 1000mm lens of a 35mm camera can be interpreted as '20x' which is meaningful if you want to photograph birdies as you only need to be 10 meters from a bird instead of 18" to get the same image.

Now perspective is a bit meaningless for all these sky things that, to all intents and purposes, appear as if they are on a flat surface at infinity.

But if you  insist on a magnification you could say a 1000mm scope with a 35mm camera is still 20x magnification.

But it is meaningless really for digital images because this ignores the resolution of the sensor. The same optics and sensor size but at 16MB or 1MB will give two pictures, the 16MB one can be use four times as large or have the middle cut out giving another 'four times magnification' with no optical magnification.

We desperately need a simple language that describes and inter-relates focal length, aperture, pixel size, sensor size and sensor sensitivity, but how can anything that covers five 'loosely correlated' variables be simple?

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ollypenrice    17,056
1 hour ago, Stub Mandrel said:

 

We desperately need a simple language that describes and inter-relates focal length, aperture, pixel size, sensor size and sensor sensitivity, but how can anything that covers five 'loosely correlated' variables be simple?

That's like asking for a single term which describes a car's top speed, fuel consumption and colour. It is not in the remit of single terms to provide such disparate information. (Not that such information wouldn't be nice!)

Although the OP doesn't like the idea, the nearest thing we can offer as a definition of 'photographic magnification' is pixel scale in arcseconds per pixel. It is, to be fair, an extremely useful definition, if not one delivered in familiar layman terms. It tells you, for instance, how many screen pixels will be given to the lunar crater Minimus (which I just made up! :icon_mrgreen:) in any specified focal length/pixel size. C'mon, be reasonable, that's not bad. If Minimus is ten arcminutes across then it will get 600 screen pixels across at 1 arcsec per pixel, 1200 screen pixels across at 0.5"PP and so on. That tells you how big the crater Minimus will be on your screen and, so far as I can see, that's as good as it gets?

Anyway the moon's back and bedtime beckons. The above post comes with a four second guarantee!

Olly

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Louis D    760
6 hours ago, Swithin StCleeve said:

I've bought a dinky little 4" mirror scope, (the Skywatcher Heritage 100), which I've been having lots of fun with.

I'll come at it as a photographer.  Since your scope has a 400m focal length, it's exactly the same as using a 400mm telephoto lens with your camera as long as it's being used at prime focus (no optics in the focuser).  Using a 2x barlow will nominally double the focal length to 800mm.  I say nominally because that would be if you could place the imaging chip at the shoulder of the barlow.  Since there's the additional distance from the shoulder to the imaging chip when using a DSLR, the actual focal length will be somewhat longer, possibly much longer with a shorty style barlow.  With photography, all you can do is calculate a field of view on the diagonal of the imager and compare it to the field of view at some other focal length to determine a "magnification" as many others have said above.  Take images of a distant object with a "normal" lens (whatever yields a non-wide/non-tele image to your eye) and then with your scope with the barlow.  On a computer, measure the size in pixels of the object in each image, and divide the smaller number in the larger to get a "magnification" number to satisfy yourself and your friends who see your images.

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