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Swithin StCleeve

A 'working out the magnification' question

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For years I used a ten inch reflector on a dob mount, then for the last three years I've used an 8 inch. Now I've bought this little four inch as a secondary, portable telescope, I've been interested in comparing what I'm seeing with through it with its bigger counterparts. The only way I can compare really is to start to work out what I call the 'magnification'. I didn't realise it was so complicated (well, it is to me), when it comes to photography. But I do understand that an image on paper, or on a screen, can be 'magnified' in so many different ways, that the term magnification becomes useless.

A layman though, will always want to know 'what magnification is that', (it's like being asked "how far can you see through that telescope" I suppose, a question that I've been asked a few times).
I suppose the best way to answer the question, would be to tell them how much bigger the image is, compared to what they'd see on a 'normal' photograph, taken with a standard lens (say, 50mm). On my screen the moon in the first shot is 3.5mm, the moon in the second shot is 70mm. So I suppose it's been magnified 20 times, from the first photograph. But the first photograph is cropped anyway, from a phone photo I took last month. So that in itself is magnified.

image.jpg

ss1.jpg

So really, magnification means nothing with an image. I think I'm starting to understand.

And yet, it gets bigger.

Edited by Swithin StCleeve

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furrysocks2    471

Google suggests angular resolution of the eye might be around 60 arc-seconds, giving another possibility for comparison between an image and the naked eye... your second photo shows the moon at approx 6.5 arc-sec/pixel, so you call it "9x" or "10x", for the sake of putting a number on it.

Edited by furrysocks2

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Stu    14,455

If there was such a thing as a standard photo and a standard distance to hold it away from you then I guess it may be possible to create a comparison? Say a 6"x4" or perhaps (more up to date) a 10" tablet held at arms length, vs naked eye? You could do a resolution comparison in this case perhaps? Still, you can magnify the image as much as you like but there a finite amount of detail visible before it just goes blurry so resolution has to be key to it all surely?

As said, not a trivial topic!

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Stub Mandrel    5,420
3 hours ago, furrysocks2 said:

Google suggests angular resolution of the eye might be around 60 arc-seconds, giving another possibility for comparison between an image and the naked eye... your second photo shows the moon at approx 6.5 arc-sec/pixel, so you call it "9x" or "10x", for the sake of putting a number on it.

That has echoes of Apple's so-called 'retinal' displays which allegedly means matching the eye's resolution.

In print we use 300 dots per inch as a standard as that matches the eyes resolution at a comfortable reading distance.

That gives a way to work out a 'magnification' for prints of photos?

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BeerMe    273
On 30/09/2017 at 15:26, ollypenrice said:

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:

I love posts like this.  They remind why I will never touch AP with a bargepole :-P

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BeerMe    273

As an aside, I always like to add this pic to posts of this nature (even though the OP is well-versed with how magnification works in respect to EPs, FL, etc...)

 

 

epcalcs.png.0b3ae4df255347510b615af403679269.png

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BeerMe    273
3 minutes ago, Stub Mandrel said:

I've actually printed out a copy of that!

I shall pin it somewhere useful...

I've got it saved on my phone, Dropbox, pc etc just in case :-)

It's very handy.

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BeerMe    273
2 minutes ago, furrysocks2 said:

One of mine, is that not? ;)

I've no idea where I first came across it but I saved it and refer to it often mate.  It's easily one of the best snippets of info I've come across 👍

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Louis D    758

That diagram is missing an arc from eyepiece focal length and focal ratio to exit pupil.  Most owners know the focal ratio of their scope from memory and thus can quickly calculate it just by dividing the eyepiece focal length by the focal ratio.  In this case, 20mm/6=3.33mm.  That's much quicker and easier than the computation shown.

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furrysocks2    471
2 hours ago, Louis D said:

That diagram is missing an arc from eyepiece focal length and focal ratio to exit pupil.  Most owners know the focal ratio of their scope from memory and thus can quickly calculate it just by dividing the eyepiece focal length by the focal ratio.  In this case, 20mm/6=3.33mm.  That's much quicker and easier than the computation shown.

It confused the graph to either have four arcs into exit pupil or two exit pupil nodes, so I had to pick one. I thought aperture/mag was a bit more intuitive, if less direct.

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ollypenrice    17,056
10 hours ago, BeerMe said:

I love posts like this.  They remind why I will never touch AP with a bargepole :-P

I wouldn't throw out the baby with the bathwater! Absolutely everything in my post applies every time you click the button on your point and shoot camera. You have an F ratio, a resolution in arcseconds per pixel, a chip size, a pixel size etc etc. But you don't need to know what they are in order to take a picture.

My post was really very simple. Visual optics multiply the size of the image on your retina by x times. What are we multiplying by x times when we take a picture?

The question cannot be answered. Nothing complicated there!

Olly

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BeerMe    273
5 hours ago, ollypenrice said:

I wouldn't throw out the baby with the bathwater! Absolutely everything in my post applies every time you click the button on your point and shoot camera. You have an F ratio, a resolution in arcseconds per pixel, a chip size, a pixel size etc etc. But you don't need to know what they are in order to take a picture.

My post was really very simple. Visual optics multiply the size of the image on your retina by x times. What are we multiplying by x times when we take a picture?

The question cannot be answered. Nothing complicated there!

Olly

It wasn't a dig Olly, I've no doubt that what you posted is entirely correct as you obviously know your stuff, it was more just a general comment that these type of posts fly right over my head every time.  I've tried to understand a few of them over the months, but all I end up with is a sore head.

I really doubt that I'll ever 'get' AP.  

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ollypenrice    17,056
8 hours ago, BeerMe said:

It wasn't a dig Olly, I've no doubt that what you posted is entirely correct as you obviously know your stuff, it was more just a general comment that these type of posts fly right over my head every time.  I've tried to understand a few of them over the months, but all I end up with is a sore head.

I really doubt that I'll ever 'get' AP.  

Honestly, if I can do it anyone can do it. You just need one idea to hook onto and understand and the rest will follow.

Olly

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JamesF    7,769

I have come to the conclusion that "magnification" is pretty much useless other than in a comparative sense.  It's perhaps helpful to say "this is how something looks at 50x magnification" and "this is how it looks at 100x" and expect the view to double in size, but the first statement alone doesn't really convey much information that's of use to anyone.

in print everything changes again.  You might take a picture of a flea, for instance, and print it in a book as 100x magnification, meaning the image of the flea is 100x larger than the actual flea.  But if you created an image of Jupiter and printed it on the same page, what magnification might you then claim?  Jupiter has a diameter 11x that of Earth (or thereabouts, from memory).  If you can see the entire planet on a single page of a book then it certainly hasn't been magnified in the same sense as the image of the flea.

Or if you display the same image of Jupiter on three monitors next to each other, one of which has a higher resolution but the same physical size as another and the third has the same resolution but larger physical size, you end up with three images of different physical sizes.  Are they the same "magnification" then?  Probably not in any obvious sense of the word.  Perhaps the only way to meaningfully compare the three images is by using some measure of how much information they contain, or, how much area is covered by a single pixel.  At which point we're right back to image scale.

James

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I think scope magnification refers to how much larger the apparent size of the object appears out of the "output" end of the telescope (eyepiece or camera sensor). Jupiter is a fair bit larger than a flea but also much further away from us than any flea on earth. If you compare a flea in your hand to jupiter in our sky, the flea is larger in apparent size. 

The above definition describes the magnification, regardless of how large you may print a resulting camera image. 

Edited by DarkAntimatter

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ollypenrice    17,056
12 hours ago, JamesF said:

I have come to the conclusion that "magnification" is pretty much useless other than in a comparative sense. 

I'd have thought that it is, by definition, a comparative term and nothing else. It compares x with x multiplied by a specified factor.

Magnification without comparison would be like the sound of one hand clapping. That's why I insist that we have to know what it is that we are going to multiply by x. 

Olly

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Stub Mandrel    5,420

The problem is that camera images are real images, so they don't subtend a consistent angle that can be compared with the "normal view" of a virtual image as in a visual telescope or bins.

Looking here:

https://en.wikipedia.org/wiki/Magnification

The answer is very simple.

For a digital camera Magnification = (focal distance -  focal length)/focal length

For stars and anything else effectively at almost infinite focal distance==>focal length, so magnification is barely above 0 and stars appear as point objects and nebulae and galaxies many, many light years across become tiny.

Even planets, (as someone pointed out for Jupiter) are reduced to tiny objects.
 

So the only 'meaningful' expression of magnification between camera setups is to compare on an 'all other things being equal' basis OR have an arbitrary benchmark.

 

One arbitrary benchmark could be that you compare angle subtended by the imaged object when displayed at 300 dpi a foot away i.e. matching the eye's resolution, (or for many 96dpi computer screens it would be reasonable to compare the angle subtended a metre away with the real angular size of the object).

This would have the benefits that:

  • It's easy to understand
  • It matches image resolution to the eye so no detail is wasted BUT undersampled pictures will be obviously less sharp, stopping people simply resampling to claim 'higher magnification'.
  • You could do an easy comparison by holding a 300dpi print of a picture up in front of you and looking from it to the actual object (obviously only works with objects bright enough to see naked eye).
  • Can be used to compare visual scope views with photos.

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Louis D    758

Or, you could go with the macrophotography definition of magnification.  If the object is the same size on the sensor as in real life, that is 1:1 magnification.  If it is half as big, 1:2.  Twice as big, 2:1.  Those are easily measured values.  So for Jupiter on a full frame 35mm DSLR sensor filling the vertical dimension, you'd have approximately 1:5825919158 magnification or about 1/6 billionths as big.  Tell that to someone the next time they ask how much magnification the telescope achieves for photography.

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