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GalileoCanon

What is the telescope's prime magnification with NO eyepiece lens?

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On ‎09‎/‎12‎/‎2017 at 22:07, GalileoCanon said:

My question is:
How much magnification will I see from the telescope's mirrors alone with no eyepiece glass and no Barlow lens in between the telescope's focuser and my camera's sensor?

Actually your question makes sense but only visually. Look at something with one naked eye and the other eye in a reflex camera's viewfinder. If the focal length is 70mm both images have the same apparent size. I checked by setting the zoom at 35mm, the camera view became exactly twice smaller than the naked eye view.

Thus a 700mm focal length scope magnifies 10x relative to the naked eye, but after a picture is taken it can be displayed in any apparent size from cell phone screen to giant living-room TV. Traditionally camera objectives have focal lengths that revolve around 70mm for that reason: the perspective at 70mm is the same as with the eye alone.

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I'm not a photography guy, but here's the way I think of it. Take a short tube and put it up to your eye. If you were to take a picture, the actual image that you see, not including the tube walls, is what will be focused onto your sensor. It will take up the entire sensor. Now take a much longer tube. Your field of view is much more narrow, right? Again, the actual image you see at the end of the tube is what will take up the entire sensor. While the image with a longer tube will appear larger, you're not really magnifying anything. You're just limiting the field of view and focusing it down to more of a point. The actual magnification happens in the eyepiece where it takes that focused point and spreads it out and magnifies it.

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Earlier on you used the term 'crop sensor.' This is a term best forgotten when you move into astrophotography and imaging with telescopes. Cameras, and particularly astronomical cameras, come in a wide range of chip sizes. In the image below I've modelled two chip sizes, large and small, in one of our telescopes. The object on the chart is M33 shown in green. The problem with the misleading term 'crop factor' is that it might lead you to believe that you are more 'zoomed in' (and capturing more detail) with the small chip than with the large. In fact you are not. If both chips have the same size pixels the image scale is literally identical as is the level of detail and the final screen size of M33 on your PC.  All that changes is that you have more sky around M33 wth the larger chip.

5a3299cc184f2_CHIPSIZE.JPG.ec4d427313b0c5a5dda3f1a989590d7e.JPG

Now let's imagine, as might well be the case, that the larger chip has smaller pixels than the smaller chip. This time the larger chip will give a larger and more detailed image of M33 than the smaller one. This is because it will put more pixels under M33 itself and when it appears as a fullsize image on your screen (meaning one camera pixel is given one screen pixel) M33 will be given more screen pixels and, therefore, be bigger.

The size of the sensor has no bearing on the scale of the image and the person who coined the term 'crop factor' should be made to listen to an entire CD of yodellers in full voice!

:icon_mrgreen:lly

 

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Just to illustrate the uselessness of the camera frame size....

Not a good frame of reference (excuse the pun!)

sensor-size-comparison.jpg

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The word telescope comes from the Greek tçleskopos, which means far-seeing... tele = 'far' and skopein = 'to look or see'. 

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On ‎12‎/‎10‎/‎2017 at 06:38, GalileoCanon said:

 If I'm not mistaken isn't a 50mm lens pretty close to what the naked eye sees? If I hold my camera to my right eye with a 50mm lens isn't that pretty close to what I'd see in size if I looked at the image directly with my left eye?

Pretty close, I think, except your eye will give a much wider FOV with peripheral vision, but your central vision would be close to what a 50mm lens gives. I think that is why 50mm is so commonly used for standard prime lenses.

On ‎12‎/‎10‎/‎2017 at 06:51, Ouroboros said:

Maybe you've said but what sort of DSLR have you got? 

Not sure if that was directed at me(with all the quotes flying around:icon_biggrin:), but my D3400 Nikon has a crop sensor.

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On 10/12/2017 at 00:46, GalileoCanon said:

 

Do I simply look at the focal length of the telescope and compare that to any normal DSLR lens' length?  So 700mm will be 20x more zoomed in than a 35mm DSLR lens?! (for example)

 

Yes that's the simple truth, but you said the dreaded word 'magnification' and that gets the assembled masses dancing with the angels on the head of a pin.

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On 10/12/2017 at 12:38, GalileoCanon said:

 

If I'm not mistaken isn't a 50mm lens pretty close to what the naked eye sees? 

I think this is said because a 50mm lens's picture will present near and distant objects at the same relative sizes as they appear naked eye. In other words there is no 'foreshortening' effect as created by zoom lenses.

Since astronomical targets are all at infinity this effect does not exist for us.

Olly

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8 minutes ago, ollypenrice said:

I think this is said because a 50mm lens's picture will present near and distant objects at the same relative sizes as they appear naked eye. In other words there is no 'foreshortening' effect as created by zoom lenses.

 

Optical illusion - you can get the same 'foreshortening' by cropping a wide angle shot :icon_biggrin: All the principles applying to digital apply once you project or print that film negative!

For a 35mm film image to have the same FOV as we see, it needs to be just 12mm F/L.

However, when looking through a film-based 35mm camera with a 50mm lens the view will be more or less identical to what you see with the naked eye - hence it being seen as 1:1.

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On 09/12/2017 at 21:15, brantuk said:

The scope just does the focusing, it's the eyepiece that does the magnification.

Its funny how something jumps out at you when you re-read something, cheers brantuk!

Now I'm far from being the astrophotographer professional, but I now know why my early images are smaller than I had anticipated, some Years ago when I first got my Dob and first attached my D5000 thinking that by using my scope I was actually using a 1200mm lens ?

As stated, the telescope captures light photons from the target and focuses this light to form a 'real' image at the focal plane, that's it!

Therefore, the scopes image at the focal plane would be the same, would it not, as if you were looking at it with the naked eye, ie. no magnification and very small.

The introduction of the eyepiece allows you to see this image in detail, and using the formula F/f=X derived from the focal length (F) of the scope and the focal length of the eyepiece (f), you apply magnification (X)  to the image..............pretty simple. 

I often mention that eyepieces alone, especially those expensive premium ones, can't improve the details beyond what the scope itself provides, but if that image is good from the start, the eyepiece can improve the field of view, and enhance the eye-relief, but to think its taken me  over three Years  to just realise what I was trying to achieve back then.........laughable!

 

Edited by Charic
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1 hour ago, Charic said:

Now I'm far from being the astrophotographer professional, but I now know why my early images are smaller than I had anticipated, some Years ago when I first got my Dob and first attached my D5000 thinking that by using my scope I was actually using a 1200mm lens ?

As stated, the telescope captures light photons from the target and focuses this light to form a 'real' image at the focal plane, that's it!

Therefore, the scopes image at the focal plane would be the same, would it not, as if you were looking at it with the naked eye, ie. no magnification and very small.

 

Sorry, no -  a 1200mm scope will work just like a 120mm telephoto.

 

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Been reading some more and think I'm sorted now, maybe not!

The Skyliner will act as a 1200mm lens, and will provide a magnification in my DSLR of about 42x? This equates to the focal length of the scope divided by the diagonal of the camera's sensor?

Therefore, I believe that if I need more magnification, a Barlow or an eyepiece is required.
Said I was no expert, but always willing to learn.

Edited by Charic

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1200mm will give an image scale 24 time larger than a 50mm lens would, or for a crop sensor with a 70mm kit lens, 1200/70 = 17 times larger.

You should only need more magnification if hunting very small targets. It will greatly increase the exposure times needed, which doesn't matter for bright planets.

M52 is a pretty small DSO, this it using my 1200mm 150Pl with no additional magnification but using 'drizzle' to get a larger image scaleM57.thumb.png.eb813b1b8a8ad226ee81203341ae1dc7.png.

 

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19 minutes ago, Stub Mandrel said:

M52

M57.... 

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12 hours ago, Charic said:

 

The Skyliner will act as a 1200mm lens, and will provide a magnification in my DSLR of about 42x? This equates to the focal length of the scope divided by the diagonal of the camera's sensor?

 

 But 42 x what? (That's why the term isn't used in imaging.)

Olly

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Sorry Olly.

I read that the image  produced in my scope with DSLR attached would equal 42 magnification, rather than 42 times something? 
It said  "divide the telescopes focal length by the the diagonal of the sensor" and for my system, this results in 42x. 
 

Edited by Charic

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Without an ep, the scope is a telephoto lens for your camera. 

So divide the scope's 700mm by whatever focal length gives 1x on your camera to get magnification. 

On a 35mm film camera, the magnification would be 700/50 = 14 as 50mm lens is 1x for that format. 

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A 1200mm scope will give a prime image that is dependant on the focal length (1200) and the subtended angle of the object. It is not a magnification simply a size in mm.

6 hours ago, 25585 said:

On a 35mm film camera, the magnification would be 700/50 = 14 as 50mm lens is 1x for that format. 

It is not "magnified" the image is simply 14x bigger in mm then the iimage is for a 50mm lens or mirror. But that is not a magnification.

What is this 1x on a camera? You say 50mm lens on a 35mm film but the DSLR's are not 35mm, so is a DSLR bigger or smaller in terms of this "magnification"? From the arguemnet it reads that a 50mm lens on a 35mm film or full frame DSLR give a different "magnification" against an APS size DSLR whereas the lens will produce exactly the same size image.

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4 minutes ago, ronin said:

It is not "magnified" the image is simply 14x bigger in mm then the iimage is for a 50mm lens or mirror. But that is not a magnification.

14x bigger is magnification where I come from :icon_biggrin:

The problem is perceived 'magnification' of a astronomical image is a relative, not absolute, term. This is because we don't compare the image to the original object by putting them side by side, but by viewing the object as we see it unaided and holding the new image close to our eyes. As we can move towards or away from the real image its angular size changes so it can appear larger or smaller than the actual object depending on where we put it so we can't use this comparison meaningfully.

When viewing a virtual image through a scope the situation is simple, as magnification is the increase in the angle subtended by the viewed object. This is because we see the virtual image at a fixed angular size.

However, real images DO have a magnification - macro photography is defined as when the real image is larger than the original object, which depends on object-lens distances being less than the focal length of the lens.

As Olly has pointed out, astronomical images are projected at vastly SMALLER sizes than the real thing - compare the size of a moon image with the real moon! The actual magnification of the real image is a tiny fraction.

Judging the perceived magnification of such an image is like saying a sound  is 20dB loud - it only has meaning if you know how loud the 0dB reference value is. A meaningful comparison is therefore between the image size given by some refernce lens (conventionally 50mm for a film SLR or 70mm for a typical APSC sensor) and the telescope.

Rather than confusing beginners by saying longer scopes don't magnify, the simple explanation is:

The size of the image is directly proportional to the focal length of the scope or lens, so all other things being equal (i.e. using the same camera or eyepiece) if you double the focal length the image will appear twice as large.

To judge the 'magnification' of a telescope, divide its focal length by the focal length of the lens you are comparing it to, but bear in mind this figure may not be meaningful in the context of someone else's setup.

 

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On 16/12/2017 at 14:14, Charic said:

Sorry Olly.

I read that the image  produced in my scope with DSLR attached would equal 42 magnification, rather than 42 times something? 
It said  "divide the telescopes focal length by the the diagonal of the sensor" and for my system, this results in 42x. 
 

The term 'magnification' is only meaningful if we know the dimensions of the thing being magnified by 42x. To know its dimensions we have to know what it is. When we look through a telescope giving 42x magnification we do know what it is we are increasing in size by 42x. It's the image perceived by the naked eye. But there is no such thing as a 'naked camera.' This might be a camera without a lens but in that case it wouldn't give an image at all. So my question' 42x what?' was perfectly serious. Of course we can certainly say that lens x will give 42x the magnification of lens y but this is only of interest to the owner of lens x. This magnification has no universal currency which is why we use the term resolution in arcseconds per pixel or, in professional circles, plate scale.

Besides, if I try two cameras in the same scope, one with small pixels and one with large, the plate scale will be unaltered but the resolution will increase, as will the size of the output image. It would be odd, though, to say, 'this chip magnifies more than that chip...' even though, in one sense, it does!

Olly

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12 hours ago, ronin said:

What is this 1x on a camera? You say 50mm lens on a 35mm film but the DSLR's are not 35mm, so is a DSLR bigger or smaller in terms of this "magnification"? From the arguemnet it reads that a 50mm lens on a 35mm film or full frame DSLR give a different "magnification" against an APS size DSLR whereas the lens will produce exactly the same size image.

On a 35mm film 50mm gives x1. On a 6x6 120/220 medium format film it is wide angle. In short depends on the size of media. 

DSLR media will be different, & medium format digital different again. 

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1 hour ago, ollypenrice said:

So my question' 42x what?' was perfectly serious.

I didn't doubt it, I thought I'd confused you from my text, but now I'm thinking that I really know little at all myself?
Ive always thought my images are small in the Skyliner anyway, but when I stick a 1200mm OTA to the DSLR I expected so much more, but not the case! Its my present understanding that F1200/sensor diagonal=42 so I'm looking at the Moon/any subject with a fixed 1200mm lens, with only 42x mag, now seems totally naff if correct, or my understanding is totally incorrect. The camera concerned is the Nikon D5000, so if you/anyone can give a final push in the right direction, I'll be so much happier and more the wiser. 

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Barlows are like teleconverters that magnify a lens prime image as I understand it. 

An eyepiece can give a larger image , so I guess is an image enlarger anyway. What matters is the area & objects you want to observe or record is filled to your satisfaction. 

Having a close focusing macro lens on a camera might be interesting to just photograph a reflected image from a diagonal or flat, not sure about other though. Has anyone tried that?

Edited by 25585

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Charic, this may help.

The image size on the sensor through your scope using a Nikon D5000 and then the view through a 15mm BST, 60 degree afov giving x80.

IMG_5339.PNG

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