Silly Question About Telescope Magnification and Distance

[Stub Mandrel]

I may not have found exactly the site you're referring to but a quick run round the internet led me to a Canon website in which I found this.

To get the binocular-like viewfinder magnification value of a lens, divide the focal length by 50.

It is important to note that they are talking about an SLR viewfinder and lens combination here and not about either a projected-onto-chip image or final image. The division by 50 simply makes the veiwfinder behave like a binocular of the same magnification. So if you use a DSLR viewfinder and a 500mm lens it will give you the same magnification as a pair of 10x binoculars, which means the apparent size of an object will be 10x larger than the naked eye view. The key thing is that this about the viewfinder image and not the final image. 

Olly

PS I don't know why the quotation copied oddly but it was a single grab from screen.

It's more subtle than that. As an old photography wonk, we always used to work on the idea that a 50mm lens on 35mm film creates the same perspective through the viewfinder as if the camera wasn't there.

That's why lenses of less than about 50mm are generally thought of as 'wide angle' and the images they produce tend to have exaggerated perspective (e.g. moon in distance of picture becomes a tiny dot beyond the trees), whilst telephoto lenses much longer than 50mm 'flatten' the perspective (telephoto picture of moon above trees makes moon look much bigger and closer).

It's also why standard lenses are usually in the range of 50-58mm focal length, as this gives a 'natural' perspective to photographs.

The 'magnification' of a lens (as viewed through the viewfinder i.e. using the camera as a monocular) is therefore lens focal length/50.

Note that you can cut out a tiny bit from a wide angle shot and blow it up to get the same effect as a telephoto (although it may be ridiculously fuzzy) .

The 42x zoom on my bridge camera only makes sense as a ratio, until it is related to a range of roughly 23-1000mm as compared to a full frame SLR.

[ollypenrice]

In my view it's useless. The way to estimate your field of view is to give your chip size and focal length in mm to one of the planetarium software packages and it will put a square on the sky chart showing what will be framed. I use SkyMap Pro but I think the free Stellarium can do it. Here's how it looks if I put my full frame CCD chip and 530mm focal length onto a chart of M45 from SkyMap. 

In meaningful numbers this covers a piece of sky 4 degrees by 2 deg 42mins at a pixel scale of 3.5 arcseconds per pixel. This is 'absolute' information and can be understood by anybody who reads it. Nobody in astrophotgraphy talks about magnification. They really don't.

Olly

[miguel87]

Thanks very useful info. I wont be trying to measure my magnification, just enjoy the views!

Themos however, you missed one word in my sentenc: 'appear'

To say that telescopes do the escape opposite is to say that telescopes make objects appear smaller. This would not be a popular telescope.

Yes, the image formed at the focal length is very small, but larger than the image formed on the retina by the actual moon.

Telescopes make things appear larger... surely this is not up for debate??

[ollypenrice]

It's more subtle than that. As an old photography wonk, we always used to work on the idea that a 50mm lens on 35mm film creates the same perspective through the viewfinder as if the camera wasn't there.

That's why lenses of less than about 50mm are generally thought of as 'wide angle' and the images they produce tend to have exaggerated perspective (e.g. moon in distance of picture becomes a tiny dot beyond the trees), whilst telephoto lenses much longer than 50mm 'flatten' the perspective (telephoto picture of moon above trees makes moon look much bigger and closer).

It's also why standard lenses are usually in the range of 50-58mm focal length, as this gives a 'natural' perspective to photographs.

The 'magnification' of a lens (as viewed through the viewfinder i.e. using the camera as a monocular) is therefore lens focal length/50.

Note that you can cut out a tiny bit from a wide angle shot and blow it up to get the same effect as a telephoto (although it may be ridiculously fuzzy) .

The 42x zoom on my bridge camera only makes sense as a ratio, until it is related to a range of roughly 23-1000mm as compared to a full frame SLR.

Agreed. I made the same point (I think!) a couple of posts back, saying there that the relative sizes of objects in a 50mm lens remained similar to the same relative sizes seen naked eye, so in this sense there's a convention that this represents a 1x magnification. Perspective is a surprizingly controvertial topic amongst perception theorists. My father was an academic in this field. I grew up amongst endless wrangling on the matter. 

Olly

[ollypenrice]

Telescopes make things appear larger... surely this is not up for debate??

Everything's up for debate on here! 

I mean, which end of your telescope do you look through when you say this????  

 lly

[jsmoraes]

From Stu

Mike, I think if you ask any of the imaging guys they will agree with acey. It's ok to talk about image scale, but magnification is meaningless in imaging terms. 

Yes, I agree with acey.

IMHO

Let remember:

With eyepiece:

If you use a camera in direct focus:

The sensor of a camera is a flat surface, as paper or wall.

The image of planet is a projection on this surface. There aren't two images: the real image and virtual image. This last image was done by any other accessory with focal length, as eyepiece.

If the image of planet is project on surface of sensor, the image of planet will have some pixels of total pixels of the sensor. And we can tell in arcsec per pixel, miles of surface  per pixel, km of surface  per pixel ...

With the same telescope with the same focal length

note 1:

If two sensors have the same resolution (horizontal x vertical pixels) and same size, when this image is viewed on monitor the image will be the same size.

If one sensor has more resolution (more pixels) and same size, when this image is viewed on monitor the image will have diferent apparent size. Bether arcsec per pixel, miles of surface per pixel, km of suraface per pixel ...

We have none magnification of image. We have only more resolution (image scale).

note 2:

If you use camera with small sensor, the total image of this planet projected may not fill fully the sensor. Maybe only a small part of planet will be projected on total area of sensor.

If you use camera with large sensor, the total image of this planet projected may fill fully the sensor. Maybe some background area will be there, also. As moons and stars.

If you see those images on a monitor, the image of small sensor will appear larger than the other. Maybe you only will see a small part of the planet.

Despite of resolution of sensor (that improves the arcsec per pixel, miles of surface per pixel, km of surface per pixel ...) we have none magnification of image. We have only a small area of planet showed on monitor. 

note 3:

If you use barlow, negative lens. The focal lenght of telescope will change. The total image of planet will be sent more away from telescope. If you use the same camera, the image of this planet projected may not fill fully the sensor. Maybe only a small part of planet will be projected on total area of sensor.

If you see on monitor the images with and without barlow they will have different aparent size. We have no magnification with barlow image. We have only more resolution. Small area of planet on the same sensor of camera. Will work as if you are using sensor with small size.

[themos]

Thanks very useful info. I wont be trying to measure my magnification, just enjoy the views!

Themos however, you missed one word in my sentenc: 'appear'

To say that telescopes do the escape opposite is to say that telescopes make objects appear smaller. This would not be a popular telescope.

Yes, the image formed at the focal length is very small, but larger than the image formed on the retina by the actual moon.

Telescopes make things appear larger... surely this is not up for debate??

It is not up for debate, it is simply wrong. Telescopes make near copies of far away things. Tiny, tiny copies. 

Eyepieces (essentially magnifying glasses) make things appear larger. 

Not confusing the two functions will help you.

[jsmoraes]

Another issue.

There aren't magnification if you use a telescope without eyepiece.

For reflector telescopes, with mirror:

spherical mirror only have magnification if the target is between the geometric center of circle and the surface of mirror. If the target is further away from this center, the mirror cause reduction of image: negative magnification.

Secondary mirror is only a flat mirror.

Why it seems to have magnification of my face when I see it on the primary mirror of my telescope ?

Check the distance between your face and the mirror. Is it between the geometric center and the surface ? If it is ... it is normal to have magnification.

The main purpose of telescopes is catch light... not to do magnification.

[miguel87]

Great discussion all, thanks.

Themos i think we agree but my use of the word appear relates to how we 'see' things. And the certainly appear bigger in a scope. Yes the image is much smaller but closer.

If i could see a van 2 miles away from a hilltop and a scooter right in front of me (sort of like moon versus projected floating image of moon). Yes the scooter is actually much smaller, but it would certainly 'appear' larger.

Brings me on to perspective olly. I guess this is the crux of the problem. A mathematical representation of human vision is near impossible. I studied perspective in my psych degree but obviously from a totally different (and not as brainy) school of thought.

[guillespiers]

Thank you all for the info

As being said, what about we use "50" in our calculations?

Since we are not using any eyepiece in my question, will this be ok?

4500 FL / 50  = 90x

[miguel87]

What do you mean? That calculation means a 50mm eyepiece is being used.

Mike

[ollypenrice]

Thank you all for the info

As being said, what about we use "50" in our calculations?

Since we are not using any eyepiece in my question, will this be ok?

4500 FL / 50  = 90x

90x what?

Olly

[Stub Mandrel]

Thank you all for the info

As being said, what about we use "50" in our calculations?

Since we are not using any eyepiece in my question, will this be ok?

4500 FL / 50  = 90x

Effectively the viewfinder on your camera acts as a ~50mm eyepiece, so yes that is a fair calculation for what you see through the finder.

But as Olly says as far as the image on computer screen, print out or live view are concerned 90x what?

[guillespiers]

Effectively the viewfinder on your camera acts as a ~50mm eyepiece, so yes that is a fair calculation for what you see through the finder.

But as Olly says as far as the image on computer screen, print out or live view are concerned 90x what?

Hi! Sorry for my late response! I left my post opened for a while, and when I pressed "post" I found like 10 more comments! I was refering to what Stub Mandrel said; trying to figure out if by using "50 mm" in the formula would help me get a magnification number, but this seems a little complex

[ollypenrice]

Hi! Sorry for my late response! I left my post opened for a while, and when I pressed "post" I found like 10 more comments! I was refering to what Stub Mandrel said; trying to figure out if by using "50 mm" in the formula would help me get a magnification number, but this seems a little complex

The 50mm business is just a way of saying that, with this lens on that chip, near and far objects will have similar relative proportions to those of the naked eye view.* It does not define a magnification of one. It just describes, rather casually, one aspect of the visual impression a daytime photo will create. You cannot use it as a definitive value to multiply by another number in order to specify a magnification.

Olly

*A further twist.   (Oh no!!)  In AP all our targets are at infinity and our optics have flat fields (we hope!) so the angular size of near and far objects is defined by their true size and their distance and not by the optical characteristics of our lenses. This means that any focal length we use preserves the relative proportions of far and near objects in view. (Is this right? I ought to think about it a bit more but I have to tidy my study. We have guests in two weeks and that may already be a bit short, given the state of it...)

[vlaiv]

You can add "eye piece" part to your DSLR image and then calculate magnification. Display DSLR image on the screen (monitor). Set the distance from which it is being viewed, calculate angular size of object on monitor from that distance, divide with actual angular size (as viewed from earth) and you get magnification