magnification

[peem]

If i attach my dslr camera without a lens to a 80mm f6.8 refactor without an eyepiece what magnification will it be? Thanks

[farunj]

It will have the same magnification as a 80*6.8 = 544mm lens.

[JamesF]

It is said that at prime focus the image will be about the same size as you'd get with a 6mm eyepiece. Magnification is a really misleading term when you're imaging at prime focus though as it doesn't make much sense. If you took two images of the same object with two different cameras, one with 3um pixels and one with 6um pixels then when you displayed them on your monitor, one would be half the size of the other. Does that mean they were taken at different magnification? Obviously not, because both camera sensors would have been at exactly the same point in the optical train. Or if you take one image and display it on two different resolution monitors so it appears different sizes, what is the magnification then?

James

[steppenwolf]

There isn't any magnification as such because so much depends on how the final image is viewed - on a computer screen? - on a large computer screen? - printed out on 6 x 4 photo paper? etc. etc.

However to try and put your question into some kind of meaningful context, if you were to attach a 50mm 'standard' camera lens to a film SLR camera (or a 31mm zoom lens to a modern DSLR camera) and look through the viewfinder, you would see a life sized image. If you used a 100mm lens on the film SLR, the magnification would be 2X through the eyepiece and so on. Your telescope would appear to have a focal length of 544mm so the 'magnification' through that old film camera's viewfinder would be 544 / 50 = 10.88 X or though a DSLR camera, 544 / 31 = 17.5 X

[steppenwolf]

Ooops, I must type faster, I've doubled!!

[JamesF]

Ooops, I must type faster, I've doubled!!

I think your explanation was very useful nonetheless, Steve.

James

[peem]

Thank you for your replies . I understand it now .Its just the same as using a telephoto lens

[LukeSkywatcher]

It is said that at prime focus the image will be about the same size as you'd get with a 6mm eyepiece

I have read that a few times on this forum.

So to work out the "magnification", simply divide the focal length of the scope by 6mm.

[LukeSkywatcher]

Thank you for your replies . I understand it now .Its just the same as using a telephoto lens

Yep. The scope becomes the telephoto lens.

[farunj]

There isn't any magnification as such because so much depends on how the final image is viewed - on a computer screen? - on a large computer screen? - printed out on 6 x 4 photo paper? etc. etc.

However to try and put your question into some kind of meaningful context, if you were to attach a 50mm 'standard' camera lens to a film SLR camera (or a 31mm zoom lens to a modern DSLR camera) and look through the viewfinder, you would see a life sized image. If you used a 100mm lens on the film SLR, the magnification would be 2X through the eyepiece and so on. Your telescope would appear to have a focal length of 544mm so the 'magnification' through that old film camera's viewfinder would be 544 / 50 = 10.88 X or though a DSLR camera, 544 / 31 = 17.5 X

Oops I stand corrected!! Sorry!!

[asteele3]

So how would I work out the magnification when I use my Canon EOS 1100d slr camera attached to a Celestron C8 OTA telescope? I believe the scopes diameter is 203mm and the focal length is 2032mm.

[LukeSkywatcher]

So how would I work out the magnification when I use my Canon EOS 1100d slr camera attached to a Celestron C8 OTA telescope? I believe the scopes diameter is 203mm and the focal length is 2032mm.

If the idea about it being the same as using a 6mm EP, then the magnification on a C8 would be 338X.

2032/6=338x

I'm sure the whole thing is a lot more technical then i have said or even understand. But on a basic level if the rule: "its similar to using a 6mm EP", is correct, then my calculations are correct.

[Cornelius Varley]

The reference to a 6mm eyepiece only works with a webcam. A DSLR has a larger sensor so the equivelant eyepiece is somewhat greater and therefore the field of view is also greater.

Peter

[LukeSkywatcher]

The reference to a 6mm eyepiece only works with a webcam. A DSLR has a larger sensor so the equivelant eyepiece is somewhat greater and therefore the field of view is also greater.

Peter

Ahhhhhhhhhhhh i did not know that. I thought it was a gerneral rule, but i can see where i was/am wrong. The DSLR certainly has a larger sensor.

So what would be the magnification of a DSLR on a scope with a focal length of 2032mm?

[JamesF]

The reference to a 6mm eyepiece only works with a webcam.

Well, yes, this is one of the reasons I don't like the comparison in the first place. For instance, it may perhaps be true for an SPC900 with a 640x480 array of 5.6um pixels, but for exactly the same reason as it's not right for a DSLR it isn't going to be true for a Lifecam Studio with a smaller pixel size and a physically larger sensor.

Plate scale seems to make far more sense.

James

[LukeSkywatcher]

You cant compare the sensor of a DRSL to the view through an EP.

It simply does not equate.

Visual astronomy and astro-imaging are two complete seperate fields.

[astrofox]

I agree with Paul. In imaging its more usefull to consider the field of view a sensor will encompass. In the initial question a standard APS-c sensor in dslr will cover a field of view of 22x26 arc min ( about half of a full moon)