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Understanding magnification

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#1 FerrariMX5

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Posted 06 June 2020 - 05:49 PM

Just making sure I'm building on a firm foundation: 

 

My SCT has an aperture of 8" or 203.2mm and more importantly, has a focal length of 2032mm

 

So, does that mean any scope with a focal length of 2032MM (without an eyepiece) should be starting out with zero magnification?

 

If so, it would follow that a scope with 1016mm focal length will also have zero magnification without an eyepiece?

A scope of 508mm focal length would like wise be zero?

 

If so, that makes sense that, 

If I use an eyepiece of 20mm with the 2032mm this is about equal to 100 times.

Just as the same eyepiece on a 1016mm would equal about 50 times,

or on a 508mm we would now be viewing at 25 times?

 

I am also assuming a full framed camera with a 50mm lens will also be at or very near zero magnification.

 

(I do understand the math will be slightly different for the camera, but just wanting a basic understanding and a workable base line).

 

Sorry for the simplicity, but just trying to build a foundation of knowledge to further my understanding.

 

Tony


Edited by FerrariMX5, 06 June 2020 - 05:50 PM.

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#2 kathyastro

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Posted 06 June 2020 - 06:10 PM

There is no such thing as zero magnification.  A scope without an eyepiece does not have a magnification at all.

 

Magnification is the focal length of the scope divided by the focal length of the eyepiece.  (Which is why magnification does not exist without an eyepiece.)  The mathematics of your examples are correct.

 

Magnification for cameras is defined differently.  The correct term is field of view.  However, as commonly interpreted, the "magnification " of a camera lens is the focal length divided by the diagonal of the sensor.  Again, zero magnification has no meaning.  A 50mm lens on a full-frame (24mm x 36mm) sensor gives a magnification of 1.16x.


Edited by kathyastro, 06 June 2020 - 06:10 PM.

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#3 SarverSkyGuy

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Posted 06 June 2020 - 06:14 PM

Forget the concept of 'zero magnification'.  

An objective ( telescope or camera lens) produces an image one focal length from the objective.  The longer the focal length, the bigger the image scale.

Focus an eyepiece on the image plane and the eyepiece magnifies the image.  The resultant power (magnification) is the objective focal length divided by the eyepiece focal length.

That's it. Simple.


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#4 WarmWeatherGuy

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Posted 06 June 2020 - 07:01 PM

Imagine you had a 3x telescope and you point it at a building off in the distance. You look at it with one eye looking through the scope and one eye unaided. You see a larger image of the building with the eye using the scope. It is in fact 3x larger. You can move the telescope so the two images line up on the left side of the building. The smaller image overlaps the larger image but only the left third of it because it is 3x smaller.

 

Use your C8 to point at the moon. Remove the eyepiece and the diagonal. Using a piece of paper you can get an image of the moon on the paper which can be in focus by moving it in and out. If the paper is very thin you can see the image from both sides of the paper. Take a magnifying glass and hold it up to your eye. While doing this get behind the paper with the focused image of the moon. Move in until you see the image of the moon on the paper be in focus. The distance between you and the paper will be the focal length of the magnifying glass. The image will be enlarged by 2032 / focal length of the magnifying glass. Now remove the paper and you will see a more clear image of the moon.

 

This is what is happening with your telescope. The main objective is projecting a focused image to a point in space and you use the eyepiece like a magnifying glass to see that focused image up close.


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#5 FerrariMX5

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Posted 06 June 2020 - 07:23 PM

 

 The main objective is projecting a focused image to a point in space and you use the eyepiece like a magnifying glass to see that focused image up close.

Thank you for this easy to follow explanation!

 

Tony


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