Image Size

[Peter B]

I am an absolute novice and this is going to be, I'm sure, an absolute novice question so I beg for some patience understanding!

 

Since I took up 'back yard astronomy'  a couple of years ago I have enjoyed viewing the night skies though either my Bresser MC 127 Maksutovor  or my Skywatcher Statravel 102 refractor  and I'm beginning to understand these two telescopes' very different characteristics now.

 

Then I got interested in astro photography so I bought a T ring and T adapter for my Nikon D5000 DSLR and spent  an afternoon on a bright but very hazy day checking the camera out on each telescope using a phone/TV repeater mast about 750m away from my house as a target.

 

I am puzzled by the results.  I can't understand the difference in magification between the two telescopes, much larger with the Mak than with the refractor, with the same camera at prime focus on each scope.  They are of course very different focal ratios with the Mak at f14.9 and the refractor at f5, but that doesn't affect image size does it?

 

So ignoring the quality of the pictures which I didn't bother with because it was an experiment with the kit, not the photography, could someone please explain the massive difference in image size between these two telescopes with the same DSLR attached at prime focus.

 

Thanks 

 

 

 

 

 

 

 

 

 

 

Edited September 6, 2023 by Peter B

[TerryMcK]

Hi Peter. The Bresser has a focal length of 1900mm whereas the SW has a focal length of 500mm. The Maksutov is like a zoom lens which gets closer to the subject. Whereas the Skywatcher is more of a zoomed out lens. The F number is just a ratio of focal length divided by aperture diameter and is nothing to do with magnification.

Forgetting the camera for now if you use your eyeball with the same eyepiece in either telescope you will see that the Mak gets closer than the SW affecting the field of view. 

[Peter B]

OK Thanks.  I understand that magnification = telescope focal length divided by eyepiece focal length and that with the same eyepiece the 1900mm Mak will give more magnification than the 500mm refractor, but I didn't expect this to apply to a camera sensor which doesn't have a focal length.  It's just a CMOS chip, in my case measuring 23.6 x 15.8mm giving 12.3 megapixels, but the same principle obviously does apply to the camera sensor which must have some kind of 'effective'  or imaginary focal length.   With this in mind I did a little experiment and put a zoom lens on the Mak and adjusted it to get about the same size image as the camera above.  Very unscientific but it it worked out at just a bit less than 25mm on the zoom lens.  Is it a coincidence that this is about the same dimension as the width of the CMOS chip in the camera I wonder?

[TerryMcK]

Have a look at this post from Cloudy Nights which explains it really well

 

https://www.cloudynights.com/topic/583308-how-is-magnification-attained-with-a-camera-connected-to-the-scope/?p=7978481

[TerryMcK]

If the post on CN disappears I reproduce it here

”The term "magnification" is meaningless in imaging.  Consider that you could take a single image and display it on either a big screen TV or a smart phone.

 

Any telescope will produce a image at the focal plane, and that image is fixed with certain characteristics that are determined by the focal length and aperture of the optics.

 

 In a visual system, you use an eyepiece which is essentially a fancy magnifying glass.  Changing the focal length of the eyepiece will change how much magnification is applied.  That's where the term "magnification" comes from and it applies there.

 

In prime focus imaging, there is no eyepiece that does magnification.  Instead, there is a sensor with an array of pixels.  Each of the pixels "sees" some amount of sky, and that is defined as the image scale.  It's expressed in arc seconds (of sky) per pixel.  For a given sensor, if you increase the focal length of the telescope, it will "see" a smaller piece of sky (the number of arc seconds per pixel goes down).  This will mean that the target object covers more pixels.  Likewise, if you reduce the focal length, each pixel will "see" more sky and it will take fewer pixels for a given object, at the cost of resolution.

 

 The second characteristic of a sensor is its overall size.  The larger the sensor, the more sky will be covered.

 

And finally, how the image is processed is important.  In the setup that I've been using so far this spring and summer, I have a moderate sized sensor with lots of very small pixels (3.5 microns each).  I'm imaging with a scope with a longish focal length (2000mm).  This gives me an image scale of about 0.39 arc seconds per pixel.  It turns out that there are limits to how much resolution the sky will give you.  On the nights of steadiest seeing that I've had this year, the best focus I could get from a star is about 1.5 arc seconds.  So my system is trying to capture more detail that the sky is delivering.  As a result, at some point in the processing, I rescale the image to combine pixels.  This reduces the number of pixels (and thus the image scale) and makes the image half as wide and half as tall in terms of pixels.  This would make the image smaller on a given screen.”

 

[Peter B]

That's excellent, thank you so much for posting this.  It explains it all so well.

[Peter B]

I've done some calculations based on the excellent Cloudy Nights post and it works out that the Nikon D5000 sensor delivers 0.60 arcseconds per pixel on my 1900mm Mak and 2.28 arcseconds per pixel on my 500mm refractor - hence the difference in image size!  Definitely helps to understand what I can expect the Nikon to 'see' though each scope.  Thanks again for clearing this up.