Magnification at prime focus

[Scorp56 (Peter)]

Hi all,

I'm new to astrophotography and just wondering how to determine magnification when using a CCD webcam at prime focus.
(The webcam lens has been removed) I am assuming when image is in focus the CCD sensor is at prime focus???

I am using a 200mm Newtonian with 1200mm focal length and CCD sensor is 4.6X4.0mm in size.

Thanks for any feedback.
Cheers,

Peter

[bmitchell82 (Brendan)]

when at prime focus its not common to talk about magnification, it is expressed as Field of View or FOV, New astronomy press has a neat little program called CCD Calc written by Ron Wodaski. I suggest you download that

Brendan

[Scorp56 (Peter)]

Thanks Brendan
I will check that program out.

Cheers

Peter

[Geoff45 (Geoff)]

Magnification is a pretty meaningless concept in this situation. First of all, you are not magnifying, you are diminishing. The size of the Andromeda galaxy at prime focus is measured in millimetres. The real thing is 100000+ light years across. It's not like taking a picture of a cockroach and pinning a real cockroach to the pic and comparing the two. Even comparing angular sizes is a bit tricky. If you hold a picture 20cm away. it's angular size is twice as big as if you hold it 40cm away. If you print it on A3, it's twice the size of an A4 print, so is the magnification twice as much? You can go round and round chasing your tail here trying to make sense of it all.

However, here is a rough guide. Print the pic on A4 paper and hold it about 30cm away. If the long side has field of view of x degrees, then the image looks about 50/x times bigger in angular size than if you looked at it in the sky with the naked eye. So if you have a half degree field, things look about 100 times bigger than with the naked eye.

[astro744]

Actual field of view (deg) = linear sensor (or film) width (mm) x 57.3 / focal length (mm).

In your case you have 4.6 x 57.3/1200 = 0.220 deg. and 4.0 x 57.3 / 1200 = 0.191 deg.

By using the same CCD on a telescope that has half the focal length will double your field of view as will doubling the CCD sensor size
Multiply by 60 to convert deg field of view to minutes, eg 0.22 x 60 = 13.2min. Multiple again by 60 to get field of view in seconds. (or multiply deg x 3600). Field expressed in minutes is useful for imaging galaxies and other deep sky objects and field expressed in seconds is useful for imaging planets.

In the days of film photographic magnification was related to that of a 50mm camera lens giving a magnification of one on 35mm (36 x 24mm) film). Therefore 1200mm focal length = 1200/50 = 24x magnification. If you had a full size CCD sensor matching film dimensions you could say your magnification is 24x compared to a standard 50mm lens. Since cameras have all different sized sensors and lenses multiplying factors are often referred to to be able to compare to a 50mm lens/35mm system. Most DSLR have a sensor 1.5x smaller than a film camera and therefore the multiplying factor is 1.5x. Referring to magnification is really meaningless for photo work and field of view can always be calculated by the above formula.

[Scorp56 (Peter)]

Thanks for the information and formula astro744.
I understand the "magnification" issue now and have a useful set of numbers to play with.

Cheers
Peter

[astro744]

No worries! In case you're wondering, 180/Pi = 57.3. See http://en.wikipedia.org/wiki/Radian for further info.

[Tallstock (Peter)]

Hi,
I connected - prime focus - my new Canon 550D DSLR (Image Sensor size 22.3 x 14.9mm with multiplying factor 1.6x) to my dob recently to look at the Moon and showed my wife the result using Liveview on the laptop. The first question she asked "what's the magnification"?
I had no answer.
We are used to looking through various eyepieces and learning about magnification so we are interested in this discussion. She and I don't want to get into discussion about a "meaningful concept".
So if my 1200 focal length/50 old film camera = 24x magnification how do I correctly calculate magnification?
Peter

[astro744]

Originally Posted by Tallstock

Hi,
I connected - prime focus - my new Canon 550D DSLR (Image Sensor size 22.3 x 14.9mm with multiplying factor 1.6x) to my dob recently to look at the Moon and showed my wife the result using Liveview on the laptop. The first question she asked "what's the magnification"?
I had no answer.
We are used to looking through various eyepieces and learning about magnification so we are interested in this discussion. She and I don't want to get into discussion about a "meaningful concept".
So if my 1200 focal length/50 old film camera = 24x magnification how do I correctly calculate magnification?
Peter

OK, I'm from the days of film and have not done any astrophotography recently other than a few afocal shots with a compact digital. I did spend a bit of time with both prime focus and eyepiece projection SLR astrophotography many years ago.

In the days of film there was 35mm format also called 135. There was also 126 (compact instamatic) and 127 (4x4cm roll format) cameras and there was also a small pocket instamatic that used 110 format film. See http://en.wikipedia.org/wiki/Film_format for a list of many others.

For the professional there was also 4.5x6cm medium format, 6x6cm medium format, 6x7cm medium format and 10x12.5cm large format film for use in medium and large format cameras.

Since 35mm film which has dimensions of 24x36mm and was the most popular and is still sold and used today; it is this format that digital DSLR CCD/CMOS sensors and equivalent lens focal lengths are compared to hence the multiplying factor. You can see that if you multiply your sensor size by 1.6x you get 14.9x1.6=24mm and 22.3x1.6=36mm,

Now forget the multiplying factor for now as it does not affect the size of the image, only focal length of the lens or telescope does. The formula for image size at the film plane is as follows:

Linear image size = angular diameter of object x focal length of telescope(or lens) / 57.3. eg. The Moon is 0.5deg in diameter. For your 1200mm focal length telescope you get an image size of 0.5x1200/57.3 = 10.5mm. If you get some plastic film transparency (or tracing paper) and hold it flat at the focal plane (film plane) of your telescope (no camera or eyepiece attached) you will get an image of the Moon that is 10.5mm in diameter (once focused). You'll even spot some detail!

Now if you have a film camera your 10.5mm Moon will be a small circle compared to the 24mm x 36mm film plane area, but still a good size to give plenty of detail. If you now remove your film camera and insert into the focuser your DSLR with 14.9 x 22.3mm 'film plane' your Moon image will still be 10.5mm but it will fill more of your smaller total area of the sensor. If you were to multiply the Moon image size by 1.6x it would equal in ratio to that of 35mm film. You would get the same result if you had a telescope of 1200x1.6=1920mm focal length and a full 24x36mm DSLR or SLR. eg,. 0.5x1920/57.3 = 16.8mm. (10.5x1.6=16.8mm).

Your magnification factor is always referred to a 50mm lens on 35mm film. Therefore 1200/50=24x and 1920/50=38x. Now you can print an image taken with a film camera on standard 10x15cm paper and get a nice picture. The image of the Moon taken with the DSLR can also be scaled up to 10x15cm paper by a factor of 1.6x and will look bigger OR printed at the same scale with the 10x15cm paper cropped to produce a smaller print (6.3x9.4cm paper) but same Moon size as on the SLR.

I personally don't like the use of the magnification factor as it is really a meaningless term since there are so many different film formats and yet we chose to use the 135 format only to apply a magnification factor to. However, since 135 format was the most popular size I can understand its use to allow the user to easily convert focal lengths to a format they are familiar with. Just look at the lenses on the compact digital cameras; they are usually less than 10mm in focal length and this means nothing to the average person. The CCD sensor size too is small compared to film but the combination gives an equivalent 35mm film format ratio.

On a typical SLR, 50mm was considered normal, 35mm wide, 28mm very wide, 70-80mm portrait, 135mm telephoto, 200mm longer telephoto. The same size lenses on a typical DSLR with either 1.5x or 1.6x smaller chips effectively operate as 1.5x or 1.6x longer lenses. This is why DSLRs's are offered with zoom lenses beginning at either 18mm or 24mm to match either 28mm or 35mm in 25mm format.

I hope I have cleared up and not clouded the magnification issue at prime focus. The main thing to remember is a given focal length telescope will always give the same image size no matter what the sensor (film) size is. How much you then want to magnify further by enlarging the image when printing or displaying is up to you.

[Tallstock (Peter)]

I am confused by your answer but thanks for trying.
The answer I am looking for seems to be 38x (1200 x 1.6 = 1920/50).
My 25mm eyepiece gives me 48x magnification.
Replacing that eyepiece with the camera shows crater details in the image on the camera's LCD screen which seem to be a bit smaller say about 40x (give or take).

[Robert9 (Robert)]

Hi Peter,
In the good old days of 35mm film photography, a "standard" 1:1 magnification was accepted as being a photo shot with a 50mm lens. A 200mm lens therefore gave a 200/50 = 4x magnification. In these terms, if you take the standard lens on your DSLR as giving a 1:1 image mag. then by dividing the focal length of your scope by the focal length of your "standard" lens, you should get a reasonable idea of the mags.
I have also heard that a truer standard lens focal length is that numerically equal to the diagonal of the image frame. In the case of a 35mm frame (24 x 36mm) this would be 43mm. Hence a 200mm lens would have, by this method, a mag of 4.3x
Hope that helps a bit.
Robert

[Tallstock (Peter)]

Thanks Robert.
I have wondered what magnification my telephoto lens (set at 200mm) gave when used on the camera. It seems 4x or thereabouts.

My Image Sensor is 14.9 x 22.3 so the diagonal would be about 25.

The scope at 1200 x no multiplying factor/43= 28x.
Maybe 1200 x no multiplying factor/ 25 = 48x
Maybe 1200 x 1.6 = 1920/25 = 76x
Maybe 1200 x 1.6 = 1920/43 = 44x
Maybe 1200 x 1.6 = 1920/50 = 38x
I am even more confused.
Peter

[Robert9 (Robert)]

Hi Peter,
Try this. If you have a zoom lens, use one eye to focus the camera on some object and compare the size of the object with th other eye. Adjust the zoom until the two images coincide. The value of the focal length on the lens will be the 1:1 value. If that comes at 50mm, then 200mm will certainly give mags. of 4x. I do have a little concern about the interpretation of the markings on a lens, (whether the multiplying factor has been incorporated or not) but maybe we won't go into that until we have the results of your test.
Robert

[Tallstock (Peter)]

Hi Robert,
My 55-250mm telephoto focused at about 70mm seems to do the trick. However, being new to this and not having perfect eyes I normally use Liveview which shows a completely different image size. I bought the camera because it was supposed to be easy to learn "entry level" and had Liveview. My learning curve is vertically upward.

The question that Scorp56 originally asked was about magnification with camera on scope at prime focus and this is where I remian confused.
Peter

[Robert9 (Robert)]

Peter,
I think there are a lot of truths within the reponses you have to your question, esp. from Brendan and Astro744.
The magnification though is what is perceived by the eye compared to what is recorded by the camera. If the image size on the film/CCD plane is the same as your eye perceives it, then you have 1:1. But when you take your pic and display it on a sheet of A4 paper and it fill the sheet, you have 'n' times that amount.
Mathematically though, your mags are f.l. of scope divided by the f.l. of your camera lens when set to 1:1. A value of 70 for the latter seems a bit high to me. I did the test on my Olympus E-620 and got 50mm, which I'm happy with. The value you got suggests a larger format camera! However, when all is said and done, it is all very subjective.
Hope I haven't confused the issue too greatly.
Robert

[Tallstock (Peter)]

Hi Robert,
The Canon 550D Manual says "35mm-equivalent focal length is approx 1.6 times the lens focal length."
My eyes might be playing tricks on me so the 1:1 at 70mm is a bit of a guess. At 55mm the ratio was not 1:1.
Peter

[Robert9 (Robert)]

Hi Peter,

If what your manual suggests, then f.l. ~ 30mm is equivalent to 50mm on 35mm format. One can only assume that the f.l. markings on the camera lens are true and not adjusted as 35mm equivalent. As I said last time, 70mm seems rather high. Check it out again - one eye against the view-finder window, the other eye peeping around the camera. The live-view is of no use.
Beyond that Peter, I think it becomes rather academic, and you should tell your wife anything that seems reasonable. I'm sure she'll be happy with that. ;-)

Just remember, 1:1 is when the image size on the sensor is the same size as the object seems to your eye. For this to happen, not disregarding all that I and others have said, the optical magnification which you would report is the f.l. of your scope divided by the focal length of the camera lens when the image it produces on the detector, (which is what you see through the view-finder) has the apparent same size as when you view the object with the naked eye.

Robert

[Tallstock (Peter)]

Hi Robert,
Thank you for taking the time and effort to help me with this.
A crucial bit of information is that "live-view is of no use".

Neither my wife nor I are able to comfortably use the tiny viewfinder so my "70mm" could be way off.

I think I will stick with the 38x mag and leave it at that.
Thanks again
Peter

[Barrykgerdes]

Hi Peter

This is a question that is often asked but I very seldom see an understansdable answer.
There are two parts to this question. The first is what physical size the object at prime focus and then how much of this will illuminate the chip.

The first size will be the ratio of the object distance to the focal distance. As an example a pole 1000 mm high at a distance of 20,000 mm will have a size in the focal plane (distance 1200 mm) of ((1200 / 20000) x 1000 ) mm = 60 mm.
This will illuminate only 4/60 of the pole image on the camera chip giving the effect of a high magnification.
If you double the focal length you will double the height of the image.
When we talk about magnification it is in reference to the visual image you see when you look at the focal plane image through another lens with a short focal length, say 10 mm. The apparent magnification then will be 1200/10 or 120x

With this brief answer do not be afraid to ask questions for further clarification

Barry