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Camera equivalence to magnification

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#1 grkuntzmd

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Posted 17 April 2021 - 08:17 AM

I know that this topic has been discussed a few times before , but I'm trying to find the flaw in my way of calculating the magnification equivalence for imaging.

 

Imagine a square postage stamp glued to a wall 10 meters away from a telescope on a tripod. You insert an eyepiece and note that when viewing the postage stamp, the stamp's diagonal exactly fills the field of view.

You then replace the eyepiece with a camera with a square sensor. The image exactly covers the postage stamp; there is no border around the stamp and there is no cropping.

 

Could you not argue that the camera "magnification" is the same as the telescope-eyepiece magnification? After all, if you look at the image at a later time, you will say, "Yes. That is exactly what I saw in the eyepiece." (Ignoring that the eyepiece view was round and that the image is square.)

 

Eyepiece:

TFoV = (AFoV / TFL) × 60

where

TFov: True Field of View in arc minutes

AFoV: Apparent Field of View

TFL: Telescope Focal Length

Mag = TFL / EFL

where

Mag: Magnification

EFL: Eyepiece Focal Length

 

Camera:

TFoV = (Diag × 3460) / TFL

where

Diag: Diagonal dimension of (square) sensor

 

So

(AFoV / Mag) × 60 = (Diag × 3460) / TFL

 

Solving for Mag

Mag = (AFoV × 60 × TFL) / (Diag × 3460)

 

Example:

Plössl with AFoV of 52°

Telescope with FL of 2000 mm

Sensor with width and height of 11.31mm (Diag = 16mm)

 

Sensor "Mag" = (52 × 60 × 2000) / (16 × 3460) = 112.7

 

If I used an 18mm Plössl eyepiece in this telescope, I would see exactly the same field of view as the camera sensor.

 

What is wrong with my logic?

 

Update: spelling correction


Edited by grkuntzmd, 17 April 2021 - 08:20 AM.


#2 spereira

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Posted 17 April 2021 - 08:55 AM

Moving to Beginning Deep Sky Imaging.

 

smp



#3 StarAlert

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Posted 17 April 2021 - 08:55 AM

I’ve thought about this, too.
The problem you run into is what eyepiece are you using for your comparison? A 26mm plössl or a 13mm Ethos? They both have the same FOV but the Ethos has twice the magnification. 


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#4 grkuntzmd

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Posted 17 April 2021 - 09:12 AM

I’ve thought about this, too.
The problem you run into is what eyepiece are you using for your comparison? A 26mm plössl or a 13mm Ethos? They both have the same FOV but the Ethos has twice the magnification. 

That occurred to me also. I have some of the 70° Orion eyepieces and my method breaks down with them also.



#5 StarBurger

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Posted 17 April 2021 - 09:21 AM

Your's is an apples to oranges comparison. +1 for StarAlert.

You could have EP's of the same FL but different fields of due to different field stops. An 82 deg EP will show more of the postage stamp than a 52 deg of the same focal length.

Then what can you say of the final image from the camera displayed on monitor screens of different widths and viewed from different distances?

If you view the monitor image from 6 inches, is that a greater magnification than viewed from 6 feet? Neither of which will match the direct view through the EP.

No amount of fancy arithmetic can demonstrate any equivalence.

Magnification is only applicable to what your eye sees in the EP/Telescope combination compared to what the eye sees without a scope at all.

We must cease speaking of magnification when it comes to sensors (except the Mk 1 eyeball). smile.gif



#6 WadeH237

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Posted 17 April 2021 - 09:29 AM

What is wrong with my logic?

The issue is that - unlike a visual system - an imaging system does not have any elements that magnify the target object, so there is quite literally no magnification happening.  Well, unless you are doing afocal imaging, but I suspect that's not what we're talking about.

 

Your telescope produces an image at the focal plane.  The size of that image is 100% determined by the focal length of the telescope - and nothing else.  When you have a visual system, you use an eyepiece that is essentially a magnifying glass.  When you put in a long focal length eyepiece, you get less magnification than you get when you put in a short focal length eyepiece.  But understand that the eyepieces are magnifying the same image from the focal plane.  Changing the eyepieces does not change the size of that image.

 

When you image, you are capturing that same image from the focal plane.  But unlike an eyepiece, the camera sensor does not magnify the image.  It just captures it.  You can then render it for viewing at any size you like.  You can also change the resolution by up or down scaling it.  But all of this happens long after the image has been captured.

 

More to the point, what use is there in trying to apply the term magnification to an imaging system?  It's not a useful concept.  The image scale and field of view tell you everything that you need to know.  Even DSLR lenses don't advertise based on magnification.  They put focal length and focal ratio (which tells you the aperture) front-and-center, which is appropriate.  I have to wonder if this notion has its roots in the fact that cheap telescopes tend to brag about ridiculous levels of magnification...


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#7 Cotts

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Posted 17 April 2021 - 09:36 AM

You're really just wanting to find an eyepiece to match the field of view of your camera it seems.  Stick with the concept of field of view and you'll be further ahead.  As others have noted magnification is not a simple concept.

 

The field of view with just your camera at prime focus  (I assume you are not using eyepiece projection) can be found here.

 

For the sake of discussion let's pretend that your camera's field is exactly one degree with your telescope. 

 

Eyepiece actual field of view requires two calculations...  Focal length of scope divided by focal length of eyepiece = Magnification.   Then Eyepiece apparent field of view divided by magnification = real field of view.. 

 

You mention a 2000mm telescope.  A 20mm plossl of 50º Apparent field would give 100x.   the 50º divided by 100x gives a 0.5º field of view...   So a 40mm Plossl would be the right eyepiece..  

 

So, yeah, we did have to use 'magnification' but only as a means to an end.  And you are still having to deal with a circular eyepiece view compared to a square camera image...

 

You'll be able to say my scope with 'this' eyepiece gives the same field of view as my camera in the same scope...   Which, I take it, is your goal.

 

Bonus - you can use a second telescope to mimic the camera field of view and mount it side-by-side with your imaging scope to find and frame your targets...

 

Dave


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#8 TOMDEY

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Posted 17 April 2021 - 09:58 AM

Same angular subtent, as viewed in the final displayed rendition is the only meaningful definition of unity magnification. And that's easy to compute, assuming that one has the final display format (and location of observer) defined --- which is often not the case!

 

I worked at Kodak, and the Graphic Arts guys (of course) take all this kinda stuff (and far more) into account. So two drivers determine the selctions and matching of Taking Lens, Printing (or projection) Lens and audience placement --- those two drivers are unity magnification and freedom from angular distortion. When those two are satisfied --- the experience takes on "Presence" --- as in Looking ~Real~    Tom



#9 BQ Octantis

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Posted 17 April 2021 - 08:15 PM

There is no equivalence to "magnification" between visual and imaging.

 

For visual, the system is afocal, with the lens-retina system placed behind the telescope-eyepiece system. Magnification is given simply as M = ftelescope / feyepiece . But have you ever wondered why you never get a magnification of M = 1? You would need an eyepiece with a focal length of the telescope! So why not place the eye at the telescope with no eyepiece? Because the exit pupil of the bare telescope is too large to enter the entrance pupil of the lens of the eye with the eye focused at its near point. Eyepieces condense the entire light column into an exit pupil that can fit into the entrance pupil of the eye to form an image on the retina. "Magnification" by the eyepiece is simply a consequence of the compound lens system. The field-of-view (FOV) of the scene through the eyepiece at its "magnification" is exclusively determined by the apparent field of view (AFOV) of the eyepiece—a design feature of the eyepiece (lens + field stop).

 

A bare sensor has no such constraint on its entrance pupil. The size of the physical image is governed exclusively by the telescope's focal length. The FOV captured by the sensor is governed by the sensor's physical size (in any dimension)—and what that length subtends of the light cone from the telescope:

 

gallery_273658_12412_8687.png

 

Sensors typically come in APS-C or full frame sizes, so the FOV for a given focal length will generally fall in one of those two buckets.

 

At a given focal length, the sample size of a pixel (i.e., the angle subtended by each pixel on the sensor) is simply a function of pixel's physical size. While the FOV calculation above could be applied to a pixel, small angle approximations allow for simplification of the calculation for a telescope (and at least at the center of the scene for most camera lenses):

 

gallery_273658_12412_7950.png

 

Pixel sizes vary greatly across sensors—typically in the 2-6 µm range. Two APS-C sensors (with the same sensor FOV) could produce very different "magnifications" (angles subtended by each pixel), which could be construed as a difference in "magnification"—indeed, a sensor with 3µm pixels would produce an image on the screen twice as large as that from a sensor with 6µm pixels. But technically, this is simply the physical sample rate (arcsec/pixel) that is a consequence of the pixel pitch (pixels/µm) applied to output light cone of a lens of a given focal length. We don't call that difference "magnification".

 

So again, "magnification" is meaningless as a currency across domains; if we use that word, we need to be very careful in clarifying what we're trying to convey. You can certainly find equivalent scenarios, with equivalent fields-of-view probably being the most meaningful:

 

  • For an eyepiece: FOV = AFOVeyepiece × ftelescope / feyepiece
  • For a sensor: FOV = 2 × atan ( Lsensor / 2ftelescope )

 

For the sensor, you have to pick which dimension you are going to make equivalent to the FOV through the eyepiece: width? height? diagonal?

 

Hope that helps.

 

BQ

 

P.S. There is another magnification concept for eyepiece projection photography that is "magnification over prime". That has its own geometry calculations. And I haven't even touched on afocal photography (like a cell phone at the eyepiece)…


Edited by BQ Octantis, 18 April 2021 - 06:04 AM.

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#10 BQ Octantis

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Posted 18 April 2021 - 03:02 AM

For what it's worth, I struggled with illustrating "what I saw through the eyepiece" with planetary last year. I do eyepiece projection planetary imaging, so I literally shoot what I see through the eyepiece. Starting in August, there were three novel storms in the Northern Tropical Zone that circumnavigated the planet. I both eyeballed them and shot them; posting a representation of them as breadcrumbs for other Cloudy Nighters depended on whether they were visual observers or planetary imagers. I took just one stab at trying to represent both; in addition to size, there are also brightness, contrast, and SNR components to what you see vs. what you can image:

 

post-273658-0-11295800-1598569352.png

 

This makes representing what you can see through the eyepiece even more challenging—even when you account for FOV and pixel pitch…

 

BQ


Edited by BQ Octantis, 18 April 2021 - 06:00 AM.


#11 WadeH237

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Posted 18 April 2021 - 08:49 AM

There is no equivalence to "magnification" between visual and imaging.

Nice!

 

That is probably the best explanation of this that I have ever seen.



#12 kathyastro

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Posted 18 April 2021 - 09:14 AM

When you calculate magnification as field of view, as in the OP, you are comparing fields of view.  Why not just say so?

 

We tend to think of magnification in fictitious units called "X"s.  What a number like 50x means is that the image is "50 times bigger than...<something else>."  And at that point the number becomes meaningless unless the basis of the comparison is clearly understood. 

 

Field of view isn't inherently a comparison, so we shouldn't tie ourselves in knots trying to make it sound like one.  The field of view of my imaging setup is 1.38 degrees, measured on the diagonal.  My guide scope system has a field of view of 3.01 degrees.  Now, with two values to compare, I can say that the FOV of the guide scope is a bit more than twice the FOV of the imaging scope.  But each number on its own is not a comparison.


Edited by kathyastro, 18 April 2021 - 09:18 AM.

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#13 BQ Octantis

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Posted 18 April 2021 - 04:13 PM

Barlows have an outright magnification declared. A 2× Barlow multiplies the focal length of the objective by two. But how much it magnifies the image depends on where you put it in the imaging train: ahead of a diagonal and it magnifies it by 3×! But attached to an eyepiece or a nosepiece for a sensor, it does what it says.

 

And what about binoculars? Is the multiplier relative to my eyesight? The FOV is declared on the body…

 

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