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Archive 2017 · Conversion ratio?

  
 
PatrickSweeney
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p.1 #1 · p.1 #1 · Conversion ratio?


OK, perhaps a silly question, but here goes;

Is there a chart, or formula, that converts the length of a telephoto, to the comparable X of a set of binoculars?

In other words, if I slap a 300mm onto a FF (crops would be different, by 1.6, obviously) what does that translate to in a pair of binoculars? 4X? 6X? 10X?



Mar 16, 2017 at 03:18 PM
jcolwell
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p.1 #2 · p.1 #2 · Conversion ratio?


I have seen it (relationship between binocs 'power' and lens focal length), but I don't remember the details, and I'm all googled-out, for today.

The best way to figure it out would probably be to match AOV, which is 8.2 deg (diagonal) for the 300mm lens on FF.

There is a slightly different way of looking at it. I have a Kenko Lens2Scope converter that essentially is a telescope eyepiece that you can attach to a camera lens. The Kenko that I have is a "10x" converter, which means that the AOV with the converter on my Tamron SP 300/5.6 lens, and this 'scope' up to my eye, is the same as a 3000mm lens on FF. IOW, I can scan the distance with this small 'scope' and get much higher detail than I would get from my 500mm lens on a FF camera, or even on a 1.6x cropper.

Other converters provide different magnification. For example, the old Pentax K-mount monocular converter was 5x.

Kenko Lens2Scope https://www.bhphotovideo.com/c/product/857279-REG/Kenko_K_LS10_CESB_LENS2SCOPE_for_CANON_EF_S.html

Pentax Monocular converter http://www.bdimitrov.de/kmp/lenses/adapters/monocular.jpg



Mar 16, 2017 at 04:19 PM
AJSJones
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p.1 #3 · p.1 #3 · Conversion ratio?


jcolwell wrote:
I have seen it (relationship between binocs 'power' and lens focal length), but I don't remember the details, and I'm all googled-out, for today.

The best way to figure it out would probably be to match AOV, which is 8.2 deg (diagonal) for the 300mm lens on FF.

There is a slightly different way of looking at it. I have a Kenko Lens2Scope converter that essentially is a telescope eyepiece that you can attach to a camera lens. The Kenko that I have is a "10x" converter, which means that the AOV with the converter on my Tamron SP 300/5.6 lens, and
...Show more
I did not know of such things - I might be inclined to add one to my kit

I've always used a rough measure that ~50mm is 1x, approximating the unaided view (so objects look neither magnified nor reduced) and divided that into my lens mm, so a 400mm is roughly 8x. Not sure how kosher this is, however.



Mar 16, 2017 at 04:39 PM
Jeff Donald
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p.1 #4 · p.1 #4 · Conversion ratio?


The 50mm = 1x is what the Zeiss and Leica Sports/Optics reps always told us to use when I was in camera retail many years ago.


Mar 16, 2017 at 08:00 PM
Spikey131
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p.1 #5 · p.1 #5 · Conversion ratio?


Divide the focal length by 50 to get the apparent magnification the viewfinder.

300mm lens: 300/50= 6x.

Magnification is not the same thing as angle of view. I have 8x binoculars with 6.5 degree angle of view and another 8x pair with 8.3 degrees.

A 300mm lens in a Canon 100% FF viewfinder has a 6x magnification and a diagonal angle of view of 8 degrees, 15 seconds.

Now imagine a 300mm lens designed for an old 6x6cm medium format camera (i.e. Hasselbad). It would have the same 6x magnification in a bigger viewfinder, so it would have a wider angle of view.

The same 300mm lens on an APS-c camera also has 6x magification. Now it is in a smaller viewfinder, so the angle of view is smaller. The image of objects projected by the lens on the sensor is the same size regardless of the size of the sensor.

The difference in angle of view in binoculars of the same magnification is determined by the design of the eyepiece. The magnification is determined by the focal length of the eyepiece, but the angle of view is determined by other features of eyepiece design.

There is no eyepiece in the camera to enlarge the image, just a flat piece of light sensitive material (film or digital). The size of this material will change the angle of view, but not the magnification.



Mar 16, 2017 at 09:12 PM
Liquidstone
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p.1 #6 · p.1 #6 · Conversion ratio?


Spikey131 wrote:
Divide the focal length by 50 to get the apparent magnification the viewfinder.



By using that formula, my old rig is about 256x. Dark, but a legitimate 256x.

http://www.pbase.com/liquidstone/image/59586426.jpg



Mar 16, 2017 at 11:24 PM
artificialyello
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p.1 #7 · p.1 #7 · Conversion ratio?


Jeff Donald wrote:
The 50mm = 1x is what the Zeiss and Leica Sports/Optics reps always told us to use when I was in camera retail many years ago.


Well 50mm is the classic "normal" but it's really an old approximation upwards to make it easier to make a good lense... The classical definition of a normal lense is the diagonal of the used frame so that would be 43mm for FF and 70mm for MF 6x4.5.



Mar 16, 2017 at 11:42 PM
Rainer
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p.1 #8 · p.1 #8 · Conversion ratio?


The magnification depends a bit on the camera used. Have a look at the specifications of the cameras: for a 50 mm lens, the magnification is
0.71x (EOS 6D)
0.76x (EOS 1DX II)
0.82x (EOS 750D)
1.0x (EOS 7D II)
If you use a different lens, just multiply these values by (focal length / 50 mm).



Mar 17, 2017 at 12:53 AM
Spikey131
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p.1 #9 · p.1 #9 · Conversion ratio?


Rainer wrote:
The magnification depends a bit on the camera used. Have a look at the specifications of the cameras: for a 50 mm lens, the magnification is
0.71x (EOS 6D)
0.76x (EOS 1DX II)
0.82x (EOS 750D)
1.0x (EOS 7D II)
If you use a different lens, just multiply these values by (focal length / 50 mm).



So what is happening here is that Canon is using eyepieces with different focal lengths in their viewfinders. This changes the apparent magnification in the viewfinder. Only in the viewfinder. So the viewfinder of a 750D will look like 5x binoculars and the 7D II will look like 6x binoculars with a 300mm lens.

But the angle of view will be the same for all cameras with the same size sensor.

And the image size on the sensor will be the same for all cameras. This is determined only by the focal length of the lens.



Mar 17, 2017 at 05:59 AM
jcolwell
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p.1 #10 · p.1 #10 · Conversion ratio?


Spikey131 wrote:
... Canon is using eyepieces with different focal lengths in their viewfinders.


Yes, and finder "coverage" (cov.) also varies between different cameras. Here's the cameras listed by Ranier, with both finder magnification and finder coverage.

6D mag. 0.71x, cov. 97%
1DX II: mag. 0.76x, cov. 100%
750D (T6i): mag. 0.82x, cov. 95%
7D II: mag. 1.0x, cov. 100%



Mar 17, 2017 at 07:31 AM
SoundHound
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p.1 #11 · p.1 #11 · Conversion ratio?


The concept in 'scopes is all about the objective diameter (read light gathering power). After which you can use various "eyepiece" magnifications to trade off exit pupil (amount of light to the eye) verses magnification.

A photographic lens Telextender "TC" is just an auxiliary "eyepiece" which increases magnification at the cost of light (and loss of OQ and a bit of extra light due to more glass) to the sensor.

To find the exit pupil divide the objective by the magnification (something like the inverse of the F stop-the higher the number the more light) Ex: A 10 power binocular with a 40mm objective yields a 4mm exit pupil which is large enough to cover the eye's pupil in most conditions.

A 85mm objective spotting scope with an eyepiece of 40x has an exit pupil of 2.1mm which is typical for "scope" use but you must be careful to center your eye's pupil. I have the Zeiss 85mm objective "Diascope" with both a 40x fixed eyepiece and a 20-60x zoom eyepiece.

I mostly use the 40x which has a better field of view than the zoom (narrows down around 35/40X to preserve the best 60x FOV). The 60x "power" also yields a 1.4mm exit pupil which I find to be too dark and narrow for comfortable viewing.

I have the Nikon "Lens Scope Adaptor" which turns a tele lens into a spotting scope (divide the objective by 10) but alas it only works with older lenses that have an F stop (not 'G" lenses). So it is useless with my 400mm F2.8 VR with its huge 142mm objective (which would also support a 70x "eyepiece" at a 2mm exit pupil).



Mar 19, 2017 at 06:07 AM
jcolwell
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p.1 #12 · p.1 #12 · Conversion ratio?


Here's 400mm on FF, with a red circle to show what's visible with my 10x Kenko Lens2scope, through the same lens.

The lens was mounted on a tripod for the photo, and then the camera was replaced by the Lens2scope. The Lens2scope 'coverage' was hand sketched, using the actual image on the camera display for guidance.

Of course, when using the scope, the image within the red circle (below) actually fills your eyeball. The magnification is very significant.



© jcolwell 2017


FF image at 400mm, red circle = Lens2Scope view, with same lens



Edited on Mar 19, 2017 at 01:40 PM · View previous versions



Mar 19, 2017 at 12:18 PM
Tapeman
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p.1 #13 · p.1 #13 · Conversion ratio?


I have no clue about the magnification, but when looking through my 5oomm/2X, I see larger subjects with better detail than when using my 10X binoculars. Even with the 1.4X on that lens the image is larger and clearer.

The field of view is smaller than the binoculars.



Mar 19, 2017 at 12:32 PM
AJSJones
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p.1 #14 · p.1 #14 · Conversion ratio?


Tapeman wrote:
I have no clue about the magnification, but when looking through my 5oomm/2X, I see larger subjects with better detail than when using my 10X binoculars. Even with the 1.4X on that lens the image is larger and clearer.

The field of view is smaller than the binoculars.


Using the rule of thumb guide above, your 500 would be ~10x and the 1.4 and 2x would yield 14 and 20x (more if you use 43mm as 1x). Your observations fit the math. In this case FoV is not relevant to sizes of objects in the image.



Mar 19, 2017 at 01:26 PM





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