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I see what you mean now. I wonder if there are any scientific studies done on this sort of thing. I'm fairly sure someone has tested this out.

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Now here's the "magical" formula that defines what you have do do to have a "normal" looking photo:
viewing distance : diagonal of print = focal length : diagonal of sensor
(Well, this is not entirely exact! But at normal magnifications this is a good approximate)

Now you can rearrange this formula to calculate e.g.:
focal length = viewing distance x diagonal of sensor : diagonal of print
Let's put some real data into this formula:
- an FF/FX sensor has a diagonal of 43mm
- a DIN A3 print (297mm × 420mm) has a diagonal of 51cm
- let's assume your viewing distance is 60cm (we'll look at this assumption later)

Then you can calculate the focal length that is needed to get a perfectly normal impression from this print as
60cm x 43mmm : 51cm = 51mm

"Bingo", I hear some of you say: "That confirms what we already knew: a 50mm lens is a normal lens for FF-format".
Yeah, but only if you print DIN A3 and view from 60cm distance

Let me give you a practical example: I'm sitting in front of a 26" monitor with a 650mm diagonal and I have a viewing distance of around 40cm from it. Now what would be the lens (on an FF/FX-body that gives me a normal viewing experience?
40cm x 43mm : 650mm = 26mm

Quite a difference, isn't it?!

Well, if you play a little with the formula you can clearly see that all depends on how close you look at an image. And before we dig deeper into assumptions and habits about viewing distances I encurrage you to test yourself:
How far away are you, when you want to take a detailed photo of a beautiful landscape in?
Please post your distance to the image and don't forget to also mention the diagonal of that image!

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Thomas (beware: Nikon-fanboy and moderator!) My Lens Reviews, My Pictures, My Photography Blog
D800+assorted lenses

I don't think its quite as simple as that.

The eye may see 20 odd mm but we only actually focus on part of that. therefore creating the 50mm, the rest is peripheral/extended vision.

That combined with a 1.0x viewfinder magnification should equal objects the same size as our eyes see them (70-90mm gets me close on my 5DMkII but thats obviously a bit more compressed)

I'd say its about 35mm for the same angle of view as when your focusing your eye on a fairly close object at least.

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I respectfully disagree, Daniel!

1. My definition of "normal" was not based on the angle of view that our eye(s) can see at all nor on the angle of view that our eye(s) can see sharp. Our eyes move around so they sweep a scene (or an image of a scene) left to right and up and down and our brain sort of "stitches" all the information together to create the impression of a larger area of optimal sharpness than one eye at any moment can produce. So I did not rely on the "sharpness" of the image. Btw. that would create another discussion about the impact of dof on "normalcy". Thus my definition of "normal" (and btw. all historic definitions of "normal") focuses on the perceived size and size-relations of objects that are mapped from a 3-dimensional scene onto a 2-dimensional image.

2. Talking about the viewfinder magnification in this context implies that you like to have a "normal" impression of the image viewed through the viewfinder. I've seen similar arguments which go like this: "What is the focal length of a lens that makes the image that you see looking through the viewfinder identical (in magnification) to the scene that you see with the naked eye?" Well yes, that's a valid question and the answer to that with my D300 is around 50mm. But: Once you've shot the image, you normally view it on a monitor or in print and your viewfinder magnification does not play any role there. Thus the formula for "normalcy" compares impressions when looking at a print/monitor with the impression you had of the scene with your naked eye, not with your eye to the viewfinder.

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Thomas (beware: Nikon-fanboy and moderator!) My Lens Reviews, My Pictures, My Photography Blog
D800+assorted lenses

Let me just elaborate a little on this remark. If you're not interested in the nitty-gritty detail of exact optics you can easily skip this post

As I already indicated the formula is wrong at any focusing distance closer than infinity
Why?
Because normally the lens-geometry changes when you focus closer. Let me give you an extreme example: Shooting 1:1 macro. You normally have to extend a macro lens by the exact same amount of its focal length to get to 1:1 magnification. So the distance of the lens from the sensor is now double that compared to focus to infinity.
So if you replace the "focal length" in the formula
viewing distance : diagonal of print = focal length : diagonal of sensor
by "lens distance" (meaning the distance of the lens from the sensor) you get
viewing distance : diagonal of print = lens distance : diagonal of sensor
You see the beauty of the equation now even better than before, don't you?

But: What exactly is lens distance? On the first look it is the focal length plus extension of the lens during focusing. Take a 50mm lens focusing at 60cm (1:10). The extension is 5mm. Now that is only 10% of the focal length so there is not much of a difference compared to the simplified formula. That catch with "lens distance" lies in the modern lens constructions, namely the IF-designs. Those IF-design don't simply extend the lens when focusing but they do some intricate internal dance of lens-groups relative to each other thus not only changing the focus but also the focal length and thus the lens geometry. See my article on the (in)famous "shrink-factor" for more details and examples.

As the shrink-factor of modern IF-designs effectively reduces the influence of focusing on the lens geometry (the obvious sign being that the length of the lens does not change through focusing) even a nerd like me wouldn't bother to go into calculating the exact influence of focus on the focal length of the perfectly "normal" lens.

This also has to do with the obvious variance in preferred viewing distances that far outshines any influence of the magnification of the lens.

And we will be discussing the "normal" viewing distance after a short break. Stay tuned
And please post your preferred distance to an image and don't forget to also mention the diagonal of that image!

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Thomas (beware: Nikon-fanboy and moderator!) My Lens Reviews, My Pictures, My Photography Blog
D800+assorted lenses

Well, what is "normal" viewing distance of an image?
There is this old rule that the diagonal of the image is the optimal viewing distance at least for printed images. With TV sets everybdy will tell you to sit much further away (like at least 3x the diagonal) but this is influenced by the resultion of the set and the quality of the material that you're viewing. But let's focus here on photographic high resolution images.
If you put this information into the formula you get as a result that the focal length of a "normal" lens should equal the diagonal of the sensor.
That gives you 43mm on a FF/FX/film camera and 28mm on a APS-C body with the ratio of 43/28 = 1.5x just reflecting the crop factor of the APS-C camera. Btw. on a four-thirds the normal lens has a focal length of 21mm.
Now this is the next surprise: 43mm is not exactly the much touted 50mm for FF/FX-bodies and 28mm is also shorter than the 35mm that many declare as "normal" lens on APS-C cameras, but at least both figures are in the ball-park.

I can only speculate about the reason that 50mm was declared "normal" and the few 45mm lenses soon became extinct: with the mirror box of full-frame SLRs behind the lens it was easier to design large-aperture normal lenses at 50-58mm then at 40mm. You can see proof of this still with e.g. the old Noct-Nikkor 58mm/1.2.
If you look at range-finder cameras, suddenly you'll find shorter normal-lenses like e.g. the Leitz Summicron 40/2.0 This seems to support my theory.

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Thomas (beware: Nikon-fanboy and moderator!) My Lens Reviews, My Pictures, My Photography Blog
D800+assorted lenses

Now there was this Q in the other forum: "how bout cropping an image"?

Well that one is easy: just replace "sensor diagonal" in the above formula by "cropped sensor diagonal" or divide the right side by the crop-factor.
This is why the "normal" lens for DX is 1.5x shorter than the normal lens for FX (all other things equal).

Btw. The formula can also be adapted for stitched panos: simply apply it to the geometry of each "tile" and make a good estimate about the viewing distance of the complete pano. The formula then reads:
viewing distance of complete pano : (diagonal of print : number of tiles) = focal length : diagonal of sensor
Let's say you're stitching a 1m x 30cm pano out of 5 portrait-oriented single shots and you love the slightly closer than image-diagonal view because there is sooo much detail in the image (let's assume arm's length = 60cm) then you should use the following

focal length = viewing distance of complete pano : (diagonal of print : number of tiles) x diagonal of sensor
= 60cm : (104 cm : 5) x 43mm = 124mm on an FX/FF-body (or 83mm on a DX/APS-C-body).
Or even easier: You use the original formula yielding 60cm : 104cm x 43mm = 24.8mm and multiply this by the number of tiles (=5) to arrive at the focal length of 124mm.

(for those of us who love to be exact: This assumes stitching without loss through overlap)

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Thomas (beware: Nikon-fanboy and moderator!) My Lens Reviews, My Pictures, My Photography Blog
D800+assorted lenses