What does "50mm" realy mean?

I know it is the length of the lens, but does it have

optical/mathematical meaning like "f" (f=lens's length/diameter)?

 

Is it the distance between the diaphragm and the film?

 

Is there a way to calc the angle of view of a lens when you know it's

length?

Some info you might find interesting in the <a href="http://www.photo.net/learn/optics/">static content</a>.

More specifically, Q2 under the "Lens FAQ" part.

It is the distance from the rear nodal point to the focal plane. This is usually an approximation to the nearest 10%. (Relative apertures may vary significantly more.) It would give the same perspective as a pinhole at the same distance.

Actually,that's pretty easy geometry and general.

<br>Take <i>d</i> the film frame diagonal(that is ~43mm for "35mm format"). Take the angle of view <i>a</i>. The tangent of the half of this angle will equal the diagonal <i>d</i> divided by the double of focal length <i>f</i> (triangle with a right angle formed by the half of the film frame diagonal, and focal distance drawn perpendicular to the film plane.)

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For 35mm format and 50mm lens, you end up with an angle of 2 x 23.4 ~ 47 degrees. While a 300mm telephoto gives you an angle of ~8.3 degrees.

<br>For the same 47-degrees "normal" angle of view, but on "6x6 format" film frame, you need a lens with a focal length <i>f</i>=98mm. [since "6x6" is not really 60mmx60mm but smaller, "normal" lenses for this format are usually considered a bit shorter than 98mm...they are around 80mm.]

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Of course, depending on whether the 50mm lens is focussed close or to infinity, the field of view does slightly change...but not that much.

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Hope it's clearer now. Cheers!

Considering the info I got, I made the calc for angle:

 

arccos((2*(f^2+12^2)-43.26^2)/(2*(f^2+12^2)))=a

 

Where 'a' is the angle, 'f' is the focal length and 43.26 is the film's diagonal length (43.26 is for 35mm film).

 

 

Thanks a lot for your help!

Leica rangefinder users will tell you that the FOV increase as the lens is focused towards infinity can be significant. In some cases 30% more can be included in the final negative than was shown in the brightlines (which show the FOV at 1 metre). This is equal to each side of the frame being 14% longer, which is a fair amount.

To make it somewhat easier:

 

a = 2 * arctan(43.27 /(2 * FL))

 

where a is the diagonal angle of view, 43.27 is the diagonal dimension of the negative (in this case 35mm), and FL is the focal length of the lens.

50 mm means that it is identical in focal length as the human eye.