Dammit!

Go to A+R to look at some nicely designed housewares. They don't allow image-linking and I am to lazy to edit my entry with all those images.

I wanted to learn more about wide angle lenses so I Googled for more information and got this crazy math equation:

Consider a 35 mm camera with a normal lens having a focal length of F=50 mm. The dimensions of the 35 mm image format are 24 mm (vertically) × 36 mm (horizontal), giving a diagonal of about 43.3 mm.

Now the angles of view are:

* horizontally, \alpha_h = 2\arctan\; h/2f \approx 39.6°
* vertically, \alpha_v = 2\arctan\; v/2f \approx 27.0°
* diagonally, \alpha_d = 2\arctan\; d/2f \approx 46.7°

Consider a rectilinear lens in a camera used to photograph an object at a distance S1, and forming an image that just barely fits in the dimension d of the frame (the film or image sensor). Treat the lens as if it were a pinhole at distance S2 from the image plane (technically, the center of perspective of a rectilinear lens is at the center of its entrance pupil.

Now α / 2 is the angle between the optical axis of the lens and the ray joining its optical center to the edge of the film. Here α is defined to be the angle-of-view, since it is the angle enclosing the largest object whose image can fit on the film. We want to find the relationship between:

the angle α
the "opposite" side of the right triangle, d / 2 (half the film-format dimension)
the "adjacent" side, S2 (distance from the lens to the image plane)

Using basic trigonometry, we find:

\tan ( \alpha / 2 ) = \frac {d/2} {S_2} .

which we can solve for α, giving:

\alpha = 2 \arctan \frac {d} {2 S_2}

To project a sharp image of distant objects, S2 needs to be equal to the focal length F, which is attained by setting the lens for infinity focus. Then the angle of view is given by:

\alpha = 2 \arctan \frac {d} {2 f} where f = F



WHAT????? I failed math you assholes! That's right. Mom was just looking through my old transcripts and confirmed my big, fat juicy F my senior year in Math Honors 2. Ugh.