Thank you, Richard.If a 50 degree photo is about 5 % of a sphere, a 50 degree retinal photo would be about 3.6 % of the retina, since the retinal area is about 72 % of the interior of the globe. Does that sound right?
Tom Quoting "Richard J. Kinch" <kinch@xxxxxxxxxxx>:
There was a question about what proportion of a sphere's surface is seen by a given field of view angle (FOV). The formula for a unit sphere is: 0.5 * (1 - cos( 0.5 * FOV)), for FOV = [ 0, 180deg ] This gives the proportion of the full sphere from an observer in the center. For example, a 50-degree FOV into this formula yields a proportion of a little less than 5 percent, or 1/20 of the whole surface. An observer at a pupil, such as in the optics of a retinal camera, instead of the center, sees a bit more due to the different perspective. And of course the eye is only approximately spherical, but this gives you an idea of the cosine relationship of FOV to area, not unlike ordinary photography. See: http://en.wikipedia.org/wiki/Solid_angle Richard J. Kinch, PhD http://www.truetex.com/range.htm