Wolfram|Alpha

Computing...

Input information:

hyperfocal distance |  \nf-number | 8\ncircle of confusion | 0.025 mm  (millimeters)\nfocal length | 50 mm  (millimeters)


Result:

hyperfocal distance | 12.5 meters\n= 41.01 feet\n= 41\' 0.126"


Equation:

h = f^2\/(N c) |  \nh | hyperfocal distance\nN | f-number\nc | circle of confusion\nf | focal length\n(nearest focused distance such that the point at infinity still appears sharp)