Optical transfer function: Difference between revisions

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The pupil function of an ideal optical system with a circular aperture is a disk of unit radius. The optical transfer function of such a system can thus be calculated geometrically from the intersecting area between two identical disks at a distance of <math>2\nu</math>, where <math>\nu</math> is the spatial frequency normalized to the highest transmitted frequency.<ref name=Williams2002 /> In general the optical transfer function is normalized to a maximum value of one for <math>\nu = 0</math>, so the resulting area should be divided by <math>\pi</math>.

The intersecting area can be calculated as the sum of thatthe areas of two identical [[circular segment]]s: <math> \theta/2 - \sin(\theta)/2</math>, where <math>\theta</math> is the circle segment angle. By substituting <math> |\nu| = \cos(\theta/2) </math>, and using the equalities <math> \sin(\theta)/2 = \sin(\theta /2)\cos(\theta /2) </math> and <math> 1 = \nu^2 + \sin(\arccos(|\nu|))^2 </math>, the equation for the area can be rewritten as <math>\arccos(|\nu|) - |\nu|\sqrt{1 - \nu^2} </math>. Hence the normalized optical transfer function is given by:

<math>\mathit{OTF}(\nu) = \frac{2}{\pi} \left(\arccos(|\nu|)-|\nu|\sqrt{1-\nu^2}\right)</math>.