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| In [[optics]], '''spatial cutoff frequency''' is a precise way to quantify the smallest object [[Optical resolution|resolvable]] by an optical system. Due to [[diffraction]] at the image plane, all optical systems act as [[low pass filter]]s with a finite ability to resolve detail. If it were not for the effects of diffraction, a 2" [[aperture]] [[telescope]] could theoretically be used to read newspapers on a planet circling [[Alpha Centauri]], over four [[light-year]]s distant. Unfortunately, the wave nature of light will never permit this to happen.
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| The spatial cutoff frequency for a perfectly corrected optical system is given by | |
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| :<math>f_o={1 \over {\lambda \times \mathrm{(f/\#)}}}\ \ \mathrm{cycles/millimeter}\ ,</math> | |
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| where <math>\lambda</math> is the [[wavelength]] expressed in millimeters and f/# is the lens' [[focal ratio]]. As an example, a telescope having an [[focal ratio|f/6]] objective and imaging at 0.55 micrometers has a spatial cutoff frequency of 303 cycles/millimeter. High-resolution black-and-white film is capable of resolving details on the film as small as 3 micrometers or smaller, thus its cutoff frequency is about 150 cycles/millimeter. So, the telescope's optical resolution is about twice that of high-resolution film, and a crisp, sharp picture would result (provided [[focus (optics)|focus]] is perfect and [[atmospheric turbulence]] is at a minimum).
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| This formula gives the best-case resolution performance and is valid only for perfect optical systems. The presence of [[Optical aberration|aberration]]s reduces image [[contrast (vision)|contrast]] and can effectively reduce the system spatial cutoff frequency if the image contrast falls below the ability of the imaging device to discern.
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| == See also == | |
| * [[Modulation transfer function]]
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| * [[Superlens]]
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| ==References== | |
| *Goodman, J.A., ''Introduction to Fourier Optics'', McGraw Hill, 1969.
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| [[Category:Optics]]
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Latest revision as of 19:56, 9 January 2015
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