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« on: January 26, 2013, 01:09:31 AM »
If the object is to compare sensors of equal number of pixels for sharpness and nothing else, why not provide a colimated light soure to adjacent pixels in a checkerboard pattern such that neighbors get alternately 0 photons and photons beyond saturation. Take the digital data and blow it up any number of times, 5, 10, 1000, or a billion. Sensor size is immaterial if you remove all the optical variables. The digital data will be identical regardless of sensor size. Prints made from either sensor will be identical.
Pick a limited number of variables and you could have the alleged photocopier.
Pick a realistic number of variables and we are beginning to talk photography. Pick your poison and make your pitch.
Generally, system resolution is calculated by taking the reciprocal of the sum of the reciprocals of the elements of the system. For simplicity, if you have film or a sensor that will resolve 100 line pairs per millimitre and a lens that will resolve 100 line pairs per millimitre, the best that can be hoped for is (1/100+1/100)^-1. That is 50 line pairs per millimitre. In reality you need to add contrast factors, paper resolution, printer resolution. Back in the day I added enlarger lens resolution to the calculations.
Take a lens that has infinite resolution and a sensor that has 100 lp/mm and you get 100 out of the system. Take a sensor with 100 billion pixels and feed it an image with 100 lp/mm lens and you get a resolution approaching 100 lp/mm out. A larger sensor has more millimitres than the small one hence a higher total number of line pairs.
Numerous factors including optical diffraction limits, lens abberations and sensor noise give larger sensor of comparable technology with identical pixel count and proportionately larger pixels an advantage in resolving detail. Mechanical accuracy of mount alignment and distance to sensor proportionately favours the larger sensor as well.
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