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You are basically talking about modeling the image chain to determine the final resolution for your optical/film system. You are right that the final resolution is a function of all of these things. In fact, while lp/mm is a common metric, it is not the best metric for performing this kind of modelling. There may be a heuristic approach to using lp/mm, but I am not familiar with it. It will probably turn out that when modeling it this way, the ultimate resolution is a result of the weakest links in the chain after each is propagated through the enlargement process. For example, the 90 lp/ mm lens applied to a 160 lp/mm film, would result in a net of slightly less than 90 lp/mm, since the 160 is much higher resolution than the 90. I'm taking an educated guess here, but you might try taking the root sum square of the reciprocols of each lp/mm value in the image chain multiplied by the enlargement factor from that value to the end of the image chain. x = 1 / sqrt( 1/x1^2 + 1/x2^2 + ... + 1/xn^2) From the example above, you would get x = 1 / sqrt( 1/160^2 + 1/90^2) = 78 lp/mm. From an engineering perspective, each optical device has a "point spread function". This is the amount of "spread" that occurs if an "infinitely small" point source is propagated through it. The spread is usually a gaussian shape whose width is defined by a sigma parameter. That gaussian is then spread by the next gaussian and the next, etc. You would also have to apply the enlargement factors at each stage as a multiplier. The calculations can be done in either the spatial domain or in the frequency domain (using Fourier analysis). There is a relationship between the gaussian and the lp/mm metrics, that I believe has to do with the distance between the centers of two gaussians when the dip in the middle is 3db below the peaks. The final visual acuity metric is turned into an arc measurement by taking the following formula: alpha = 2 * arctan ( r / 2f ) Where alpha = the visual acuity angle r = the resolution in mm/lp or mm between resolvable points (so it would be the reciprocol of your lp/mm value) f = the focal length in mm Good luck with your calculation. Todd Jamison Chief Scientist Observera, Inc.

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