These resolution charts are devised to be printed on a 300 dpi laser printer. The number of lines per unit chart length was chosen based on this 300 dpi. The resolution of what comes out of a laser printer can not be as good as what we are used to in the way of a commercial lens-test target. Therefore, you will find that these targets have rather coarse lines on them and hence must be used at great distances. Test Chart 1.25 has 20, 25, 31.25, 41.66, and 50 lines per chart unit and has a chart unit which is 254 mm (10 inches) long. These spatial frequencies are approximately in the ratio of 1.25:1. Test Chart 1.6 has 20.83, 31.25, 50, 83.3, and 125 lines per chart unit and has a chart unit which is 127 mm (5 inches) long. These spatial frequencies are approximately in the ratio of 1.6:1. A formula for the distance of the chart from the "front" of the lens is given on the chart. If the chart is placed at the formula distance, the lines per unit chart length become lines per millimeter on the film. How is this formula found? Taking Test Chart 1.6 as an example, we must make the 127 mm chart unit become 1 mm on the film so the magnification required is 1 mm/127 mm = 1/127. One definition of magnification is: m = Si / So where: Si = distance of the image (the film) from the rear principal point of the lens. So = distance of object (the chart) from the front principal point of the lens. The principal points are within the lens in the case of normal lenses, tend to be in front of the lens for telephotos (in front of meaning towards the object), and tend to be behind the lens for retrofocus and wide-angle lenses. From high school physics: 1/F = 1/So + 1/Si where F = the focal length of the lens since Si = m*So (from above) in the present case: 1/Si = 127/So combining these two equations: 1/F = 1/So + 127/So = 128/So solving: So = 128*F As you can see from this example, the algorithm for chart distance is to figure the magnification you want, invert it, add 1, and multiply times the focal length of the lens under test. For instance, if you wanted half the spatial frequency of the chart 1.6 example given above, the new distance of the chart from the lens would be (127/2)+1= 64.5 times the focal length of the lens under test. John Bercovitz (JHBercovitz@lbl.gov)