The parameter called resolution is always declared by the manufacturer in the scanner’s technical specification. The real optical resolution is, however, determined not only by the number of the elements of the light-sensitive sensor, but also by the quality of the optical system of the scanner. Good resolution means the scanner’s ability to catch the minutest details in the original. But a scanner with a higher optical resolution doesn’t necessarily produce scans of a higher quality.
We can check this parameter out in practice by digitizing special test targets. Such a target is a set of patterns of white and black lines (“dark-line/white-space” combination). The spatial frequency characteristic of the pattern is determined by the number of pairs of such lines per length unit – line pairs per inch (lppi) in our case. The test target is shown in the next picture:
There are five patterns at the top of the target. They differ in their line density: 30, 75, 95, 140 and 180 lppi. I will scan them at the maximum optical resolution of the scanner.
The lines of the pattern are square-wave signals
The high-contrast transitions get blurred in the scan
Besides square-wave patterns there can also be sine-wave patterns in which the lines are originally “blurred”.
The image’s contrast degenerates as the line density increases. We can describe this with the modulation (M) parameter:
Here, Smax and Smin are the max and min levels of the pixels of the image (0 ≤ M ≤ 1).
Having measured the modulation for five areas of the target we can build a diagram that would show the dependence of the image contrast on the line density of the pattern. The results of the Scanjet 8200 are compared to the Epson Perfection 4870:
As you see, the graphs are almost identical, but the Scanjet 8200 is a little worse than its competitor. The similarity of the graphs suggests that the effective resolution of the two scanners is almost the same.