## Resolution, visual acuity and lp/mm

In order to understand some of the concepts, it might be necessary to refer to the Wikipedia articles that correspond to them. Just click on the links and you will be able to follow my line of thought.

According to the Snellen fraction at 20/20, a ‘normal’ eye subtends 5 arc minutes of an Optotype. The most commonly used optotype (probably because it has 5 lines) is the letter ‘E’. Each line therefore subtends 1 arc minute. For a ‘perfect human’, the angle would be 0.4 arc minutes, regardless of distance. If the object gets closer, the line width gets smaller, but the angle never changes. Even under magnification (microscopy or telescopy), the perfect human eye can only read at 0.4 arc minutes. Practically not very useful (which is why they use E), but makes calculations easier. It would be unfair to use line pairs on sensors.

The sensor does not have the field of vision like an eye. The lens compresses the image it to a single plane – the back of the lens. It is as if the sensor is almost watching a movie in a dark hall. Sure, there’s light scattering, and other ‘leaks’, but for our purposes, we can assume our camera lens to be perfect, and treat the back of the lens like an image plane. Therefore, it is not important to consider what is in focus and what isn’t, etc.

I assumed the distance to be the focal flange distance (back of the lens to sensor). EF (44mm) and PL (55mm). I’m from the motion picture background. Some cameras like the RB67 (which I own) has a flange focal distance of 112mm. Obviously this figure varies as the lens zooms, focuses, etc. For simplicity sake, I have assumed it to be constant for now. For a list of flange distances for various mounts, click here.

The smallest readable line that leaves the lens must subtend an angle of 0.4 arc minutes to a point behind the pixel – because the pixel is not a point object but a line itself. However, a pixel never has to resolve anything greater than 0.4 arc minute, so we can safely assume that a pixel must be smaller than the width of the line that is subtended on the back of the lens. Treating the sensor like the eye, and because the angle is too small which makes the pixel size roughly the same as the lens line width, and going by the fact that a pixel can always have one and only one value, it can be treated like a point object if only for theoretical calculation. In this case, if the angle is 0.4 arc minute and the distance is 44mm (for EF), say, then the width of the line is 5.112 microns.

For the EF mount:
@ 0.4 arc min, minimum line/pixel/dot width is 5 microns
@ 1 arc min, minimum line/pixel/dot width is 12.76 microns

For the RB67:
@ 0.4 arc min, minimum line/pixel/dot width is 13 microns
@ 1 arc min, minimum line/pixel/dot width is 32.5 microns

For the Pentax Q mount (the smallest I could find) (9.2mm flange)
@ 0.4 arc min, minimum line/pixel/dot width is 1 micron
@ 1 arc min, minimum line/pixel/dot width is 2.7 microns

Assuming a viewing distance of:

@ 2 inches (50mm) (minimum focus distance for a good child’s eye): @ 0.4 arc min, the ppi is 4370 ppi and @ 1 arc min the ppi is 1160 ppi.

@ 4 inches (100mm) (minimum focus distance for a good adult’s eye): @ 0.4 arc min, the ppi is 2185 ppi and @ 1 arc min the ppi is 2320 ppi.

@ 12 inches (305mm): the line width @ 0.4 arc min must be 35.5 microns or 0.035mm. This corresponds to 726 ppi at 0.4 arc min and about 290 ppi (very close to 300 dpi isn’t it?) for 1 arc min.

@ 2.5 feet (762mm) (normal computer monitor viewing distance): @ 0.4 arc min, the ppi is 287 ppi and @ 1 arc min the ppi is 115 ppi. Right now, the average screen is only 100 ppi. Professional monitors aim for higher ppi to justify this resolution.

@ 1000mm the width of one line is 0.12mm or 116.3 microns. This corresponds to 212 ppi @ 0.4 arc min, and 88 ppi @ 1 arc min.

@ 6 feet or 1830mm – the ‘normal’ viewing distance for home television. The ppi required is 120 @ 0.4 arc min. @ 1 arc min the ppi required is only 48 ppi. My LCD screen (42″) has a horizontal ppi of 64. This is why 2 MP is okay for home viewing, even on a large monitor. Smaller TV sets need lesser ppi, and larger sets will be viewed farther away. The 72 dpi standard for TV is okay. The same standard is used for film projection (where half the resolution is lost in projection).

Line pairs per mm for EF mount:

Since the minimum required line width is 5 microns for an EF mount lens, a line pair would be 10 microns. Therefore, the lp/mm required would be 100 lp/mm corresponding to 0.4 arc min. @ 1 arc min, it is 40 lp/mm.

The highest possible Canon EF lens lp/mm (according to DXOMark) is the 85mm 1.8L which has a lp/mm of 67. Of course, it all depends on the testing and cameras used, etc. But if a lens is above 40 lp/mm, it is good enough. But to match human acuity, it must top 100 lp/mm.

Printing and Scanning:

I’m not an expert on printing, but the maximum 720 ppi limit seems justified, if all things are perfect @ 1 feet. However, the Epson 9900 is capable of 2880 ppi. Paper resolutions are easily greater than 60 lp/mm or around 720 ppi+ to 1440 dpi for Epson Fine Art papers. Of course, it is important to avoid interpolation when making these calculations.

A drum scanner is capable of 3000 dpi to 24,000 dpi, and a flatbed scanner can top 5000 dpi.

Very important: My calculations make many assumptions and are amateurish. I did them just to understand how the standards that are being used have come about. I probably have made mistakes (hopefully not grave ones). But, I’m also secretly glad they correspond to the standards currently used in photography and cinema today.