The eye has a visual acuity threshold below which
an object will go undetected. This threshold varies from person
to person, but as an example, the case of a person with normal
20/20 vision can be considered. As light enters the eye through
the pupil, it passes through the lens and is projected on the
retina at the back of the eye. Muscles called extraocular muscles,
move the eyeball in the orbits and allow the image to be focused
on the central retinal or fovea.
The retina is a mosaic of two basic types of photoreceptors:
rods, and cones. Rods are sensitive to blue-green light with peak
sensitivity at a wavelength of 498 nm, and are used for vision
under dark or dim conditions. There are three types of cones that
give us our basic color vision: L-cones (red) with
a peak sensitivity of 564 nm, M-cones (green) with a peak sensitivity
of 533 nm, and S-cones (blue) with a peak sensitivity of 437 nm.
Cones are highly concentrated in a region near the
center of the retina called the fovea region. The maximum concentration
of cones is roughly 180,000 per square mm in the fovea region
and this density decreases rapidly outside of the fovea to a value
of less than 5,000 per square mm. Note the blind spot caused by
the optic nerve which is void of any photoreceptors.
The standard definition of normal visual acuity
(20/20 vision) is the ability to resolve a spatial pattern separated
by a visual angle of one minute of arc. Since one degree contains
sixty minutes, a visual angle of one minute of arc is 1/60 of
a degree. The spatial resolution limit is derived from the fact
that one degree of a scene is projected across 288µm of the retina by the eye's lens.
In this 288µm, there are 120 color sensing
cone cells packed. Thus, if more than 120 alternating white and
black lines are crowded side-by-side in a single degree of viewing
space, they will appear as a single gray mass to the human eye.
With a little trigonometry, it is possible to calculate the resolution
of the eye at a specific distance away from the lens of the eye.
For the case of normal visual acuity the angle Theta
is 1/60 of a degree. By bisecting this angle we have a right triangle
with angle Theta/2 that is 1/120 of a degree. Using this right
triangle it is easy to calculate the distance X/2 for a given
X/2 = d (tan Theta/2)
When visually inspecting an object for a defect such as a crack,
the distance (d) might be around 12 inches. This would be a comfortable
viewing distance. At 12 inches, the normal visual acuity of the
human eye is 0.00349 inch. What this means is that if you had
alternating black and white lines that were all 0.00349 inch wide,
it would appear to most people as a mass of solid gray.