Hello Art: Don't forget that even though I used a 1/4 hole, most of that hole space is occupied by a 1/8 inch bead. So, in effect, what we have is a small 1/16 th inch donut-shaped ring of light around the bead.
John Carmichael Tucson >John Carmichael wrote: > >> The design which worked the best was a 1/8 inch spherical bead, suspended >by >> thin brass crosswires, in the exact center of a 1/4 inch round hole. (The >> style was about 24 inches from the analemma). > >> A very curious thing happens with this type of style. The bead alone, by >> itself, casts a shadow that was twice as big as the bead; but when the >1/8th >> in. bead is in the center of a 1/4" hole, with a space of 1/16th of an >inch >> between the bead's edge and the hole edge, the bead's shadow miraculously >> sharpens into a tight, dark shadow that is only 1/16th of an inch in >> diameter, smaller than the bead itself!!!! The wires which keep the bead >> suspended in the middle of the hole are so thin that they don't cast a >> visible shadow onto the analemma. > >And Richard M. Koolish calculated: > >> The linear diameter of the diffraction spot (Airy disk) produced by >> a pinhole of a given diameter is: >> >> spot = (2.44 * wavelength * focal_length) / diameter >> >> The optimal size is where spot = diameter, so: >> >> diameter * diameter = (2.44 * wavelength * focal_length) >> >> diameter = sqrt (2.44 * wavelength * focal_length) >> >> An example of a pinhole for a distance of 100 mm and a wavelength of >> 550 nm is: >> >> diameter = sqrt (2.44 * .000550 * 100) = sqrt (.01342) = .366 mm > >Using a distance of 24 inches = 610 mm, this becomes 0.9 mm = 1/32 inch, >still several times smaller than John's hole. I think the explanation lies >in simple geometrical optics. Imagine putting your eye where the shadow is >being cast and looking back toward the style and the sun. I would like to >suppose that the distance to the style was something closer to 14" (subject >to objection and correction from John), so that the image of the sun would >be just eclipsed by the 1/4 inch bead, giving a black shadow at the center. >Just a little off-center, an arc of the sun would show around the bead, so >the brightness would grow, but only until the disk of the sun runs into the >edge of the hole. Thereafter the brightness would decrease slowly until the >sun is entirely outside the hole. This would lead to a shadow with a >diameter-at-half-brightness of about 1/16 inch, within a diffuse bright >field with diameter on the order of 1/4 inch. The size of the shadow is >reduced at the cost of reducing the contrast with the surrounding lighted >area. The principle is much the same as a sundial that images the sun >through a pinhole: a sharper image is a dimmer image. > >--Art Carlson > >
