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level for daylight unaided exposure with the Sun at the zenith. Thusthe threshold exposure duration for photochemical retinal damagefor observing Znamya is only slightly longer than that required forthreshold damage from unaided sungazing.While the retinal image of Znamya is extremely small, it has avery high associated retinal irradiance. It could be a very seriousretinal hazard when viewed through binoculars or telescopes.On every clear day, near the beginning of sunrise and the endof sunset, most of the Sun’s disk is obscured by the Earth’s limb, andhumans would therefore face an equivalent ocular hazard, exceptthat the large intervening air mass which is present in such situationsprotects us. An astronaut on an airless moon or planet would facesuch a hazard at these times, or even if most of the Sun’s disk wereobscured by an opaque object such as a large rock, unless his spacesuitprovided visual protection greater than that from air mass 1. Equivalently,if an astronaut were to travel substantially farther than 1 AU fromthe Sun, so that the resulting decrease in solar irradiance caused hiseyes’ entrance pupils to expand significantly, he would be subject toeye damage unless he were farther than roughly 40 AU from the Sun,so that it was no longer fully resolved on his retinas.We are grateful to P.A. Delaney, M.M. DeRobertis, M.L. McCall, andthe referees, for valuable discussions and comments. This work wassupported by the Natural Sciences and Engineering Research Councilof Canada under grant A-4638, and by the Canadian OptometricEducation Trust Fund.James G. LaframboisePhysics and Astronomy DepartmentYork University4700 Keele StreetToronto, Ontario, M3J 1P3Therefore, as viewed by an observer at the centre of the illuminatedspot on the ground, every part of the mirror would be seen to bereflecting some portion of the Sun’s surface: so for this observer, themirror would be seen as fully illuminated.For an observer located at a distance r from the centre of theFig.1 — All 3 km disks inside a 6 km circle will overlap at its centre.illuminated spot, we calculate the fractional irradiance as follows.We note that the centres of the 3 km disks all lie within a radius of 1.5km from the centre of the 6 km illuminated spot (Figure 2). We assumethat within this 1.5 km radius, these centres are distributed randomly(uniformly), implying that the tilts of all the small mirrors in theB. Ralph ChouSchool of OptometryUniversity of WaterlooWaterloo, Ontario, N2L 3G1AppendixVariation of Irradiance Acrossthe Illuminated SpotAs noted in Section 1, the announced width of the spot of light onthe ground was to have been 6 to 8 km in the second Znamya attempt.We again assume 6 km, because this gives us the “worst case.” Also, 8km may correspond to nonvertical incidence of the beam. Again, 6km corresponds to an angular width of the spot on the ground of aboutone degree, as seen from the spacecraft. In comparison, the anglesubtended by the Sun’s disk, at the Earth, is about 1/2 degree, andtherefore, a perfectly flat mirror would produce a spot of light about3 km wide on the ground (1/2 degree ×π/180 radians per degree ×360 km = 3.14 km). Also, tilting such a mirror by half a degree woulddisplace such a spot about 3 km. Since 3 km + 3 km = 6 km = theassumed width of the spot on the ground, the Znamya mirror, asannounced, can be considered as a mosaic of smaller flat mirrorswhose maximum tilt relative to each other is 1/2 degree. We note thatall 3 km disks inside a 6 km circle will overlap at its centre (Figure 1).Fig. 2 — An observer at radius r sees an amount of illumination proportionalto the area of overlap between the two 3 km disks shown. The disk whoseboundary is shown as dashed contains the central points of all the overlappingilluminated spots. The other 3 km disk contains all such points located within1.5 km of the observer.December/ décembre 2000 JRASC239