Identifier: americanjournroen08ameruoft (find matches)
Title: The American journal of roentgenology, radium therapy and nuclear medicine
Year: 1906 (1900s)
Authors: American Radium Society American Roentgen Ray Society
Subjects: Radiotherapy X-rays
Publisher: Springfield, Ill. C.C. Thomas
Contributing Library: Gerstein - University of Toronto
Digitizing Sponsor: University of Toronto

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. To find how much of the total radiationof the source will fall on surface ABD ifthe total radiation of our source is O. With S as a center and R as a radius, cir-cumscribe the sphere, EFG. The cone ofrays going to our surface intersects the sur-face of our sphere in the circle HCL. Theplane of circle HCL cuts off a portion of thesurface of our sphere HML. Such a surfaceso cut off is known as a zone of one base. Itsarea is equal to (27rR) X (CM). (The area of a zone is equivalent to thecircumference of a great circle by the alti-tude of the zone.) If R.S., R.Z., and R.X. represent the ra-diation on the sphere, the zone and the sur-face respectively, then R.Z. (27rR) X (CM) . r:s: = 4^R^— ^ (27rR) X (CM) is the area of the zone, and47rR2 is the area of the sphere.R.Z. CM Q ~ R.Z. = 2R 392 Total Radiation from Point Sources CM (R —SC) But —5— is equal to ^^ , which in SCturn is equal to i — -:^ , which in its turn is equal to (i — Cos. a), so that QR.Z. = ^ X (I —Cos. a).
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But R.Z. is equal to R.X., and (i — Cos. a)is equal to Versin a, or OR.X. = ^ X Versin a.2 From the above expressions, we may ex-press the following law: The total radiation on any surface receiv-ing a cone of rays from a point source isequal to half the total radiation at the sourceby the Versin of the cone angle.^ If the surface exposed be a flat surface,then the expression 2A cone angle is that plane angle which wouldgenerate the cone if revolved about one of its sidesas an axis. R.X. = R.X. = Q Q X Versin a, becomesd X (I- ) (d+r^)5 in which d is equal to the distance of thesurface from the source, and r is its radius.If we were to expose a flat surfacewhose radius was 8 inches to a point sourceat a distance of 10 inches, we would findfrom the above expression that our surfacereceived o. 11 per cent of our total radiation.If now we removed this surface to a distanceof 100 inches it would receive .0016 per centof the total radiation, and not .0011 per centwhich it should receive

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