T.O. 33B-1-1Where:T1 = Original Exposure (MAS).T2 = New Exposure (MAS).D1 = Original Distance (SFD).D2 = New Distance (SFD).For example, a technique calls for exposing a part at 36 inches using 300 mas. However, the tube head needsto be moved to make a 48 inch SFD, what would the new exposure be?Substituting:Cross multiplying gives (T2) * (1296) = (300) * (2304) or T2 = (300 * 2304)/1296.Solving, T2 = 533 MAS, which would be our new exposure.6.7.2.9InverseSquareLaw.When the X-ray tube output is held constant or when a particular radioactive source is used, the radiationintensity reaching the specimen is governed by the distance between the tube (or source) and the specimen,varying inversely with the square of this distance. The explanation below is in terms of X-rays and visiblelight but applies with equal force to gamma rays as well. Since X-rays conform to the laws of light, theydiverge when they are emitted from the anode and cover an increasing larger area with lessened intensity asthey travel from their source. This principle is illustrated by Figure 6-26.Figure 6-26.Inverse Square Law Diagram6.7.2.9.1In this example it is assumed that the intensity of the X-rays emitted at the anode (A) remains constant, andthat the X-rays passing through the aperture (B) cover an area 4 square inches on reaching and recordingsurface (C1), which is 12 inches (D) from (A). Then when the recording surface (C1) is moved 12 inches fartherfrom the anode to (C2), so that the distance between (A) and (C2) is 24 inches (2D) or twice the distancebetween (A) and (C1); the X-rays will cover l6 square inches, an area of 4 times as great as that at (C1). ItChange 36-57
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