A mobile radiographer images an isolation patient at 40-inch SID with a fixed technique, then repeats the projection at 80-inch SID to reduce magnification and maintain detector coverage. Scatter and filtration remain constant and collimation is unchanged. Which inverse square law prediction best describes the change in primary beam quantity at the receptor?
Select an answer to reveal the explanation.
Short Explanation
You've got this doubling-distance math down once you see the pattern. Doubling SID cuts intensity to a quarter, so you'll need a square-law bump in mAs to match density.
Full Explanation
Primary beam quantity leaving the tube follows the inverse square law because photons diverge from a point source, so intensity is proportional to 1/d2 and doubling distance from 40 to 80 inches reduces intensity to one-fourth if output and filtration are constant. Beam quality, determined mainly by kVp and filtration, changes negligibly over this air distance, so the hardening explanation confuses quantity with energy distribution and misses the stated constant filtration. The linear halving explanation applies proportional reasoning instead of square-law divergence and therefore underestimates the loss and would underexpose a compensated image. The remnant radiation explanation reverses physics because remnant photons exiting the patient decrease, not increase, when fewer primary photons arrive, and tissue attenuation does not fall with SID. Exam caveat: ARRT expects you to calculate 1/4, 4x, or 1x for double, half, or unchanged distance, and practical check: quadruplicate mAs or use the exposure maintenance formula when doubling SID.