An acceptable chest radiograph is obtained at source distance 0.75 m using 50 kVp and 50 mAs. The subject moves to 1 m while kVp stays unchanged. Find the new mAs needed for the same exposure, to two decimals.
At unchanged beam quality, exposure scales with mAs divided by distance squared. Set mAsnew/1²=50/.75². Thus mAsnew=50×(1/.75)²=88.8889 mAs, rounded to 88.89 mAs. Increasing distance requires greater tube-current–time product in this ideal inverse-square model.
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