If SID is reduced by one half, what must be done to mAs to maintain a constant receptor exposure?

If SID is reduced by one half, what must be done to mAs to maintain a constant receptor exposure?

• A. Reduce mAs to one fourth its original value
• B. Reduce mAs to one half its original value
• C. Increase mAs by four times its original value
• D. Increase mAs by two times its original value

Answer: (A)

The concept being tested is how receptor exposure changes with distance according to the inverse square law. Receptor exposure depends on mAs and on the distance from the source; when SID changes, the beam intensity at the receptor changes by the inverse square of the distance. If you halve the SID, the exposure at the receptor would increase by a factor of four with the same mAs. To keep the exposure constant, you must reduce the mAs by the same factor that distance increased exposure, which is a quarter. So the new mAs should be one quarter of the original value. The other options would not correctly offset the fourfold increase in exposure.

The concept being tested is how receptor exposure changes with distance according to the inverse square law. Receptor exposure depends on mAs and on the distance from the source; when SID changes, the beam intensity at the receptor changes by the inverse square of the distance. If you halve the SID, the exposure at the receptor would increase by a factor of four with the same mAs. To keep the exposure constant, you must reduce the mAs by the same factor that distance increased exposure, which is a quarter. So the new mAs should be one quarter of the original value. The other options would not correctly offset the fourfold increase in exposure.