Question

an x - ray exposure is made using 80 kilovoltage peak (kvp) and 100 milliampere - seconds (mas). at which of the following distances would the image receptor (ir) receive the most exposure?
30 inches (76 cm) source - to - image distance (sid)
40 inches (102 cm) source - to - image distance (sid)
48 inches (122 cm) source - to - image distance (sid)
72 inches (183 cm) source - to - image distance (sid)

Explanation:

Step1: Recall inverse - square law

The intensity of x - ray exposure at the image receptor is inversely proportional to the square of the source - to - image distance (SID). That is, $I=\frac{k}{d^{2}}$, where $I$ is the intensity of exposure, $d$ is the SID, and $k$ is a constant.

Step2: Compare distances

The smaller the SID, the greater the exposure at the IR. Among 30 inches (76 cm), 40 inches (102 cm), 48 inches (122 cm), and 72 inches (183 cm), 30 inches is the smallest distance.

Answer:

30 inches (76 cm) source - to - image distance (SID)