Question

question 10 / 10
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 the inverse - square law

The intensity of radiation received by the image receptor (IR) is inversely proportional to the square of the source - to - image distance (SID). Mathematically, $I\propto\frac{1}{d^{2}}$, where $I$ is the intensity of radiation and $d$ is the SID.

Step2: Analyze the relationship

As the SID decreases, the intensity of radiation received by the IR increases.

Step3: Compare the given distances

We have SIDs of 30 inches (76 cm), 40 inches (102 cm), 48 inches (122 cm), and 72 inches (183 cm). The smallest SID among them is 30 inches (76 cm).

Answer:

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