Decreasing the field size from 14x17 inches to 8x10 inches will have which effect?

Decreasing the field size from 14x17 inches to 8x10 inches will have which effect?

• A. SID increases, exposure rate decreases and radiographic density decreases
• B. SID increases, exposure rate increases and radiographic density decreases
• C. SID remains the same, exposure rate remains the same, and radiographic density decreases
• D. SID decreases, exposure rate increases and radiographic density decreases

Answer: (A)

Reducing the beam’s field size by collimation changes how much of the patient is irradiated, not the distance from the tube to the image receptor. So the distance (SID) stays the same. The exposure rate per unit area of the beam reaching the patient is determined by tube output and distance, and with the same SID and technique factors, this rate doesn’t change. However, because the field is smaller, the total number of photons reaching the receptor decreases, leading to less overall exposure on the image and thus lower radiographic density. In short: SID remains unchanged, exposure rate per unit area stays the same, but radiographic density decreases due to the smaller irradiated area. If density needs to be maintained, mA–s would need to be increased.

Reducing the beam’s field size by collimation changes how much of the patient is irradiated, not the distance from the tube to the image receptor. So the distance (SID) stays the same. The exposure rate per unit area of the beam reaching the patient is determined by tube output and distance, and with the same SID and technique factors, this rate doesn’t change. However, because the field is smaller, the total number of photons reaching the receptor decreases, leading to less overall exposure on the image and thus lower radiographic density. In short: SID remains unchanged, exposure rate per unit area stays the same, but radiographic density decreases due to the smaller irradiated area. If density needs to be maintained, mA–s would need to be increased.