If SID is halved, how does the intensity at the image receptor change?

If SID is halved, how does the intensity at the image receptor change?

• A. It increases two times
• B. It remains the same
• C. It decreases
• D. It increases four times

Answer: (D)

The main idea here is the inverse square law: the intensity at the image receptor varies with 1/distance^2. If you halve the source-to-image distance, the distance becomes d/2, and the intensity becomes 1/(d/2)^2 = 4/d^2. That fourfold increase means the image receptor would receive four times as much exposure, assuming the tube output (mA and kVp) stays the same and nothing else changes. In practice, technicians adjust exposure factors to compensate for changes in distance to keep the image properly exposed.

The main idea here is the inverse square law: the intensity at the image receptor varies with 1/distance^2. If you halve the source-to-image distance, the distance becomes d/2, and the intensity becomes 1/(d/2)^2 = 4/d^2. That fourfold increase means the image receptor would receive four times as much exposure, assuming the tube output (mA and kVp) stays the same and nothing else changes. In practice, technicians adjust exposure factors to compensate for changes in distance to keep the image properly exposed.