At which source-to-image distance (SID) would the image receptor receive the most exposure when using 80 kVp and 100 mAs?

The source-to-image distance (SID) significantly influences the exposure received by the image receptor in radiographic procedures. At shorter SIDs, the intensity of radiation that reaches the image receptor is higher due to the inverse square law. This law states that as the distance from the source of radiation increases, the intensity of the radiation decreases proportionally to the square of the distance. Therefore, when the SID is lower, the image receptor is exposed to more radiation, resulting in a higher exposure.

In this scenario with 80 kVp and 100 mAs, the shortest distance of 30 inches (76 cm) allows for the maximum intensity of radiation to reach the receptor compared to longer distances. As the distance increases to 40 inches, 48 inches, or 7 inches, the amount of radiation reaching the image receptor decreases, leading to a lower exposure. While shorter distances generally yield higher exposure levels, an extremely short distance like 7 inches (183 cm) could introduce additional factors that complicate image quality, such as increased magnification and potential for distortion; hence the selection of 30 inches (76 cm) is ideal for maximizing exposure under the given conditions.

By understanding how distance affects radiation intensity, practitioners can better manipulate the SID to optimize