Question

given a diagnostic radiograph, describe how the exposure technique would be adjusted for individual changes in mAs or exposure time, kVp, grid ratio, beam collimation, SID, film-screen RS, and patient thickness

Answer 1

Diagnostic Radiography produces high-quality images for diagnosis of injury or disease.

1.There is a directly proportional relationship between mAs and quantity of radiation; if the mAs is doubled, the amount of radiation is doubled.
2.Increasing kVp more photons produces with greater energy (more penetrating). Although increasing kVp increases density of photons, the effect changes (is not proportional) throughout the range of kilovoltage used in diagnostic imaging. Low kVp produced a greater variation of x-ray intensities, producing a high or short-scale contrast image. Low kVp produces higher radiographic contrast, and high kVp produces lower radiographic contrast.
3.The grid is the device which is used to absorb scatter radiation in the radiation, leaving the patient before it reaches the image receptor. Using the formula, mAs1/mAs2 = grid conversion factor1/ grid conversion factor2.
4.Increasing collimation less scatter produced, leading to fewer photons reaching the IR (less density) and higher contrast. Although the effect is minimal, but when increased tube filtration results in a beam with higher energy photons, producing a lower-contrast image.
5.When SID is increased, the beam intensity and radiation that is reaching the IR is decreased. To maintaining the original exposure to the IR, the mAs must be increased.
mAs1/mAs2 = (SID1)2/(SID2)2 This formula describing the relationship between the mAs and SID. The mAs/distance compensation formula describes the relationship between SID and mAs needed to compensate for changes in SID. When the SID is increased from 36 inches SID to 72 inches SID, the mAs must be increased to maintain exposure to the IR (but not by a factor of 2). Decreasing the SID by a factor of 2 requires a decrease in mAs by a factor of 4.