Principles of Exposure and Image Quality



Principles of Exposure and Image Quality




This chapter explains the prime factors of radiographic exposure and their radiographic effects. You have already been introduced to some of them. In addition, it introduces the four primary factors of radiographic quality and the principal methods for controlling them. You will begin to observe the effects of exposure on radiographs and to understand how the various factors controlled by the limited operator affect the final image.




Prime Factors of Radiographic Exposure


Exposure is a broad term used to describe the x-rays that the patient is exposed to, the amount of x-rays in the primary beam, and also the amount of x-rays that reach the image receptor (IR). The x-ray beam is often described in terms of its quantity and its quality. The prime factors that affect x-ray quantity are milliamperage-seconds (mAs), kilovoltage (kVp), source–image receptor distance (SID), and filtration. The factors that affect x-ray quality are kilovoltage and filtration. Note that kilovoltage and filtration affect both quantity and quality (Box 7-1). Filtration was discussed in Chapter 5.



The quantity and quality of the x-ray beam are controlled by four prime factors. These factors are under the direct control of the limited operator. The prime factors of exposure are milliamperage (mA), exposure time (S), kVp, and SID.



Milliamperage


As explained in Chapter 5, changes in mA affect the rate of exposure, that is, the number of photons produced per second during an exposure. For this reason, a change in mA will alter the quantity of exposure to the IR. An increase in mA will increase the quantity of exposure; decreased mA will reduce the quantity of exposure. Exposure is directly proportional to mA; that is, if the mA doubles, the quantity of exposure also doubles. Technically, when the mA is doubled, the number of electrons at the filament doubles. During the exposure, the number of photons emitted from the tube doubles as well. The opposite is true if the mA decreases by 50%. The electrons at the filament and the photons emitted will be halved. Dose to the patient is also directly proportional. For example, if the mA is doubled, the dose to the patient is doubled.




Milliampere-Seconds


As stated in Chapter 5, the unit used to indicate the total quantity of x-rays in an exposure is milliampere-seconds, abbreviated mAs. This unit is the product of mA and exposure time (mA × time = mAs). For example, if the control panel were set at 200 mA and 0.2 second, the mAs would equal 200 × 0.2:


200mA×0.2sec=40mAs


image

A desired quantity of exposure may be obtained by any combination of mA and time that, multiplied together, equals the desired mAs. For example, 40 mAs could be obtained using any of the following combinations:



The quantity of exposure and the patient dose are directly proportional to the mAs. For each of the four exposure techniques listed earlier, the volume of photons emitted and the dose to the patient will be equal. The density or blackening effect on the image will also be equal. The unit mAs is the primary controller of radiographic density.


It is important to understand that the mA, exposure time, or the equivalent, mAs, all follow the same directly proportional rule in terms of exposure and dose. If any of these three factors are doubled, both the exposure and the dose are doubled. If any of the three are cut in half, the exposure and dose are cut in half.



Kilovoltage


The kVp controls both the quality and the quantity of the x-ray beam. As the kVp is increased, the energy of the photons in the beam is increased, changing its quality. As the photon energy increases, the penetrating ability of the photons increases. When larger or denser body parts are x-rayed, the kVp is increased so that the photons can get through the part and reach the IR. If the kVp is set too low, the photons may not all get through the body part to form the image. Setting the correct kVp for each body part is very important so that the correct number of photons reach the IR. The kVp is the primary controller of the penetration of x-rays.


The kVp also has an effect on the quantity of exposure to the IR. When the kVp is increased, the electrons from the filament reach the anode with more energy. More interactions then occur in the anode and more x-rays are emitted. When kVp is increased, density is increased; however, mAs is the primary controller of density. Unlike the effects of mA, exposure time, or mAs, changes in exposure are not directly proportional to kVp. The kVp is never doubled. A doubling of the kVp would result in four times more photons being emitted! Conversely, the kVp would never be halved because four times fewer photons would result. These would be extreme changes in exposure. Although kVp will affect density, kVp should not be used to control radiographic density.


The contrast of the image is directly affected by kVp. High kVp produces a low-contrast image and low kVp produces a high-contrast image. Each body part that is radiographed will have a kVp assigned to it via the exposure technique chart. This kVp is predetermined based on the penetration needed for the part and the contrast required. Often the physician may request that an additional radiograph be taken at a different contrast level in order to see the anatomy differently. Therefore kVp is the primary controller of radiographic contrast.



Source–Image Receptor Distance


The distance between the tube target and the IR is called the source–image receptor distance, abbreviated SID. Because the x-ray beam diverges, forming the shape of a cone, the photons get farther apart as they get farther from the target (Fig. 7-1). Thus the SID affects the intensity of the x-ray beam and the quantity of x-rays.



The relationship between the SID and the intensity of the beam is expressed in the inverse square law, which states that the intensity is inversely proportional to the square of the distance. The inverse square law is expressed mathematically as a formula:


I1I2=D22D12


image

In this formula, I represents radiation intensity and D represents SID. For all SID calculations, the distance is always squared. As the distance increases, the intensity decreases and vice versa. For example, if the distance were doubled, the intensity would decrease to one fourth of the original intensity. If the distance were reduced 50%, the intensity would increase by four times. In Fig. 7-1, suppose D1 is 40 inches and D2 is 80 inches, or twice as great. If the original intensity at D1

Mar 7, 2016 | Posted by in GENERAL RADIOLOGY | Comments Off on Principles of Exposure and Image Quality
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