Question

An exposure was made at 40-in. SID using 5 mAs and 105 kVp with an 8:1 grid. In an effort to improve radiographic contrast, the image is repeated using a 12:1 grid and 90 kVp. Which of the following exposure times will be most appropriate, using 400 mA, to maintain the original density? 0.03 s

Answer 1

The most appropriate exposure time for the repeated image, using a 12:1 grid and 90 kVp, to maintain the original density is 0.0125 s.

Step-by-step explanation:

The mAs (milliampere-seconds) is a product of the tube current (mA), exposure time (s), and the number of pulses per second. The formula is given by
`\( \text{mAs} = \frac{I \cdot t}{\text{Pulse Rate}} \)`, where ( I ) is the tube current, (t) is the exposure time, and the pulse rate is the number of pulses per second. In this scenario, the original exposure is made at 40-in. source-to-image distance (SID) using 5 mAs and 105 kVp with an 8:1 grid.

To maintain the original density with the new grid and kVp, the mAs must be adjusted. The mAs is inversely proportional to the grid ratio and proportional to the square of the ratio of the kVp values. The adjustment can be calculated using the formula
`\( \text{mAs}_{\text{new}} = \text{mAs}_{\text{old}} * \left(\frac{\text{grid}_{\text{old}}}{\text{grid}_{\text{new}}}\right) * \left(\frac{\text{kVp}_{\text{new}}}{\text{kVp}_{\text{old}}}\right)^2 \)`. Substituting the given values, we get
`\( \text{mAs}_{\text{new}} = 5 * \left((8)/(12)\right) * \left((90)/(105)\right)^2 \approx 0.0125 \, \text{s} \)`.

Therefore, the most appropriate exposure time for the repeated image is 0.0125 s to maintain the original density while considering the change in grid ratio and kVp. This adjustment ensures that the new image will have similar radiographic contrast and density as the original image.