The first image was taken with a SID of 100 cm and centered perpendicular to the midpoint of the IR. The kVp was 60 and the mAs was 8.
* The next image was taken with an SID of 80 cm and since we were using a new distance we had to change our mAs based on the direct square law. mAs1/mAs2 * distance 12/ distance 22 . Our math obtained got us a new mAs of 5. As you can see the 2 images are very similar density. As I looked up in Papp p.296 it mentioned how an increase in SID causes a decrease in optical density which a little show in this image. Papp however mentions that if the SID is changed by a factor of 2 the optical density changes by 22 of 4 times the density.
*So in next image we had mAs of 3,6 and an SID of 40 instead of 80 we would see a much greater degree of density in beautiful fashion. Also the mAs plays the most significant role in density. As you know mAs depicts the quantity of x-ray photons that are produced. Therefore a greater mAs increases our optical density. A 25 – 30% change in mAs is required to change the optical density because of the human eye only being able to distinguish specific structures.
* The next image we did was with a 100 cm SID and a lateral decentering by moving the centering point. Our original mAs was used here because we weren’t changing our distance. This type of grid problem is called OFF CENTERING. This causes the image to have a decrease in exposure over the entire film (GRID CUT OFF) therefore having less optical density across the entire film which can be seen by the bright whiteness in the image with hardly any density levels.
* The last image we did was a combination of the 80 cm and lateral decentering. The 5 mAs was used here again because we were changing our distance. As you can see this image truly displays the grid cut off. The blurring of the image and terrible contrast levels indicate that the primary beam wasn’t positioned correctly and is the most noticeable off all the grid cut off exposures that we took.