According to the inverse square law, if the source-to-image distance (SID) decreases, the receptor exposure will:

The most easily adjusted radiographic beam-restricting device is the:

If an initial exposure is made using 800 mA, 0.02 sec, 16 mAs, 70 kVp, 58” SID, 4” OID, 5:1 grid ratio, 0.2 mm focal spot and 3.0 mm of filtration, with a 14 x 17 in field size, and the technologist changes the SID to 40” for a second exposure without changing any other factors, what differences would you expect to see in the second exposure?1) Increased IR exposure2) Decreased IR exposure3) Increased contrast4) Decreased contrast5) Increased spatial resolution6) Decreased spatial resolution7) Increased distortion8) Decreased distortion

Which one of these would produce the least magnification?

During a fluoroscopy procedure, the x-ray technologist stands at 5 meters from the x-ray tube for 1 hour and acquires an equivalent dose of 0.1 mSv. If the technologist had stood at a distance of 1 meter, what would their dose have been?

Ten kVp is considered a 15% change in which of these kVp ranges?

A radiograph is acquired with a source-to-skin distance (SSD) of 50 inches. If the radiograph is repeated using the same exposure factors, but the SSD decreases to 35 inches, the patient’s dose will:

The difference between two adjacent densities is

A radiograph is to be made with a 40 inch source-to-image distance (SID) using 12 miliampere-seconds (mAs). If the SID is increased to 72 inches, how could the mAs be changed to maintain the same receptor exposure??

Which kVp selection would produce the greatest relative percentage of scattered radiation?