If a radiograph is produced with 4 mAs and 70 kVp giving an exposure of 0.001 mGy, what will be the exposure with 8 mAs at the same kVp?

To determine the exposure with 8 mAs at the same kVp, it is important to understand the relationship between milliamperage-seconds (mAs) and exposure. Doubling the mAs will approximately double the exposure, given that kVp remains constant.

In this scenario, the initial exposure produced with 4 mAs is 0.001 mGy. If the mAs is increased to 8, which is double the original 4 mAs, we can expect the exposure to also double.

Calculating the expected exposure with 8 mAs:

0.001 mGy (initial exposure) multiplied by 2 (since 8 mAs is double 4 mAs) results in an exposure of 0.002 mGy.

This reasoning directly supports the choice indicating that the exposure will be 0.002 mGy when the mAs is increased to 8 while keeping the kVp constant.