Final answer:
To maintain optimal image receptor exposure when doubling the distance from 40" to 80", the mAs must be quadrupled from 20 mAs to 80 mAs, following the Inverse Square Law of radiography.
Step-by-step explanation:
The subject of this question is related to the principles of x-ray imaging and radiation physics, specifically the concept known as the Inverse Square Law in radiography. According to this law, the intensity of radiation exposure to the image receptor is inversely proportional to the square of the distance from the x-ray source. Therefore, when the distance is doubled (from 40" to 80"), the area of exposure is quadrupled (since 2 squared is 4). To maintain the same image receptor exposure at double the distance, we must also quadruple the mAs. Hence, if 20 mAs is optimal at 40", then at 80" (which is twice the distance) we need 4 times the original mAs, which is 20 mAs x 4 = 80 mAs.
The correct answer to the student's question, "If 20 mAs provide an optimum image receptor exposure at 40", how much mAs is necessary at 80"?" is C. 80 mAs.