If a beam intensity of 1200 photons/unit area is achieved at a source to image receptor distance of 80", what would the beam intensity be at a 40" distance using the same factors?

To determine the beam intensity at a different source to image receptor distance while using the same exposure factors, the inverse square law must be applied. This law states that the intensity of radiation is inversely proportional to the square of the distance from the source.

If the original beam intensity is 1200 photons/unit area at an 80" distance, we want to find the intensity at a 40" distance. According to the inverse square law, doubling the distance will reduce the intensity by a factor of four, meaning that halving the distance will increase the intensity by a factor of four.

Starting with the original intensity of 1200 photons/unit area at 80", if the distance is halved to 40", the calculation for the new intensity is as follows:

1. The distance decreases from 80" to 40", which is a factor of 2.
2. Since the intensity increases by the square of that distance reduction (2^2 = 4), the new intensity can be found by multiplying the original intensity by 4.
3. Therefore, 1200 photons/unit area * 4 = 4800 photons/unit area.

This calculation rationalizes the choice of 4800 photons/unit area as the correct answer