Question

5 multiple choice 1 point if distance is adjusted, what prime factor needs to be adjusted? mas kvp filtration distance 6 fill in the blank 2 points if distance is doubled, how much is the beam intensity affected? type your answer... would image density increase or decrease? type your answer...

Explanation:

Step1: Recall the inverse - square law for radiation

The intensity of a radiation beam is inversely proportional to the square of the distance ($I=\frac{k}{d^{2}}$, where $I$ is intensity, $d$ is distance and $k$ is a constant).

Step2: Analyze question 5

When distance is adjusted in radiation - related procedures (such as in X - ray imaging), mAs (milliampere - seconds) is the prime factor that needs to be adjusted to maintain image quality. It controls the quantity of X - rays produced.

Step3: Analyze question 6

If the distance $d$ is doubled ($d' = 2d$), then the new intensity $I'=\frac{k}{(2d)^{2}}=\frac{k}{4d^{2}}=\frac{1}{4}I$. So the beam intensity is reduced to one - fourth of its original value. And since image density is related to the intensity of the radiation reaching the image receptor, a decrease in beam intensity will cause a decrease in image density.

Answer:

Question 5: mAs
Question 6:
The beam intensity is reduced to one - fourth of its original value.
Decrease