RTBC Advanced Exposure Factors Practice Test

To determine the new intensity for an exposure at a distance of 40 inches when the intensity at 50 inches is 200 μGy, the Inverse Square Law is applied. This law states that the intensity of radiation is inversely proportional to the square of the distance from the source. The relationship can be expressed mathematically as:

\[ I_1/I_2 = (D_2^2)/(D_1^2) \]

In this scenario, let \( I_1 \) be the intensity at 40 inches, and \( I_2 \) be the intensity at 50 inches, which is given as 200 μGy. The distances \( D_1 \) and \( D_2 \) are 40 inches and 50 inches, respectively. Plugging the values into the formula gives:

\[ I_1/200 = (50^2)/(40^2) \]

Calculating the squares of the distances:

\[ 50^2 = 2500 \]

\[ 40^2 = 1600 \]

Thus, the equation becomes:

\[ I_1/200 = 2500/1600 \]

This simplifies to:

\[ I_1/200 = 2