If a radiographic unit has an output intensity of 42 uGya/mAs at an 80cm distance, what would be the intensity when the distance is reduced to 60 cm?

To determine the new intensity at a reduced distance, 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 of radiation.

Given the initial intensity at 80 cm is 42 µGya/mAs, we can use the formula from the inverse square law:

( I_1/I_2 = (D_2^2)/(D_1^2) )

Where:

• ( I_1 ) is the initial intensity (42 µGya/mAs)

• ( I_2 ) is the new intensity we want to find

• ( D_1 ) is the original distance (80 cm)

• ( D_2 ) is the new distance (60 cm)

By inserting the values into this formula:

• ( D_1^2 = 80^2 = 6400 )

• ( D_2^2 = 60^2 = 3600 )

Plugging these into the formula gives:

( 42/I_2 = 3600/6400 )

Cross-multiplying to solve for ( I_2 ):

( 42 * 6400 = 360