Question

When taking a radiograph of an equine limb, you take a radiograph where the portable X-ray unit source is 90 cm away from the cassette. You then move closer so that the X-ray source is now 30 cm away from the cassette. If no changes were made to the settings of the machine, what will the beam intensity be at the second radiograph relative to the first one?

Answer 1

Answer:

$\frac{I_2}{I_1} = 9$

Explanation:

Intensity of the beam inversely depends on the distance from the cassette

so here we can say it is

$I = \frac{Power}{distance^2}$

so we have

$\frac{I_1}{I_2} = \frac{d_2^2}{d_1^2}$

so we will have

$d_1 = 90 cm$

$d_2 = 30 cm$

so we have

$\frac{I_2}{I_1} = (\frac{d_1}{d_2})^2$

$\frac{I_2}{I_1} = (\frac{90}{30})^2$

$\frac{I_2}{I_1} = 9$