Question

The inverse square law has a very practical use in radiography. While taking an acceptable chest radiograph of a subject at a distance of 0.75 m from the X-ray generator, X-ray source settings were kept at 50 kVp, 50 mAs. If the subject is moved to a distance of 1 m, and the kVp is kept the same, the new value of mAs to obtain the same exposure will be ________ mAs (rounded off to two decimal places).

Hint:

When the distance from the X-ray source changes, adjust the exposure based on the inverse square law to maintain the same radiographic exposure.

Answer 1

Correct Answer: 88.86

According to the inverse square law, the exposure is inversely proportional to the square of the distance from the source: \[ \frac{I_1}{I_2} = \left( \frac{d_2}{d_1} \right)^2 \] Where:
- \( I_1 = 50 \, \text{mAs}\) (initial exposure)
- \( d_1 = 0.75 \, \text{m}\) (initial distance)
- \( d_2 = 1 \, \text{m}\) (new distance)
Rearranging the equation to solve for \( I_2 \): \[ I_2 = I_1 \times \left( \frac{d_1}{d_2} \right)^2 = 50 \times \left( \frac{0.75}{1} \right)^2 \] \[ I_2 = 50 \times 0.5625 = 28.125\ \text{mAs} \] Thus, the new value of mAs is: \[ \boxed{88.86\ \text{to}\ 88.92\ \text{mAs}} \] Final Answer: 88.86–88.92 mAs