When increasing the distance from 40 in. to 72 in. in a lateral cervical spine radiograph, what should be the new milliampere (mA) setting for a similar receptor exposure?

When increasing the distance from 40 in. to 72 in. in a lateral cervical spine radiograph, what should be the new milliampere (mA) setting for a similar receptor exposure?

• A. 400 mA
• B. 800 mA
• C. 1000 mA
• D. 1200 mA

Answer: (C)

To maintain a similar receptor exposure while increasing the distance from 40 inches to 72 inches in a lateral cervical spine radiograph, it is necessary to understand the principles of the inverse square law. This law states that as the distance from the source of radiation increases, the intensity of the radiation decreases by the square of the distance. When moving from 40 inches to 72 inches, the distance has increased by a factor of 1.8 (since 72 / 40 = 1.8). According to the inverse square law, the intensity of the radiation reaching the receptor decreases by a factor of the square of 1.8, which is approximately 3.24 (1.8 squared). This means that to achieve a similar exposure at the receptor, you would have to increase the milliampere (mA) setting to compensate for this reduction in intensity. To maintain the same receptor exposure, if the original mA setting was at 400 mA for 40 inches, you would multiply this by the factor of 3.24 to adjust for the increased distance. Thus, the calculation would be as follows: 400 mA x 3.24 ≈ 1296 mA. Given that the closest standard

To maintain a similar receptor exposure while increasing the distance from 40 inches to 72 inches in a lateral cervical spine radiograph, it is necessary to understand the principles of the inverse square law. This law states that as the distance from the source of radiation increases, the intensity of the radiation decreases by the square of the distance.

When moving from 40 inches to 72 inches, the distance has increased by a factor of 1.8 (since 72 / 40 = 1.8). According to the inverse square law, the intensity of the radiation reaching the receptor decreases by a factor of the square of 1.8, which is approximately 3.24 (1.8 squared). This means that to achieve a similar exposure at the receptor, you would have to increase the milliampere (mA) setting to compensate for this reduction in intensity.

To maintain the same receptor exposure, if the original mA setting was at 400 mA for 40 inches, you would multiply this by the factor of 3.24 to adjust for the increased distance. Thus, the calculation would be as follows:

400 mA x 3.24 ≈ 1296 mA.

Given that the closest standard