A pelvis radiograph is taken with 80 kVp, 30 mAs, and at a distance of 44 inches. If the distance is reduced to 36 inches while maintaining 80 kVp, what will the new intensity be? What mAs should be used to keep the radiographic exposure consistent

To solve this problem, we'll use two key concepts in radiography:

1. **Inverse Square Law:** This law states that the intensity of radiation is inversely proportional to the square of the distance from the source. Mathematically, it can be expressed as:

   \[
   \frac{I_1}{I_2} = \left(\frac{D_2}{D_1}\right)^2
   \]

   where:
   - \( I_1 \) is the initial intensity,
   - \( I_2 \) is the new intensity,
   - \( D_1 \) is the initial distance,
   - \( D_2 \) is the new distance.

2. **mAs and Exposure Relationship:** The exposure is directly proportional to the product of mA (milliamperes) and time (seconds). Therefore, to keep the radiographic exposure consistent, the product of mA and time (mAs) needs to be adjusted according to changes in intensity.

### Step 1: Calculate the New Intensity
Using the inverse square law:

\[
\frac{I_1}{I_2} = \left(\frac{D_2}{D_1}\right)^2
\]

Plugging in the values:

\[
\frac{I_1}{I_2} = \left(\frac{36 \text{ inches}}{44 \text{ inches}}\right)^2 = \left(\frac{36}{44}\right)^2
\]

Now calculate the square:

\[
\frac{I_1}{I_2} = \left(\frac{36}{44}\right)^2 \approx \left(\frac{9}{11}\right)^2 \approx \left(0.8182\right)^2 \approx 0.6694
\]

Thus:

\[
I_2 = \frac{I_1}{0.6694} \approx 1.494 \times I_1
\]

This means the new intensity \( I_2 \) is about 1.494 times the initial intensity \( I_1 \).

### Step 2: Adjust mAs for Consistent Exposure
To maintain consistent exposure, the mAs should be adjusted inversely to the change in intensity:

\[
\text{New mAs} = \frac{\text{Old mAs}}{\text{Change in Intensity}} = \frac{30 \text{ mAs}}{1.494} \approx 20.1 \text{ mAs}
\]

### Final Answers:

- **New Intensity:** The new intensity will be approximately 1.494 times the original intensity.
- **New mAs:** To keep the radiographic exposure consistent, you should use approximately **20.1 mAs**.

Would you like more details or have any other questions?

Here are 5 related questions:
1. What is the significance of kVp in radiography, and why is it kept constant in this problem?
2. How does changing the distance affect the sharpness and contrast in a radiograph?
3. What are some potential consequences of not adjusting mAs when the distance is changed?
4. Can you explain how changing the mAs affects the patient dose?
5. How would the calculations change if the kVp were also adjusted?

**Tip:** Always double-check your calculations when using the inverse square law, as small errors in distance can lead to significant changes in intensity.

A pelvis radiograph is taken with 80 kVp, 30 mAs, and at a distance of 44 inches. If the distance is reduced to 36 inches while maintaining 80 kVp, what will the new intensity be? What mAs should be used to keep the radiographic exposure consistent?

Explore how to calculate the new intensity and adjust mAs for a pelvis radiograph when the distance is changed from 44 inches to 36 inches, using the Inverse Square Law. This problem involves key radiography concepts and is suitable for students in Grades 11-12.

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