Final answer:
To maintain the same exposure when decreasing the kVp from 85 to 70 in a radiograph, an increase in exposure time is needed. Assuming an inverse proportionality, the new time would be calculated to be approximately 0.182 seconds.
Step-by-step explanation:
If a radiograph is exposed using 85 kVp at 0.15 seconds and the kVp is decreased to 70 kVp, to maintain the same exposure we would have to increase the time since the intensity of the X-rays decreases with the decrease in kVp. Assuming the relation between exposure time and kVp is inversely proportional, which is a simplification that ignores other factors that might play a role in a real-world scenario, we can set up the following proportion to find the new time required:
Old kVp × Old time = New kVp × New time
85 kVp × 0.15 s = 70 kVp × New time
New time = (85 kVp × 0.15 s) / 70 kVp
New time ≈ 0.182 s
Therefore, the new exposure time would be approximately 0.182 seconds. This calculation assumes linearity in the relationship between kVp and exposure time, which may not be precise in actual clinical settings due to factors like changes in contrast and the exact behavior of the X-ray machine at different voltages.