Final answer:
The student's question involves calculating the new exposure time needed when the kVp is reduced from 90 to 75 while producing the same quality of diagnostic film. This is done using the relationship between kVp and exposure time, which implies that if the kVp is lowered, the exposure time must increase to compensate.
Step-by-step explanation:
The question involves finding the required exposure time to produce a diagnostic film using a different kilovoltage peak (kVp) while maintaining image quality. The original exposure settings were 90 kVp for 0.25 seconds. To equal the density of the diagnostic film at a reduced kVp of 75, one must compensate for the reduction in beam penetration by increasing the exposure time. The relationship between kVp and exposure time is governed by the following equation:
I1 \( T1 = I2 \( T2
Where I is the intensity provided by the voltage (kVp) and T is the exposure time. Assuming that the intensity is proportional to the square of the kVp (since x-ray production efficiency varies with the square of the voltage), we can represent the initial and desired conditions as:
(90 kVp)2 \( 0.25 s = (75 kVp)2 \( T2
To find the new exposure time (T2), we need to solve for T2:
T2 = (902 \( 0.25) / (752)
When you calculate this, you will find the new required exposure time to achieve the same film density at the reduced kVp setting. Remember to use consistent units and follow safe radiation practices for patient protection.