Answer:
the new exposure time would be 0.049 seconds
Step-by-step answer:
Given that:
Distance 1 = D1 = 42 inch
Current = 300 mA
Time 1 = T1 = 0.06 seconds
Max voltage = 80 kVp
Distance 2 = D2 = 38 inches
Time 2 = T2 = ?
Inverse square law states that:
The intensity of radiation becomes weaker as it spreads out from the source since the same about of radiation becomes spread over a larger area. The intensity is inversely proportional to the distance from the source.
Mathematically,
I1/ I2 = D1^2/ D2^2 ------------ eq1
Intensity 1 = I1 = Current * time = 300 * 0.06 = 18 mAs
Let Intensity 2 as I2
Putting values in eq1
18/ I2 = (42)^2/(38)^2
By simplifying:
18/ I 2 = 1764/1444
By cross multiplying we get:
1764 * I2 = 25,992
Dividing both sides by 1764 we get:
Intensity 2 = I2 = 14.7 mAs
Now we will find exposure time.
We know that:
Intensity = Current * Time
As Intensity 2 = 14.7 , Current = 300 mA and time is T2
So by putting values:
14.7 = 300 * T2
By simplifying,
T2 = 0.049 seconds
So the new exposure time required would be 0.049
i hope it will help you!
