QUESTION IMAGE
Question
using the exposure maintenance formula (direct square law), calculate the missing factor.
| 1st exposure | 2nd exposure | |
|---|---|---|
| b | 72\ sid 75 mas | 36\ sid type you |
| c | 40\ sid 160 mas | 56\ sid type you |
| d | 40\ sid 12 mas | type you 27 mas |
| e | 72\ sid 20 mas | type you 10 mas |
Step1: Recall the direct - square law formula
The direct - square law formula for exposure maintenance is $\frac{mAs_1}{D_1^{2}}=\frac{mAs_2}{D_2^{2}}$, where $mAs_1$ and $mAs_2$ are the milli - ampere seconds values for the first and second exposures respectively, and $D_1$ and $D_2$ are the source - to - image receptor distances (SIDs) for the first and second exposures respectively. We can solve for the unknown $mAs$ or $SID$.
Step2: Solve for part a
Given $D_1 = 40$ inches, $mAs_1=40$ mAs, $D_2 = 72$ inches. Rearranging the formula to solve for $mAs_2$ gives $mAs_2=\frac{mAs_1\times D_2^{2}}{D_1^{2}}$. Substitute the values: $mAs_2=\frac{40\times72^{2}}{40^{2}}=\frac{40\times5184}{1600}=129.6$ mAs.
Step3: Solve for part b
Given $D_1 = 72$ inches, $mAs_1 = 75$ mAs, $D_2 = 36$ inches. Using the formula $mAs_2=\frac{mAs_1\times D_2^{2}}{D_1^{2}}$, we have $mAs_2=\frac{75\times36^{2}}{72^{2}}=\frac{75\times1296}{5184}=18.75$ mAs.
Step4: Solve for part c
Given $D_1 = 40$ inches, $mAs_1 = 160$ mAs, $D_2 = 56$ inches. Then $mAs_2=\frac{mAs_1\times D_2^{2}}{D_1^{2}}=\frac{160\times56^{2}}{40^{2}}=\frac{160\times3136}{1600}=313.6$ mAs.
Step5: Solve for part d
Given $D_1 = 40$ inches, $mAs_1 = 12$ mAs, $mAs_2 = 27$ mAs. Rearranging the formula to solve for $D_2$ gives $D_2^{2}=\frac{mAs_2\times D_1^{2}}{mAs_1}$. Then $D_2=\sqrt{\frac{27\times40^{2}}{12}}=\sqrt{\frac{27\times1600}{12}}=\sqrt{3600}=60$ inches.
Step6: Solve for part e
Given $D_1 = 72$ inches, $mAs_1 = 20$ mAs, $mAs_2 = 10$ mAs. Solving for $D_2$ using $D_2^{2}=\frac{mAs_2\times D_1^{2}}{mAs_1}$, we have $D_2=\sqrt{\frac{10\times72^{2}}{20}}=\sqrt{\frac{10\times5184}{20}}=\sqrt{2592}\approx50.91$ inches.
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a. 129.6 mAs
b. 18.75 mAs
c. 313.6 mAs
d. 60 inches
e. Approximately 50.91 inches