A radiographic technique, using an exposure of 10mAmin, produces a film density of 1.5 where a film density of 2.3 is required. A film charactaristic curve is available which gives a log relative exposure value of 1.2 for [density = 1.5] with a correspondingvalue of 1.8 for [density = 2.3]the acceptable minimum film density should be achieved with an exposure of ?
+33661 I'm guessing here, but:
log(10/k) = 1.2 so 10/k = 101.2 where k is a reference level exposure.
log(E/k) = 1.8 so E/k = 101.8 where E is the required exposure.
Divide the second of these by the first and rearrange to get:
$${\mathtt{E}} = {\frac{{\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{1.8}}}}{{{\mathtt{10}}}^{{\mathtt{1.2}}}}} \Rightarrow {\mathtt{E}} = {\mathtt{39.810\: \!717\: \!055\: \!349\: \!725\: \!1}}$$
So E ≈ 39.8 mA.min
.
+33661 I'm guessing here, but:
log(10/k) = 1.2 so 10/k = 101.2 where k is a reference level exposure.
log(E/k) = 1.8 so E/k = 101.8 where E is the required exposure.
Divide the second of these by the first and rearrange to get:
$${\mathtt{E}} = {\frac{{\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{1.8}}}}{{{\mathtt{10}}}^{{\mathtt{1.2}}}}} \Rightarrow {\mathtt{E}} = {\mathtt{39.810\: \!717\: \!055\: \!349\: \!725\: \!1}}$$
So E ≈ 39.8 mA.min
.
