Page:Calculus Made Easy.pdf/181

the time-constant being ${\displaystyle 0.01}$. This means that it takes ${\displaystyle 0.01}$ sec. for the variable term to fall by ${\displaystyle {\dfrac {1}{\epsilon }}=0.3678}$ of its initial value ${\displaystyle 10\epsilon ^{-{\frac {0}{0.01}}}=10}$.

To find the value of the current when ${\displaystyle t=0.001}$ sec., say, ${\displaystyle {\dfrac {t}{T}}=0.1}$, ${\displaystyle \epsilon ^{-0.1}=0.9048}$ (from table).

It follows that, after ${\displaystyle 0.001}$ sec., the variable term is ${\displaystyle 0.9048\times 10=9.048}$, and the actual current is ${\displaystyle 10-9.048=0.952}$.

Similarly, at the end of ${\displaystyle 0.1}$ sec.,

the variable term is ${\displaystyle 10\times 0.000045=0.00045}$, the current being ${\displaystyle 9.9995}$.

(2) The intensity ${\displaystyle I}$ of a beam of light which has passed through a thickness ${\displaystyle l}$ cm. of some transparent medium is ${\displaystyle I=I_{0}\epsilon ^{-Kl}}$, where ${\displaystyle I_{0}}$ is the initial intensity of the beam and ${\displaystyle K}$ is a “constant of absorption.”

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by ${\displaystyle 18\%}$ in passing through 10 cms. of a certain transparent medium, this means that ${\displaystyle 82=100\times \epsilon ^{-K\times 10}}$ or ${\displaystyle \epsilon ^{-10K}=0.82}$, and from the table one sees that ${\displaystyle 10K=0.20}$ very nearly; hence ${\displaystyle K=0.02}$.

To find the thickness that will reduce the intensity to half its value, one must find the value of ${\displaystyle l}$ which satisfies the equality ${\displaystyle 50=100\times \epsilon ^{-0.02l}}$, or ${\displaystyle 0.5=\epsilon ^{-0.02l}}$.