Page:Calculus Made Easy.pdf/181

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

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

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

Similarly, at the end of $0.1$ sec.,

${\frac {t}{T}}=10;\quad \epsilon ^{-10}=0.000045$;

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

(2) The intensity $I$ of a beam of light which has passed through a thickness $l$ cm. of some transparent medium is $I=I_{0}\epsilon ^{-Kl}$, where $I_{0}$ is the initial intensity of the beam and $K$ 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 $18\%$ in passing through 10 cms. of a certain transparent medium, this means that $82=100\times \epsilon ^{-K\times 10}$ or $\epsilon ^{-10K}=0.82$, and from the table one sees that $10K=0.20$ very nearly; hence $K=0.02$.

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