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.,

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 $$0.5 = \epsilon^{-0.02l}$$.