Question

When a vertical beam of light passes through a transparent medium, the rate at which its intensity Idecreases is proportional to$\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, where trepresents the thickness of the medium (in feet).In clear seawater, the intensity 3feet below the surface is 25% of the initial intensity${\mathbf{I}}_{\mathbf{0}}$of the incident beam. What is the intensity of the beam 15feet below the surface?

Short answer

The intensity of the beam fifteen feet below the surface is ${\text{0.00098I}}_0\text{W/ft}$.

Step-by-step solution

Define growth and decay

The initial-value problem, $\frac{\text{dx}}{\text{dt}}{\text{= kx, x(t}}_{\text{0}}{\text{) =x}}_{\text{0}}$ where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. This is in the form of a first-order reaction (i.e.) a reaction whose rate, or velocity,$\text{dx/dt}$ is directly proportional to the amount x of a substance that is unconverted or remaining at time t.

Solve for first order growth and decay equation.

Let the linear equation with the amount of radioactive substance as be,

$\frac{\text{dI}}{\text{dt}}\text{= - kI}$… (1)

And with the conditions,${\text{I(t = 0 feet) =I}}_0$ and ${\text{I(t = 3 feet) = 0.25I}}_{\text{0}}\text{W/ft}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$$\begin{gathered} \frac{\text{dI}}{\text{I}}\text{= - kdt} \\ \int \frac{\text{1}}{\text{I}}\text{dI = - k}\int \text{dt} \\ {\text{lnI = - kt +c}}_{\text{1}} \\ {\text{e}}^{\text{ln(I)}}{\text{=e}}^{\left({\text{- kt +c}}_{\text{1}}\right)} \end{gathered}$$

Then, the equation becomes,

$$\begin{gathered} {\text{I =e}}^{\text{- kt}}{\text{e}}^{c_1} \\ {\text{= ce}}^{\text{kt}} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point in the equation (2), then

$$\begin{gathered} {\text{I}}_{\text{0}}{\text{= ce}}^{\text{0}} \\ {\text{c =I}}_{\text{0}} \end{gathered}$$

Substitute the value of in the equation (2).

$\mathrm{I}={\mathrm{I}}_0{\mathrm{e}}^{-\mathrm{kt}}$… (3)

Again, apply the other point${\text{(I,t) = (0.25I}}_{\text{0}}\text{,3)}$ in the equation (3).

$$\begin{gathered} {\text{0.25I}}_{\text{0}}{\text{=I}}_{\text{0}}{\text{e}}^{\text{- 3k}} \\ {\text{0.25 =e}}^{\text{- 3k}} \\ \text{ln(0.25) = - 3k} \\ \text{k = -}\frac{\text{(ln0.25)}}{\text{3}} \\ \text{=}\frac{\text{- 1.3863}}{\text{- 3}} \\ \text{= 0.462} \end{gathered}$$

Substitute the value of k in the equation (3).

${\text{I =I}}_{\text{0}}{\text{e}}^{\text{- 0.462t}}$… (4)

Obtain theintensity of the light beam.

Substitute the value $\text{t = 15}$into the equation (4).

$\begin{array}{rcl}\mathrm{I}& =& {\mathrm{I}}_0{\mathrm{e}}^{-0.462\times 15}\\ & =& {\mathrm{I}}_0{\mathrm{e}}^{-6.93}\\ & \approx & 0.00098{\mathrm{I}}_0\mathrm{W}/\mathrm{ft}\end{array}$