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The rate (per foot) at which light is absorbed as it passes through water is proportional to the intensity, or brightness, at that point.

a. Find the intensity as a function of the distance the light has traveled through the water.

b. If 50% of the light is absorbed in 10 feet, how much is absorbed in 20 feet? 25 feet?


for part a: $y^{\prime}=ky$ then its $y=Ce^{kd}$

for part b: I am guessing for when $d=0$ the intensity that is absorbed is 0 thus meaning $C = 1$

$0.50=1e^{d\left(10\right)} \Rightarrow d = \dfrac{\ln \vert 0.5 \vert}{10}$

thus for 20ft or 25ft the equation are $y_{25\mathrm{ft}} = 1e^{\dfrac{\ln \vert 0.5 \vert}{10}\left(20\right)} = 0.25$

and for 25ft the answer is $\dfrac{\sqrt{2}}{8}$

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1 Answer 1

up vote 1 down vote accepted

Your calculations are correct. The answers for (b) are not quite right because of misinterpretation of what is asked for.

If $y$ is the intensity at depth $x$, then $$\frac{dy}{dx}=-ky.$$ We have used $-k$ because it is nice to have our constants, in this case $k$, positive.

Solving the differential equation, we find that $y=Ce^{-kx}$, where $C=y(0)$.

We have $y(10)=\dfrac{1}{2}y(0)$. Substituting in our solution, and cancelling $y(0)$, we find that $e^{-10k}=\dfrac{1}{2}$. Taking logarithms, we find that $-10k=\ln(1/2)=-\ln 2$.

Thus $k=\dfrac{\ln 2}{10}$.

For $20$ feet, we don't need much of the above. If the intensity is reduced by the factor $\dfrac{1}{2}$, it is reduced by the factor $\dfrac{1}{4}$ in $20$ feet. So the fraction absorbed is $\dfrac{3}{4}$.

For $25$ feet, if the intensity at the surface ($x=0$), then the intensity at depth $25$ is $y(0)e^{-(\ln 2)(25/10)}$. So the proportion of the light that remains is $e^{-(\ln 2)(25/10)}$.

This can be expressed in various other ways, such as your $\dfrac{\sqrt{2}}{8}$. Thus the proportion of the light that is absorbed is $1-\dfrac{\sqrt{2}}{8}$.

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