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Supposing the mean absorption coefficient of seawater in the visible region is $\pu{0.8 m-1}$, calculate the depth at which a diver will experience half the surface light intensity.

I know that you have to use $A = -\log \left(\dfrac{I}{I_o}\right) = \epsilon cl$ but I'm not sure how to proceed

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    $\begingroup$ The answer depends on whether you use log to base 10 or base e, i.e whether you are taught $I/I_0=10^{-\epsilon c l}$ or $e^{-\epsilon c l}$. As the absorption coefficient is $\epsilon c=0.8 $ (which you can see by looking at units) you can calculate the answer by making $I=I_0/2$ $\endgroup$
    – porphyrin
    Commented Aug 1, 2020 at 16:58

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In my opinion, Beer–Lambert law is no use. I would rather write

$$\frac{\mathrm dI}{I\,\mathrm dl} = \pu{0.8 m^-1},\tag{1}$$

which leads to the integrated formula

$$\ln\left(\frac{I_0}{I}\right) = 0.8 l.\tag{2}$$

And, if $\displaystyle\frac{I}{I_0} = 0.5,$ the corresponding depth is

$$l = \frac{\ln 2}{0.8} = \pu{0.846 m}.\tag{3}$$

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