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January 17th, 2015, 01:11 PM   #1
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Post I cant figure this out! please help

In some murky harbors, the intensity of light d feet below the surface is given approximately by: I = I0e^(-0.23d)
What percentage of the surface light will reach a depth of:
a. 10 feet?

b. 20 feet?

Last edited by skipjack; January 17th, 2015 at 07:15 PM. Reason: to provide the exponentiation presumably intended
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January 17th, 2015, 01:29 PM   #2
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I=Io*exp(-0.23*d)
we suppose that the unity of the term (0.23 are inverse of a foot)

for d=10 ft
I(10 ft)=Io * 0.10032

Percentage is 10%


for d=20 ft
I(20 ft)=Io * 0.01005

percentage is 1%
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January 17th, 2015, 05:13 PM   #3
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Thank you so much, but I still don't understand how you get the answer.
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January 17th, 2015, 06:10 PM   #4
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Quote:
Originally Posted by alistrawberry View Post
I=Io*exp(-0.23*d)
we suppose that the unity of the term (0.23 are inverse of a foot)

for d=10 ft
I(10 ft)=Io * 0.10032

Percentage is 10%


for d=20 ft
I(20 ft)=Io * 0.01005

percentage is 1%
I'm not sure exactly what alistrawberry did, either... I don't know much about light penetration in water, but I didn't get about 0.1 for exp(-.23*10). Not sure what line 2 was supposed to mean (unity?).
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January 17th, 2015, 07:10 PM   #5
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At d = 0, I = 10. We want a percent so we want "intensity at a given depth"/"intensity at d = 0".

With d= 10: $\displaystyle \frac{10e^{-0.23\cdot10}}{10}=e^{-0.23\cdot10}\approx0.1002$

That's pretty close to 10%.

With d= 20: $\displaystyle \frac{10e^{-0.23\cdot20}}{10}=e^{-0.23\cdot20}\approx0.0101$

That's pretty close to 1%.
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January 17th, 2015, 07:18 PM   #6
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Quote:
Originally Posted by alistrawberry View Post
. . . for d=10 ft
I(10 ft)=Io * 0.10032
$e^{-2.3}$ = 0.10026 approximately, not 0.10032.

The question defines $d$ to be dimensionless, so inverse foot units are not required for the constant 0.23.
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