How Is Light Intensity Affected by Water Depth?

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Homework Help Overview

The discussion revolves around the effect of water depth on light intensity, specifically how intensity decreases as light travels through water. The original poster presents a scenario involving a light ray passing through a transparent surface and its intensity at various depths, with a focus on calculating the intensity at 15 feet below the surface based on given conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between light intensity and depth, with one suggesting an exponential decay model and another proposing a simpler multiplicative approach based on observed intensity reduction.

Discussion Status

Multiple approaches are being discussed, with some participants offering guidance on finding the decay constant and others suggesting a straightforward method based on the intensity reduction factor. There is an acknowledgment of different interpretations of the problem, but no consensus has been reached.

Contextual Notes

Participants note the original poster's uncertainty with the language and the mathematical concepts involved, which may influence the clarity of the discussion. The problem also involves assumptions about the properties of light in water and the specific conditions under which the intensity measurements are made.

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Homework Statement



When a light ray goes trough a transparent surface, its intensity rate of change decreases proportionaly to I(t), where t is the thickness. On a specific kind of water, intensity 3 feet from below the surface is 25% its original intensity. What is the intensity from 15 ft below the surface? (sorry for my english lol)

Homework Equations



dI/dt = kI, I(3) = 0.25I_0

The Attempt at a Solution


My attempt is here: http://img69.imageshack.us/img69/4237/zill2.png

The correct answer is I = 0.0009765625 I_0, but I don't know how to get there! Any hint please?
 
Last edited by a moderator:
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supermiedos said:

Homework Statement



When a light ray goes trough a transparent surface, its intensity rate of change decreases proportionaly to I(t), where t is the thickness. On a specific kind of water, intensity 3 feet from below the surface is 25% its original intensity. What is the intensity from 15 ft below the surface? (sorry for my english lol)

Homework Equations



dI/dt = kI, I(3) = 0.25I_0

The Attempt at a Solution


My attempt is here: http://img69.imageshack.us/img69/4237/zill2.png

The correct answer is I = 0.0009765625 I_0, but I don't know how to get there! Any hint please?

You know that I(t) = I_0 e^{-kt} for some k > 0. The first step is to find k from the condition I(3) = 0.25I_0, so that
<br /> I_0 e^{-3k} = 0.25 I_0<br />
Then you can find I(15) = I_0 e^{-15k}.
 
Last edited by a moderator:
pasmith said:
You know that I(t) = I_0 e^{-kt} for some k &gt; 0. The first step is to find k from the condition I(3) = 0.25I_0, so that
<br /> I_0 e^{-3k} = 0.25 I_0<br />
Then you can find I(15) = I_0 e^{-15k}.

Ah of course! Because I(0) = I_0, and thus, c = I_0. The rest is easy. Thank you!
 
Supermiedos,
there is a much simpler approach: since the light intensity decreases by a factor of 4 every time it goes through a layer of 3 meters of water, the light intensity after 5 layers is just ##(1/4)^5=1/1024=0.0009765625##. No need for an ODE :wink:
 
Coelum said:
Supermiedos,
there is a much simpler approach: since the light intensity decreases by a factor of 4 every time it goes through a layer of 3 meters of water, the light intensity after 5 layers is just ##(1/4)^5=1/1024=0.0009765625##. No need for an ODE :wink:

Hehe, that's actually right. Thank you :D
 

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