How Is Light Intensity Affected by Water Depth?

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Light intensity decreases exponentially with water depth, following the equation I(t) = I_0 e^{-kt}. Given that the intensity at 3 feet is 25% of the original, the decay constant k can be determined from this condition. The intensity at 15 feet can then be calculated using the same formula, resulting in I = 0.0009765625 I_0. A simpler method involves recognizing that intensity decreases by a factor of 4 for every 3 feet, leading to the same result without complex calculations. Understanding these principles is crucial for solving problems related to light intensity in water.
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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?
 
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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}.
 
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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