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How to solve this problem using diff equations?

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    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)

    2. Relevant equations

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

    3. The attempt at a solution
    My attempt is here: http://img69.imageshack.us/img69/4237/zill2.png [Broken]

    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: May 6, 2017
  2. jcsd
  3. Feb 28, 2013 #2

    pasmith

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    Homework Helper

    You know that [itex]I(t) = I_0 e^{-kt}[/itex] for some [itex]k > 0[/itex]. The first step is to find [itex]k[/itex] from the condition [itex]I(3) = 0.25I_0[/itex], so that
    [tex]
    I_0 e^{-3k} = 0.25 I_0
    [/tex]
    Then you can find [itex]I(15) = I_0 e^{-15k}[/itex].
     
    Last edited by a moderator: May 6, 2017
  4. Feb 28, 2013 #3
    Ah of course!!! Because I(0) = I_0, and thus, c = I_0. The rest is easy. Thank you!
     
  5. Mar 2, 2013 #4
    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:
     
  6. Mar 3, 2013 #5
    Hehe, that's actually right. Thank you :D
     
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