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Intensity with mirrors

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data

    The EPM has finally crossed the fine line between party tricks and felonies
    with the genocide of the pigmy marmoset. You are now determined to make him '
    pay the price. You fashion a makeshift sherical mirror/laster beam our of a
    roll of aluminum foil and a stick of bubblegum. Considering the power radiated
    from the sun lands on earth with a maximum flux of almost a kilowatt per square
    meter:


    How large should the surface of the mirror be, if you are to deliver over a
    megawatt per square meter over a 1 square centimeter monkey-butt surface?



    2. Relevant equations


    I = P / A
    3. The attempt at a solution

    I have tried this several times but I don't see the link between everything. I obviously have to use I = P / A and solve for A(mirror)
     
  2. jcsd
  3. Mar 20, 2007 #2
    Just to clarify, this is a mirror and not a lens...right?
     
  4. Mar 20, 2007 #3
    correct.......
     
  5. Mar 20, 2007 #4
    The question as i am reading isnt very clear. You would have to have a parabolic mirror to focus the light...so it would depend...unless im missing something
     
  6. Mar 20, 2007 #5
    it's a spherical mirror...specifically a concave sphere.......

    I know i have to use intensity but i don't know how i can relate the suns intensity to that of the mirror ; and / or power
     
  7. Mar 20, 2007 #6
    Well, it would depend on the focal length of the mirror and the distance between obects
     
  8. Mar 20, 2007 #7
    does the problem want a numerical answer, or an equation as a function of focal length and distance?
     
  9. Mar 20, 2007 #8
    You have as much information as I do. It is all stated above. It wants to know how much area the mirror must have. hence i find the Intensity possibly very useful, since I = P / A .... but how can P be related from the sun to the mirror
     
  10. Mar 20, 2007 #9
    Well, a mirror of size A=1000m^2 would obviously give you that amount of energy...but im not sure about other parameters and how they would affect...sorry
     
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