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Intensity with mirrors

  • Thread starter stringa
  • Start date
  • #1
7
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Homework Statement



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?



Homework Equations




I = P / A

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)
 

Answers and Replies

  • #2
110
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Just to clarify, this is a mirror and not a lens...right?
 
  • #3
7
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correct.......
 
  • #4
110
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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
 
  • #5
7
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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
 
  • #6
110
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Well, it would depend on the focal length of the mirror and the distance between obects
 
  • #7
110
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does the problem want a numerical answer, or an equation as a function of focal length and distance?
 
  • #8
7
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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
 
  • #9
110
0
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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