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Plane Mirror Problem

  1. Mar 18, 2009 #1
    I'm having problems doing the problem below. The equation in the hint is: I = Ps/(4*PI*r^2) which I have not clue how it relates to this question except for that it's an equation for intensity.

    I know that there will be a image produced a distance d behind the mirror, so there will be another "light source" 3d away from the screen.

    Last edited: Mar 18, 2009
  2. jcsd
  3. Mar 18, 2009 #2
    We define intensity as


    If we assume that the source is pointlike, the intensity will spread like a sphere from the source. [tex]4\pi \ r^2[/tex] is the area of a sphere with the radius r. The intensity at the distance r from a pointlike source is thus

    [tex]I=\frac{P}{4\pi \ r^2}[/tex]

    Does that help?

    (I noticed that in your formula you don't have the radius squared. Are you sure that it isn't squared in your book? This is how I would interpert the formula, might be I am wrong)
  4. Mar 18, 2009 #3
    oops thanks for pointing out the typo but I still don't understand what power has to do with this problem.
  5. Mar 18, 2009 #4


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    Hi DANIELLYMA! :smile:

    It doesn't really matter what power has to do with the problem …

    you're only asked to say how the intensity changes

    use the formula to calculate by what proportion it goes up because of the "extra" source :wink:
  6. Mar 18, 2009 #5
    r is the radius of the source, how do I take into account the the 3d distance from the screen?
  7. Mar 18, 2009 #6


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    I don't follow :confused: … r is distance.
  8. Mar 18, 2009 #7


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    Yes, r is "the 3d distance from the screen".
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