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Determining light intensity with a mirror

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    You put a point source of light (S) a distance (d) in front of screen (A). How is the light intensity at the center of the screen changed if you put a completely reflecting mirror (M) a distance (d) behind the source?

    M-----d-----S-----d-----A

    2. Relevant equations

    I=Power/Area ??
    3. The attempt at a solution

    The answer it is giving in the book is 10/9 of the original intensity but i have no clue at all how to get this.
     
  2. jcsd
  3. Mar 23, 2010 #2

    kuruman

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    When you put the mirror in place, the image constitutes a second source in addition to the original one. How far from the screen is this second source? How does intensity depend on distance?
     
  4. Mar 23, 2010 #3
    i would think intensity in this case would be power/distance, so S/d originally. then with the mirror S/d + S/2d ... am i on the right track??
     
  5. Mar 23, 2010 #4

    kuruman

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    No. Answer me this, if you double the distance between the source and the screen, by what factor is the intensity on the screen reduced?
     
  6. Mar 23, 2010 #5
    decreases by the square of the distance from the source?? i think.. just re-reading my text book trying to figure it out
     
  7. Mar 23, 2010 #6

    kuruman

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    Correct. Now call the original screen-to-source distance one "unit". How many "units" is the image-to-source distance?
     
  8. Mar 23, 2010 #7
    image to source would be 2 units, so intensity would be decreased by a factor of 4 for the light comming from the image on the screen compared to light comming from the source on the screen? .. argh sorry if im clueless..
     
  9. Mar 23, 2010 #8

    kuruman

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    Two units is the mirror-to-source distance. The image is behind the mirror by an additional how many units?
     
  10. Mar 23, 2010 #9
    argh forgot about the whole "virtual" image thing..

    virtual image to screen would be 3 units ... as source to mirror is 1 unit.. then virtual image to mirror another 1 unit.. then 1 unit to screen..

    sooo..

    3 units from virtual image to screen... 1 unit from source to screen..???
     
  11. Mar 23, 2010 #10

    kuruman

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    Correct. So if the intensity of the primary source to the screen is "1" intensity unit, what is the intensity of the secondary source that is 3 distance units away?
     
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