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Light intensity model

  1. Nov 28, 2014 #1
    Can someone assist with a simple model for the following question? I have looked on the net but cannot find anything on it.

    Say you are sat at a desk with the light bulb directly above you switched off and the rest on; so:

    on - on - on
    on - off - on
    on - on - on

    Each at distance r apart with the light intensity at each point the same (say I)

    I wish to model the intensity of light at the off point and show that just switching the bulb at this point will have very little difference to the intensity of light here point regardless if the bulb at that piont was switched on or off.

    I wish to double check my working to see if I am on the right track.


    Many thanks,
     
  2. jcsd
  3. Nov 28, 2014 #2

    mathman

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    It depends on how far the desk is from the array of lights. If far away (compared to r) turning off 1 light reduces the illumination by 1/9. If close to array it will be more. Essentially you need to use the inverse square law.
     
  4. Nov 29, 2014 #3
    Thank you for your reply. How did you come to this result? I am very rusty on my Physics at the moment.

    Lets say the distance between the each light is fixed and at r distance apart. The height distance is h; if I have understood your reply and I failed to consider that aspect. I do not think that would be a major issue though.

    Can you refer me to a link or a book that explains the above scenario. I cannot find one at present that explains what I am trying to solve,

    Best,
     
  5. Nov 29, 2014 #4

    berkeman

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    Staff: Mentor

    Just search on "Inverse Square Law" for light. mathman's point is that if the ceiling with the lights is 10cm over your head, switching on the middle bulb 10cm over your head will increase the light for you quite a bit. If the ceiling is several meters away, you will see a smaller increase in intensity.
     
  6. Nov 29, 2014 #5

    mathman

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    Intensity of each light source is proportional to [itex]\frac{1}{d^2}[/itex]. Total illumination is sum over all sources. Overhead [itex]d^2=h^2[/itex], four nearest to overhead - each [itex]d^2=h^2+r^2[/itex], four corners - each [itex]d^2=h^2+2r^2[/itex].
     
  7. Nov 30, 2014 #6
    Thanks both for your help.
     
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