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Inverse Square Law (srs)

  1. Feb 10, 2010 #1

    Nul

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    1. The problem statement, all variables and given/known data
    A source of light is used to illuminate a screen. 1 m from the source, the intensity of the light falling on the screen is 12 000 lux. Calculate the intensity on the screen at distances of:
    (a) 2 m
    (b) 3 m
    (c) 4 m


    2. Relevant equations
    I 1/I 2 = d^2 2/d^2 1
    or something


    3. The attempt at a solution
    I don't understand at all. I'm hoping you guys would be able to explain what I am meant to do here because the teacher fails at teaching.
    Just give me the equation for (a) so that I have a better understanding of what the is going on which would hopefuly enable me to do (b) and (c).
    Thank you.
     
  2. jcsd
  3. Feb 10, 2010 #2
    I'm going to assume that the source is either a point source or a spherical, isotropic one.
    The inverse square law applies for intensity, in that the intensity of the light at a point away from the source is inversely proportional to the square of the distance of that point from the source.
    Mathematically,
    [tex]I \varpropto \frac{1}{r^2}[/tex]​

    Thus, problems can be solved using this proportionality:
    [tex]I_{1} = \frac{k}{{r_{1}}^2}[/tex]

    [tex]I_{2} = \frac{k}{{r_{2}}^2}[/tex]

    [tex]\frac{I_{1}}{I{_2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}}[/tex]
     
  4. Feb 10, 2010 #3

    Nul

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    yes... i know the formula
    i have it in my book
    but i don't know HOW to use them
    can you just show me where the numbers should go for (a) then I would probably understand
    please
     
  5. Feb 10, 2010 #4
    Using the final equation I posted, substitute the values:
    [tex]I_{1} = 12 000, r_{1} = 1, r_{2} = 2[/tex]​

    Then it follows that
    [tex]I_{2} = I_{1}\,\frac{{r_1}^2}{{r_2}^2} = 12 000\,\frac{1^2}{2^2} = 3 000 \,lux[/tex]​

    From this, the inverse square law becomes apparent: at double the distance, intensity decreases to a quarter, and so on.
     
  6. Feb 10, 2010 #5

    Nul

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    ok thanks man
    ^^
     
  7. Feb 10, 2010 #6

    Nul

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    so if i were doing (b) would the answer be 1333.333 lux
     
  8. Feb 10, 2010 #7
    Yes, but you need to round off appropriately.
     
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