Inverse Square Law (srs)

  • Thread starter Nul
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  • #1
Nul
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Homework Statement


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


Homework Equations


I 1/I 2 = d^2 2/d^2 1
or something


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.
 

Answers and Replies

  • #2
954
117
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]
 
  • #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
 
  • #4
954
117
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.
 
  • #5
Nul
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ok thanks man
^^
 
  • #6
Nul
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so if i were doing (b) would the answer be 1333.333 lux
 
  • #7
954
117
so if i were doing (b) would the answer be 1333.333 lux
Yes, but you need to round off appropriately.
 

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