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Inverse Square Law concerning Light

  • #1
[a]1. Homework Statement [/b]
How does the distance from a point source of light affect the visibility of that light from the said source?


Homework Equations


This is a bit tricky without sub points, but here goes:

I1 x d12=I2 x d22
where:
I=intensity d2=distance per sq. unit

[c]3. The Attempt at a Solution [/b]

By using the ISL formula above, I plan to find the intensity of light at predefined intervals (i.e. every 10 cm) by substituting variables in the equation. After which, I will use a voltmeter and CdS cell to verify my calculations. I'm pretty sure I'm doing everything right, but some suggestions would be nice. :biggrin:
 
Last edited:

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
You might want to be careful defining "visibility"
What you are measuring is how the power received by a fixed area (the lightmeter) = the irradiance of the light varies with distance.
 
  • #3
did i say visibility? I meant intensity. Ty.
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
12
I meant intensity. Ty.
intensity is the power of the source
the correct term is irradiance = power/area
 

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