Inverse Square Law concerning Light

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SUMMARY

The discussion centers on the application of the Inverse Square Law (ISL) in understanding how the distance from a point light source affects light intensity. The formula I1 x d1² = I2 x d2² is utilized to calculate light intensity at specific intervals, with measurements taken using a voltmeter and a CdS cell. The conversation highlights the importance of distinguishing between "visibility" and "irradiance," with irradiance defined as power per unit area. The participants confirm the correct application of the ISL in their calculations.

PREREQUISITES
  • Understanding of the Inverse Square Law (ISL)
  • Familiarity with light intensity and irradiance concepts
  • Basic knowledge of using a voltmeter and CdS cell
  • Ability to perform calculations involving power and area
NEXT STEPS
  • Research the practical applications of the Inverse Square Law in optics
  • Learn how to accurately measure irradiance using light meters
  • Explore the relationship between light intensity and distance in various mediums
  • Study advanced concepts in photometry and radiometry
USEFUL FOR

Students in physics, educators teaching optics, and professionals in fields related to lighting design and photometry will benefit from this discussion.

MangoOverlord
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[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:
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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.
 
did i say visibility? I meant intensity. Ty.
 
MangoOverlord said:
I meant intensity. Ty.
intensity is the power of the source
the correct term is irradiance = power/area
 

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