Inverse Square Law concerning Light

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Homework Help Overview

The discussion revolves around the relationship between distance from a point light source and the visibility or intensity of that light, specifically through the lens of the inverse square law (ISL). Participants are exploring how light intensity changes with distance and the implications of these changes in practical measurements.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the inverse square law to calculate light intensity at various distances, while others question the definitions of terms like "visibility" and "intensity," suggesting a focus on irradiance instead.

Discussion Status

The discussion is active, with participants clarifying terminology and exploring the correct application of concepts related to light intensity and measurement. Suggestions for careful definition and understanding of terms have been provided, indicating a productive dialogue.

Contextual Notes

There appears to be some confusion regarding the terminology used to describe light intensity and visibility, with participants addressing the need for precise definitions in the context of their calculations and measurements.

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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