How Does Illuminance Change With Distance in Physics Experiments?

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Homework Help Overview

The discussion revolves around the relationship between illuminance and distance in a physics lab context. Participants are exploring how illuminance (measured in Lux) changes with varying distances from a light source, specifically looking at data points collected during an experiment.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an equation from a graph of illuminance versus distance but expresses confusion over the use of the formula E = I/d², questioning why d² is used instead of d. Other participants suggest comparing the equation to the standard form of a straight line to facilitate graphing.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to replot the data to achieve a linear representation. There is an ongoing exploration of the mathematical relationships involved, particularly regarding the derivation of the equation and its implications.

Contextual Notes

Participants are discussing the assumptions behind the formula for illuminance and the implications of light spreading from a point source, including the geometry of the situation and the surface area of a sphere.

garet122
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Hello, We had a lab in physics about the relationship between illuminance and distance. Our teacher gave use data so we make a graph of them. E is the illuminance in Lux and d is the distance in cm. (d,E) (50cm, 94 lux) (60cm, 64 lux) (70 cm, 49 lux) (80cm, 40 lux) (90 cm, 32 lux) (100cm, 26 lux)
It the graph, when I connect the points it give me an inverse proportion. My objective is to find an equation.


What I first tried is to find an equation directly from this graph but after searching on the internet I found this formula : E=I/d^2 and this explanation : intensity is inversely proportional to the square of the distance from the source. I do not understand, in my graph d is my abscissa not d^2 . I do not understand how we can use d^2 instead of d.
 
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Are you familiar with the equation for a straight line? Compare the standard equation for a straight line...

y = mx + c

with your equation which I have rewritten slightly..

E = I(1/d2) + 0

Then replot your data suitably modified so you get a straight line. Come back if you get stuck.

PS: It's not d2 you use.
 
CWatters said:
Are you familiar with the equation for a straight line? Compare the standard equation for a straight line...

y = mx + c

with your equation which I have rewritten slightly..

E = I(1/d2) + 0

Then replot your data suitably modified so you get a straight line. Come back if you get stuck.

PS: It's not d2 you use.


E = I(1/d2) + 0


my question why is the equation not E=I(1/d)

where do the 2 come from
 
Imagine a point source of light surrounded by a spherical glass surface of radius d. The light from the source spreads out uniformly in all directions and passes through the sphere. The "density" of the light (aka illuminance) passing through the glass will be...

The total amount of light emitted / Surface area of the sphere

In other words the bigger the sphere the less light will pass through each square meter.

What is the equation for the surface area of a sphere?
 

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