SOLUTION: Using the function, what happens to the intensity if you double the distance from the light
source? (Consider a certain distance and use the function to determine the intensity. T
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Question 979943: Using the function, what happens to the intensity if you double the distance from the light
source? (Consider a certain distance and use the function to determine the intensity. Then
double the distance and determine the intensity. Do this for several pairs of distances. You
should see a pattern develop. Express that pattern. Hint: you should be able to multiply
the first intensity by some number and get the second intensity.)
Function: f(x) =800/x^2
Distance from sensor (cm) 20 30 40 50 60 70 80 90 100
Intensity Gauge Reading 1 .86 .48 .39 .27 .19 .16 .13 .03
I doubled the distance but I kept getting a differing pattern each time. Any help would be appreciated.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Picking a number and then trying it, or even picking several numbers and trying them is an ok strategy to get an idea of what is happening, but it is a poor strategy for proving a mathematical relationship. For this you need to choose an arbitrary value and operate on that abstract quantity.
Let 
represent the original distance, so 
must be the doubled distance. Compare )
and )
\ =\ \frac{800}{x^2})
\ =\ \frac{800}{(2x)^2}\ =\ \frac{200}{x^2})
And therefore
\ =\ \frac{1}{4}f(x))
Hence, for any physically possible distance value x, the intensity is divided by 4 when the distance is doubled.
Extra credit. What happens to the intensity when the distance is multiplied by some number n? In other words, compare and )
John
My calculator said it, I believe it, that settles it
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