Question 354·Medium·Math

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The intensity $I$ , in arbitrary units, of light from a point source is inversely proportional to the square of the distance $r$ , in meters, from the source, so that $I = \dfrac{k}{r^2}$ for some constant $k$ .

If the intensity of the light is 36 units at a distance of 2 meters, at what distance, in meters, from the source is the intensity 4 units?

Light intensity that follows an inverse-square relationship depends on the square of the distance from its source. Use the first reading to find the constant, then enter the target intensity in a Desmos regression and keep the positive distance. A drop in intensity changes squared distance by the matching factor, not distance itself. Multiplying the starting distance by the full intensity ratio gives a distance that is too large.

Hints

Hint 1
A constant of proportionality is the value of $k$ that stays fixed for one light source. Put the first intensity and its distance into the formula. What value of $k$ does that reading give?
Hint 2
The target intensity gives a new equation with the same $k$ . A distance can't be negative, so restrict the unknown to positive values when you solve that equation in Desmos.

Step-by-step

Approach 1: Find the constant, then the distance
Step 1
Use the first reading
In the given inverse-square formula, substitute the 36-unit intensity and $2$ -meter distance: $36=\frac{k}{2^2}$ . This equation fixes the constant for this light source.
Step 2
Find the constant in Desmos
Type $36\sim k_1/2^2$ . The $\sim$ tells Desmos to fit $k_1$ , which stands for the constant of proportionality $k$ . Under PARAMETERS, Desmos shows $k_1=144$ , so $k=144$ .
Step 3
Write the equation for the target intensity
The source hasn't changed, so use the same constant. Substitute $I=4$ and $k=144$ into the formula: $4=\frac{144}{r^2}$ . Here $r$ is a distance, so the solution must be positive.
Step 4
Find the positive distance
Type $4\sim144/x_1^2\{x_1>0\}$ , where $x_1$ stands for $r$ . The restriction in braces excludes negative distances. Desmos shows $x_1=6$ under PARAMETERS, so the intensity is $4$ units at $6$ meters. Choice B.

Approach 2: Compare the two readings
Step 1
Find the intensity factor
Compare the readings: $36/4=9$ . The first intensity is nine times the target intensity. That gives an intensity factor, not a distance factor.
Step 2
Turn it into a distance factor
Because $I=k/r^2$ and $k$ stays the same, a ninefold decrease in intensity means $r^2$ grows by $9$ . A square root undoes a square, so the distance grows by $\sqrt{9}=3$ .
Step 3
Scale the starting distance
Multiply the starting $2$ meters by the distance factor $3$ : $2\times3=6$ . So the intensity is $4$ units at a distance of $6$ meters. Choice B.

Lessons that teach this

SAT Advanced AlgebraIntermediateCoreRearrange nonlinear formulas

Question 354·Medium·Math

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