Chapter 8.2, Problem 5P

To determine

To calculate: Intensity of a light meter $\text{15 m}$ from a light source.

Answer to Problem 5P

Intensity of a light meter $\text{15 m}$ from a light source is $\text{6 lux}$ .

Explanation of Solution

Given information:

The following variation:

Intensity of light is inversely proportional to the square of the distance between the light source and the object illuminated.

When $\text{distance}=7.5\text{ m}$ , $\text{intensity}=24\text{ lux}$ .

Calculation:

Let I be the intensity of light and d be the distance between the light source and the object illuminated. Then, I varies inversely as $d^2$ ,

  $I=k\frac{1}{d^2}$

where k is a constant.

Substituting $I=24\text{ lux}$ and $d=7.5\text{ m}$ ,

  $\begin{array}{l}I=k\frac{1}{d^2}[\text{Equation]}\\ 24=\frac{k}{{\left(7.5\right)}^2}[\text{Substitute]}\\ 24=\frac{k}{56.25}[\text{Find square]}\\ k=1350[\text{Multiply both sides by 56}\text{.25]}\end{array}$

Therefore, an equation of the given variation is $I=1350\frac{1}{d^2}$

To find I when $d=15\text{ m}$ , substitute in $I=1350\frac{1}{d^2}$ .

  $\begin{array}{l}I=1350\frac{1}{d^2}[\text{Equation]}\\ I=\frac{1350}{{\left(15\right)}^2}[\text{Substitute]}\\ I=\frac{1350}{225}[\text{Find square]}\\ I=6[\text{Simplify]}\end{array}$