Chapter 8.2, Problem 6P

To determine

To calculate: Intensity of a light meter at the edge of a table.

Answer to Problem 6P

Intensity of light at the edge of the table is 16 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.

A light hangs $\text{4}\text{.8 ft}$ above the center of a circular table, $\text{7}\text{.2 ft}$ in diameter. Intensity of light at center of the table is $\text{25 lux}$ .

Formula used:

The following Pythagorean formula:

  $\text{hypotenuse = }\sqrt{{\text{adjacent}}^2{\text{ + opposite}}^2}$

Calculation:

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

  $I=k\frac{1}{l^2}$

where k is a constant.

Substituting $I=25\text{ lux}$ and $l=4.8\text{ ft}$ ,

  $\begin{array}{l}I=k\frac{1}{l^2}[\text{Equation]}\\ 25=\frac{k}{{\left(4.8\right)}^2}[\text{Substitute]}\\ 25=\frac{k}{23.04}[\text{Find square]}\\ k=576[\text{Multiply both sides by 23}\text{.04]}\end{array}$

Therefore, an equation of the given variation is $I=576\frac{1}{l^2}$

Since diameter is $\text{7}\text{.2 ft}$ , $\text{radius}=\frac{\text{diameter}}{2}=\frac{7.2\text{ ft}}{2}=3.6\text{ft}$ .

Therefore, $r=3.6\text{ ft}$ .

Now, to find the distance between the source of light and edge of the table, use the Pythagoras theorem,

  $\begin{array}{l}\sqrt{r^2+l^2}[\text{Pythagoras theorem]}\\ \text{=}\sqrt{{\left(3.6\right)}^2+{\left(4.8\right)}^2}[\text{Substitute]}\\ =\sqrt{12.96+23.04}[\text{Find square]}\\ =\sqrt{36}[\text{Simplify]}\\ =6[\text{Find square root]}\end{array}$

To find I when $l=6\text{ ft}$ , substitute in $I=576\frac{1}{l^2}$ .

  $\begin{array}{l}I=576\frac{1}{l^2}[\text{Equation]}\\ I=\frac{576}{{\left(6\right)}^2}[\text{Substitute]}\\ I=\frac{576}{36}[\text{Find square]}\\ I=16[\text{Simplify]}\end{array}$