Chapter 11, Problem 40CR

(a)

To determine

Find the intensity I of light 5 ft from the bulb.

Answer to Problem 40CR

Intensity of the bulb is 9 units when the distance is 5 ft.

Explanation of Solution

Given:

For a 225 − watt bulb, the intensity I of light in lumens at a distance of x feet is $\text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}$ .

What is the intensity of light 5 ft from the bulb?

Concept Used:

Intensity of the bulb given by $\text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}$

Now substitute x = 5 ft and calculate for I the intensity.

Calculation:

  $\begin{array}{l}\text{x}\text{=}\text{5ft;}\text{Find}\text{I}\\ \text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}=\frac{\text{225}}{{\text{5}}^2}=\frac{\text{225}}{25}=9\\ \text{I}\text{=}\text{9}\end{array}$

Thus, Intensity of the bulb is 9 units when the distance is 5 ft.

(b)

To determine

Find the intensity of bulb when the distance doubles.

Answer to Problem 40CR

Intensity of the bulb is 2.25 units when the distance is 10 ft.

Explanation of Solution

Given:

For a 225 − watt bulb, the intensity I of light in lumens at a distance of x feet is $\text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}$ .

Suppose your distance from the bulb doubles. How does the intensity of light change? Explain.

Concept Used:

Intensity of the bulb given by $\text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}$

Now distance is double, so substitute x = 10 ft and calculate for I the intensity.

Calculation:

  $\begin{array}{l}\text{x}\text{=}\text{10ft;}\text{Find}\text{I}\\ \text{I}\text{=}\frac{\text{225}}{{\text{x}}^2}=\frac{\text{225}}{{\text{10}}^2}=\frac{\text{225}}{100}=2.25\\ \text{I}\text{=}\text{2}\text{.25}\end{array}$

The intensity of the bulb will be one − fourth when distance doubles.

Thus, Intensity of the bulb is 2.25 units when the distance is 10 ft.