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Chapter 14: Problem 51

Light intensity \(3.3 \mathrm{m}\) from a lightbulb is \(0.73 \mathrm{W} / \mathrm{m}^{2} .\) Find the bulb's power output, assuming it radiates equally in all directions.

Short Answer

Expert verified
The power output of the lightbulb is approximately 100.25 Watts.

Step by step solution

01

Identify given values

The light intensity \(I\) is given as 0.73 W/m虏, and the distance from the bulb (which is also the sphere's radius) \(r\) is given as 3.3 m.
02

Use the area of sphere formula

Calculate the area of the sphere which is \(A = 4\pi r^2\). Substituting the given radius \(r = 3.3m\), we find \(A \approx 137.35 m^{2}\).
03

Use the intensity of light formula

The intensity of light \(I\) is power \(P\) divided by area \(A\), or \(I=\frac{P}{A}\). We are looking for the power, so rearrange the formula to get \(P = I \cdot A\). Substituting the known values, we find \(P = 0.73 W/m^{2} \cdot 137.35 m^{2} = 100.25W\).

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