To solve this problem, we need to determine the average power emitted by an isotropic source given that the light intensity at a distance of 13.7 meters from the source is [tex]\(2.50 \times 10^{-12} \, \text{W/m}^2\)[/tex].
First, let's understand the relationship between the intensity of light and the power of the source. The intensity ([tex]\(I\)[/tex]) is defined as the power per unit area and for an isotropic source, the power is distributed uniformly in all directions over the surface of a sphere.
The formula for the intensity of light at a distance [tex]\(r\)[/tex] from the source is:
[tex]\[ I = \frac{P}{A} \][/tex]
where
- [tex]\(I\)[/tex] is the intensity of the light,
- [tex]\(P\)[/tex] is the total power emitted by the source,
- [tex]\(A\)[/tex] is the surface area of a sphere with radius [tex]\(r\)[/tex].
The surface area of a sphere is given by:
[tex]\[ A = 4\pi r^2 \][/tex]
Substituting this into the intensity formula, we get:
[tex]\[ I = \frac{P}{4\pi r^2} \][/tex]
To find the power ([tex]\(P\)[/tex]), we need to rearrange this formula:
[tex]\[ P = I \cdot 4\pi r^2 \][/tex]
Given that:
- [tex]\(I = 2.50 \times 10^{-12} \, \text{W/m}^2\)[/tex],
- [tex]\(r = 13.7 \, \text{m}\)[/tex],
we can calculate [tex]\(P\)[/tex] as follows:
[tex]\[ P = 2.50 \times 10^{-12} \times 4\pi \times (13.7)^2 \][/tex]
Plugging in the values:
[tex]\[ r^2 = (13.7)^2 = 187.69 \, \text{m}^2 \][/tex]
Then:
[tex]\[ P = 2.50 \times 10^{-12} \times 4\pi \times 187.69 \][/tex]
[tex]\[ P \approx 2.50 \times 10^{-12} \times 4 \times 3.14159 \times 187.69 \][/tex]
[tex]\[ P \approx 2.50 \times 10^{-12} \times 2357.52 \][/tex]
[tex]\[ P \approx 5.89 \times 10^{-9} \, \text{W} \][/tex]
Therefore, the average power emitted by the source is approximately
[tex]\[ 5.896455251522681 \times 10^{-9} \, \text{W} \][/tex]
So, the average power emitted by the isotropic light source is [tex]\( 5.896 \times 10^{-9} \)[/tex] watts.
