Question

2 formula 1 point how much fainter is your light on your phone if its 15 m away if at 1 m away it measures 1 w/m²? whole numbers only answer

Explanation:

Step1: Recall the inverse square law for light intensity.

The intensity \( I \) of light (or any spherical wave) follows the inverse square law, which is \( I \propto \frac{1}{r^2} \), where \( r \) is the distance from the source. So the ratio of intensities at two distances \( r_1 \) and \( r_2 \) is \( \frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} \). Here, \( I_1 = 1\space W/m^2 \) at \( r_1 = 1\space m \), and we want to find \( I_2 \) at \( r_2 = 15\space m \), and then find how much fainter \( I_2 \) is compared to \( I_1 \) (i.e., \( \frac{I_1}{I_2} \)).

Step2: Calculate the ratio of intensities.

First, find \( \frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} \). Substitute \( r_1 = 1 \) and \( r_2 = 15 \):
\( \frac{I_2}{I_1}=\frac{1^2}{15^2}=\frac{1}{225} \)
Then, the factor by which \( I_2 \) is fainter than \( I_1 \) is \( \frac{I_1}{I_2}=\frac{1}{\frac{1}{225}} = 225 \).

Answer:

225