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inicialmente, estamos a 20 metros y medimos $i1= 7.0 \\times 10^{-4}$ w…

Question

inicialmente, estamos a 20 metros y medimos $i1= 7.0 \times 10^{-4}$ watt/ $m^2$,
si ahora nos alejamos a 100 metros,
¿cuál es la intensidad en el nuevo lugar (i2)?

Explanation:

Step1: Recall inverse square law

The intensity of a spherical wave follows $I \propto \frac{1}{r^2}$, so $\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$

Step2: Plug in given values

Substitute $I_1=7.0 \times 10^{-4}\ \text{Watt/m}^2$, $r_1=20\ \text{m}$, $r_2=100\ \text{m}$
$$I_2 = I_1 \times \frac{r_1^2}{r_2^2} = 7.0 \times 10^{-4} \times \frac{(20)^2}{(100)^2}$$

Step3: Calculate the ratio

$$\frac{(20)^2}{(100)^2} = \frac{400}{10000} = 0.04$$

Step4: Compute final intensity

$$I_2 = 7.0 \times 10^{-4} \times 0.04$$

Answer:

$2.8 \times 10^{-5}\ \text{Watt/m}^2$