Question

A pilot is preparing his private jet for a short trip. During take-off, the sound intensity level of a jet engine is 140 dB at a distance of 30 m. What is the sound intensity level at a distance of 1.0 km?

Answer 1

The sound intensity of point source drops proportionally to distance squared:

$$\dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2}$$

where $r_1 = 30m$, $r_2 = 1km = 1000m$.

In order to express intensity in dB, one should take the log of both sides of this expression and multiply it by 10:

$$10\lg I_1 - 10\lg I_2 = 10\lg\left( \dfrac{r_2^2}{r_1^2} \right)$$

Now, taking into account that $10\lg I_1 = 140dB$ and expressing $10\lg I_2$, obtain:

$$10\lg I_2 = 10\lg I_1 - 10\lg\left( \dfrac{r_2^2}{r_1^2} \right)\\ 10\lg I_2 = 140 - 10\lg\left( \dfrac{1000^2}{30^2} \right) \approx 110dB$$

Answer. 110 dB.