The problem is about calculating the decibel level $D$ of a sound given its intensity $I$ using the formula:

$D = 10 \log \left( \frac{I}{10^{-12}} \right)$

Given Data:

• Intensity $I = 10^{-8} \, \text{W/m}^2$

Solution:

1. Substitute $I = 10^{-8}$ into the formula: $D = 10 \log \left( \frac{10^{-8}}{10^{-12}} \right)$

2. Simplify the fraction inside the logarithm: $\frac{10^{-8}}{10^{-12}} = 10^{4}$

3. Substitute $10^4$ into the logarithm: $D = 10 \log(10^4)$

4. Use the property of logarithms $\log(10^4) = 4$: $D = 10 \times 4 = 40$

Answer:

The decibel level $D$ is 40 dB.

Would you like further details on the logarithmic properties used, or do you have any questions?

Here are some related questions to deepen your understanding:

1. How does changing the intensity $I$ affect the decibel level $D$?
2. What would be the decibel level for an intensity of $10^{-9} \, \text{W/m}^2$?
3. Can the decibel scale go below zero? If so, under what conditions?
4. How does the decibel formula relate to human perception of sound?
5. What other fields use logarithmic scales similar to the decibel scale?

Tip: Remember that decibels are a logarithmic measure, which means every 10 dB increase represents a tenfold increase in intensity!