To calculate the decibel level $N$ of a sound given its intensity $I$, we use the formula:

$N = 10 \log_{10} \left( \frac{I}{I_0} \right)$

where:

• $I$ is the intensity of the sound (given as $3.5 \times 10^{-11} \, \text{watts/cm}^2$),
• $I_0$ is the reference intensity level, usually taken as $10^{-16} \, \text{watts/cm}^2$ for air.

Step-by-Step Calculation

1. Insert the values: $N = 10 \log_{10} \left( \frac{3.5 \times 10^{-11}}{10^{-16}} \right)$

2. Simplify inside the logarithm: $\frac{3.5 \times 10^{-11}}{10^{-16}} = 3.5 \times 10^{5}$

3. Calculate the logarithm: $N = 10 \log_{10}(3.5 \times 10^{5})$

4. Split the logarithm: $N = 10 \left( \log_{10}(3.5) + \log_{10}(10^{5}) \right)$

5. Solve: $\log_{10}(3.5) \approx 0.544$ $\log_{10}(10^{5}) = 5$ $N = 10 \times (0.544 + 5) = 10 \times 5.544 = 55.4 \, \text{dB}$

Answer

The decibel level $N$ is approximately 55.4 dB.

Would you like more details, or do you have questions?

1. What is the formula for calculating decibel levels from intensity?
2. How does changing the reference intensity $I_0$ affect the decibel calculation?
3. What are common reference intensities in different media?
4. How does intensity in watts/cm² relate to watts/m²?
5. What does a change in decibel level represent perceptually in sound?

Tip: Remember that every 10 dB increase represents a tenfold increase in intensity.