To compute the decibel intensity level (dB IL) for a sound wave, we use the following formula:

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

Where:

• $L$ is the sound intensity level in decibels (dB),
• $I$ is the sound intensity in watts per square meter (W/m²),
• $I_0$ is the reference intensity, which is typically $1.0 \times 10^{-12} \, \text{W/m²}$, the threshold of hearing.

Given:

$I = 5.0 \times 10^{-6} \, \text{W/m²}$

Step 1: Set up the formula

Substitute the given intensity $I = 5.0 \times 10^{-6} \, \text{W/m²}$ and the reference intensity $I_0 = 1.0 \times 10^{-12} \, \text{W/m²}$ into the formula:

$L = 10 \cdot \log_{10} \left( \frac{5.0 \times 10^{-6}}{1.0 \times 10^{-12}} \right)$

Step 2: Simplify the ratio

$\frac{5.0 \times 10^{-6}}{1.0 \times 10^{-12}} = 5.0 \times 10^{6}$

Step 3: Calculate the logarithm

$L = 10 \cdot \log_{10}(5.0 \times 10^{6})$

The logarithm of a product can be split into the sum of logarithms: $L = 10 \cdot \left( \log_{10}(5.0) + \log_{10}(10^{6}) \right)$

$L = 10 \cdot \left( \log_{10}(5.0) + 6 \right)$

From logarithm tables or using a calculator, $\log_{10}(5.0) \approx 0.6990$.

Step 4: Final calculation

$L = 10 \cdot (0.6990 + 6) = 10 \cdot 6.6990 = 66.99 \, \text{dB}$

Result:

The sound intensity level is approximately $67 \, \text{dB}$.

Would you like more details on any part of the process?

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Here are some related questions:

1. What is the significance of the reference intensity $I_0$?
2. How does the decibel scale relate to human hearing perception?
3. Can you explain the relationship between sound intensity and sound pressure level?
4. How does doubling the intensity affect the decibel level?
5. What are some common examples of sound intensities in decibels?

Tip: When working with logarithmic scales, small changes in intensity can result in large changes in decibel levels.