To find the intensity of the sound $I$, we can use the following formula that relates sound intensity to loudness in decibels (dB):

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

Where:

• $L$ is the loudness in decibels (dB).
• $I$ is the intensity of the sound.
• $I_0$ is the reference intensity, typically $I_0 = 1 \times 10^{-12} \, \text{W/m}^2$ (the threshold of hearing).

We are given: $L = 74.3 \, \text{dB}$

To solve for the intensity $I$, we rearrange the formula:

$I = I_0 \times 10^{\frac{L}{10}}$

Now we can plug in the values:

$I = 1 \times 10^{-12} \times 10^{\frac{74.3}{10}}$

Let me calculate that for you.The intensity of the sound is approximately $2.69 \times 10^{-5} \, \text{W/m}^2$.

Would you like further details or have any questions about this calculation?

Here are 5 related questions for further exploration:

1. How does the intensity of sound change with a 10 dB increase in loudness?
2. What is the intensity level of a sound that measures 60 dB?
3. How does distance affect the intensity of sound as perceived by a listener?
4. What is the loudness in decibels of a sound with intensity $1 \times 10^{-6} \, \text{W/m}^2$?
5. How does the logarithmic scale affect the perception of sound loudness?

Tip: A small increase in dB (like 10 dB) represents a significant change in intensity, often perceived as twice as loud!