To solve for the intensity $I$ of the sound given the decibel level $D = 90 \, \text{dB}$, we can use the given formula for the decibel level:

$D = 10 \log \left( \frac{I}{I_0} \right)$

where:

• $D = 90 \, \text{dB}$ (decibel level),
• $I_0 = 10^{-12} \, \text{W/m}^2$ (reference intensity).

Step 1: Plug in the given values.

Substitute $D = 90$ and $I_0 = 10^{-12}$ into the equation:

$90 = 10 \log \left( \frac{I}{10^{-12}} \right)$

Step 2: Isolate the logarithmic term.

Divide both sides of the equation by 10 to simplify:

$9 = \log \left( \frac{I}{10^{-12}} \right)$

Step 3: Solve for the ratio $\frac{I}{10^{-12}}$.

Rewrite the equation in exponential form (since $\log$ is the inverse of exponentiation with base 10):

$10^9 = \frac{I}{10^{-12}}$

Step 4: Solve for $I$.

Multiply both sides by $10^{-12}$ to solve for $I$:

$I = 10^9 \times 10^{-12} = 10^{-3} \, \text{W/m}^2$

Final Answer:

The intensity of the sound is $I = 10^{-3} \, \text{W/m}^2$.

Would you like any further details or clarifications?

---

Here are 5 related questions to further explore:

1. How does the intensity of sound change with a 10 dB increase?
2. What would be the intensity of a sound at 100 dB?
3. How would you convert between dB levels and intensity for sounds in everyday life?
4. What role does the reference intensity $I_0 = 10^{-12}$ play in this formula?
5. How does this formula apply to different sources of noise, such as a conversation or machinery?

Tip: Decibels use a logarithmic scale, meaning each 10 dB increase corresponds to a tenfold increase in sound intensity.