The decibel (dB) scale is logarithmic, and the relationship between sound intensity levels in dB and the ratio of intensities is given by the following formula:

$\Delta L = 10 \cdot \log_{10} \left( \frac{I_2}{I_1} \right)$

where:

• $\Delta L$ is the change in sound level (in decibels, dB),
• $I_1$ is the original intensity,
• $I_2$ is the new intensity.

If the sound intensity decreases to $\frac{1}{10}$ of the original level, we plug this into the formula:

$\Delta L = 10 \cdot \log_{10} \left( \frac{1}{10} \right)$

The logarithm of $\frac{1}{10}$ is:

$\log_{10} \left( \frac{1}{10} \right) = -1$

So:

$\Delta L = 10 \cdot (-1) = -10 \text{ dB}$

Thus, if the sound intensity decreases by a factor of $\frac{1}{10}$, the sound level decreases by 10 dB.

Would you like more details or have any questions?

Here are 5 related questions:

1. How is sound intensity level measured in decibels?
2. What is the formula to calculate sound level changes for different intensity ratios?
3. What does a negative change in dB indicate about the sound intensity?
4. How would the dB change if the sound intensity decreases to $\frac{1}{100}$?
5. How does the logarithmic nature of the dB scale affect perception of sound?

Tip: When comparing sound intensities, small changes in decibels can represent large differences in actual sound intensity.