If you install special sound-reflecting windows that reduce the sound intensity level by 33.0 dBdB , by what factor have you reduced the sound intensity

To determine by what factor the sound intensity is reduced when the sound intensity level decreases by 33.0 dB, we can use the formula that relates decibels to intensity:

\[ \text{Decibel Level (dB)} = 10 \times \log\left(\frac{I}{I_0}\right) \]

In this formula: - \( I \) is the intensity of the sound. - \( I_0 \) is the reference intensity, usually the threshold of hearing at \( 1 \times 10^{-12} \text{W/m}^2 \).

To find out by what factor the intensity is reduced, we can set up the equation as follows with \( I' \) being the new intensity:

\[ -33.0 = 10 \times \log\left(\frac{I'}{I}\right) \]

Dividing both sides by 10 gives us:

\[ -3.3 = \log\left(\frac{I'}{I}\right) \]

To find \( \frac{I'}{I} \), we need to use the inverse of the logarithm, which is the antilog or 10 raised to the power of the value:

\[ 10^{-3.3} = \frac{I'}{I} \]

\[ \frac{I'}{I} = 10^{-3.3} \approx 0.0005012 \]

The factor by which the sound intensity is reduced is therefore approximately 0.0005012. This means that the sound intensity is reduced to roughly 0.05012% of its original intensity.