How many times more intense is 90, dB sound t | vedclass

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(D) Let $I$ represent the intensity of sound.The loudness of sound is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.For two di

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$10^5$

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(D) Let $I$ represent the intensity of sound.The loudness of sound is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.For two different sound levels $L_1$ and $L_2$ with intensities $I_1$ and $I_2$,the difference in loudness is:$L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.Given $L_2 = 90\, dB$ and $L_1 = 40\, dB$:$90 - 40 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.$50 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.$5 = \log_{10} \left( \frac{I_2}{I_1} \right)$.Therefore,$\frac{I_2}{I_1} = 10^5$.

How many times more intense is $90\, dB$ sound than $40\, dB$ sound?

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