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A sound absorber attenuates the sound level by $20\, dB$. The intensity decreases by a factor of
$100$
$1000$
$10000$
$10$
Solution
We have, $L_{1}=10 \log \left(\frac{I_{1}}{I_{0}}\right)$
$\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
$\mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
or, $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}} \times \frac{\mathrm{I}_{0}}{\mathrm{I}_{2}}\right)$
or, $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$
or, $\quad 20=10 \log \left(\frac{I_{1}}{I_{2}}\right)$
or, $\quad 2=\log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$
or, $\quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=10^{2}$
or, $\quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{100}$
$\Rightarrow$ Intensity decreases by a factor $100 .$
