A sound absorber attenuates the sound level by $20\ dB$. The intensity decreases by a factor of
AIEEE 2007,AIIMS 2012, Medium
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We have, $\mathrm{L}_{1}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right) ; \mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
$\therefore \mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
or, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}} \times \frac{\mathrm{I}_{0}}{\mathrm{I}_{2}}\right)$ or, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$
or, $20=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ or, $2=\log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$
or, $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=10^{2}$ or, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{100}$
$\Rightarrow$ Intensity decreases by a factor $100 .$
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