Question

A sound absorber attenuates the sound level by $20dB$. The intensity decreases by a factor of

• A. $100$
• B. $200$
• C. $10000$
• D. $10$

Answer 1

The correct option is A $100$

We have, $L_1=10log\left(\frac{I_1}{I_0}\right)$
$L_2=10log\left(\frac{I_2}{I_0}\right)$
$\therefore L_1-L_2=10log\left(\frac{I_1}{I_0}\right)-10log\left(\frac{I_2}{I_0}\right)$
or, $\Delta L=10log(\frac{I_1}{I_0}\times \frac{I_0}{I_2})$
or, $\Delta L=10log\left(\frac{I_1}{I_2}\right)$
or, $20=10log\left(\frac{I_1}{I_2}\right)$
or, $2=log\left(\frac{I_1}{I_2}\right)$
or, $\frac{I_1}{I_2}={10}^2$
or, $I_2=\frac{I_1}{100}$
$\Rightarrow$ Intensity decreases by a factor $100$