Question

An increase in intensity level of $1dB$ implies an increase in the intensity by $($given $antilo{g}_{10}0.1=1.2589)$

• A. $1\%$
• B. $3.01\%$
• C. $26\%$
• D. $0.1\%$

Answer 1

The correct option is C $26\%$

Given, $I{L}_2-I{L}_1=1dB$

$1=10{\log }_{10}\left[\frac{I_2}{Io}\right]-10{\log }_{10}\left[\frac{I_1}{Io}\right]$

where $I_1$ and $I_2$ are the initial and final intensity of sound

respectively such that the increase in

intensity level is $1dB$.

$\therefore 1=10[{\log }_{10}\left[\frac{I_2}{I_o}\right]-10{\log }_{10}\left[\frac{I_1}{I_o}\right]]$

$\therefore 1=10\left[{\log }_{10}\left[\frac{\frac{I_2}{I_o}}{\frac{I_1}{I_o}}\right]\right]$

$\therefore 1=10\left[{\log }_{10}[\frac{I_2}{I_o}\ast \frac{I_o}{I_1}]\right]$

$\therefore 1=10{\log }_{10}\left[\frac{I_2}{I_1}\right]$

$\therefore 0.1=10{\log }_{10}\left[\frac{I_2}{I_1}\right]$

$antilog0.1=\frac{I_2}{I_1}$

$I_2=\left(antilog0.1\right)\times I_1$

$I_2=1.26\times I_1$

$\therefore$ Increase in intensity that produces an increase in intensity level by $1dB$

$=I_2-I_1=1.26I_1-I_1=(1.26-1)I_1=0.26I_1$ or $26\%$ of $I_1$

Hence, when the intensity of sound increases by $26\%,$ the intensity level increases by $1dB$ .