# Question bae89

Nov 21, 2017

IL_1 = IL_2 - (10 dB)\log_{10}(I_2/I_1); \qquad IL_2 = 85 dB; \qquad I_2 = 2I_1; 
$I {L}_{1} = 85$ $\mathrm{dB} - 3.01$ $\mathrm{dB} = 81.99$ $\mathrm{dB}$

#### Explanation:

The Intensity Level of a sound of intensity $I$, in $\mathrm{dB}$, is defined as follows:
$I L \setminus \equiv \left(10 \mathrm{dB}\right) \setminus {\log}_{10} \left(\frac{I}{I} _ 0\right)$ ...... (1)
where ${I}_{0}$ is the intensity at the threshold of hearing.

Comparing the intensity levels of two different sounds of intensities ${I}_{1}$ and ${I}_{2}$,

$I {L}_{2} - I {L}_{1} = \left(10 \mathrm{dB}\right) \left[\setminus {\log}_{10} \left({I}_{2} / {I}_{0}\right) - \setminus {\log}_{10} \left({I}_{1} / {I}_{0}\right)\right]$

But, $\setminus q \quad \setminus {\log}_{10} \left(a\right) - \setminus {\log}_{10} \left(b\right) = \setminus {\log}_{10} \left(\frac{a}{b}\right)$

$I {L}_{2} - I {L}_{1} = \left(10 \mathrm{dB}\right) \setminus {\log}_{10} \left({I}_{2} / {I}_{1}\right)$

$I {L}_{1} = I {L}_{2} - \left(10 \mathrm{dB}\right) \setminus {\log}_{10} \left({I}_{2} / {I}_{1}\right)$

The intensity of sound wave produced by two identical fire crackers is twice the intensity of one : I_2 = 2I_1; \qquad I_2/I_1 = 2#

$I {L}_{2} - I {L}_{1} = \left(10 \mathrm{dB}\right) \setminus {\log}_{10} \left(2\right) = 3.01$ $\mathrm{dB}$

$I {L}_{1} = I {L}_{2} - 3.01$ $\mathrm{dB}$ = $85$ $\mathrm{dB} - 3.01$ $\mathrm{dB}$ = $81.99$ $\mathrm{dB}$