A baby's mouth is 30 cm from her father's ear and 3.0 m from mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother. (answer in dB)

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Solution

Here $r_{1} = 30 c m, r_{2} = 3 m$
Now, $B_{1} = 10 l o g (\frac{I_{1}}{I_{0}})$ and $B_{2} = 10 l o g (\frac{I_{2}}{I_{0}})$
$B_{1} - B_{2} = 10 l o g (I_{1} / I_{0}) - 10 l o g (I_{2} / I_{0})$
$= 10 l o g (I_{2} / I_{1})$
But $\frac{I_{1}}{I_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$
$B_{1} - B_{2} = 10 l o g (\frac{r_{2}^{2}}{r_{1}^{2}})$
$= 20 l o g (\frac{r_{2}}{r_{1}})$
$= 20 l o g (3 / 0.3)$
$= 20 l o g 10$
$= 20 d B$

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