A baby’s mouth is 30 cm from her father’s ear and 1.50 m from her mother’s ear. What is the difference between the sound intensity levels heard by the father and by the mother?

The sound intensity level is given by ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1=\left(10\mathrm{dB}\right)\log \left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\right)$Where, ${\mathrm{\beta }}_2-{\mathrm{\beta }}_1$is the difference in the sound intensity level, ${\mathrm{l}}_2$is the intensity of the sound heard by the father, ${\mathrm{l}}_1$is the intensity of the sound heard by the mother.

The ratio of the sound intensity heard by the father to the sound heard by the mother

with respect distance is, $\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}$Where, ${\mathrm{r}}_1$is the distance of the mother from the baby,${\mathrm{r}}_2$ is the distance of the father from the baby.

Substitute 30 cm for ${\mathrm{r}}_2$and 1.5 m for ${\mathrm{r}}_1$in the above equation to find$\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}$

$$\begin{gathered} \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}=\frac{\left(1.5\mathrm{m}\right)\left({\displaystyle \frac{100{\mathrm{cm}}^2}{1\mathrm{m}}}\right)}{\left(30{\mathrm{cm}}^2\right)} \\ =25 \end{gathered}$$

Thus, ratio of the sound intensity heard by the father to the sound heard by the mother with respect distance is 25

Substitute 25 for$\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}$ to find${\mathrm{\beta }}_2-{\mathrm{\beta }}_1$

$$\begin{gathered} {\mathrm{\beta }}_2-{\mathrm{\beta }}_1=\left(10\mathrm{dB}\right)\log \left(25\right) \\ =\left(10\mathrm{dB}\right)1.4 \\ =14\mathrm{dB} \end{gathered}$$

Therefore, the difference in the sound intensity heard by the father and by the mother is 14dB