Intensity level 200 cm from a source of sound is 80 dB. If there is no loss of acoustic power in air and intensity of threshold hearing is $10^{-12} \mathrm{Wm}^{-2}$ then, what is the intensity level at a distance of 400 cm from source
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$I \propto \frac{1}{r^2} \Rightarrow \frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} = \frac{2^2}{(40)^2} = \frac{1}{400} \Rightarrow I_1 = 400 I_2$ Intensity level at point 1, $L_1 = 10 \log_{10} \left(\frac{I_1}{I_0}\right)$ and intensity at point 2, $L_2 = 10 \log_{10} \left(\frac{I_2}{I_0}\right)$ $\therefore L_1 - L_2 = 10 \log_{10} \frac{I_1}{I_2} = 10 \log_{10} (400)$ $\Rightarrow L_1 - L_2 = 10 \times 2.602 = 26$ $L_2 = L_1 - 26 = 80 - 26 = 54 \text{ dB}$
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