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Q. The ratio of intensities between two coherent sound sources is $4 : 1$. The difference of loudness in decibels between maximum and minimum intensities, when they interfere in space, is

AIIMSAIIMS 2009

Solution:

Given $ \frac{I_{1}}{I_{2}}=\frac{4}{1}$
We know
$I \propto a^{2} $
$\therefore \frac{a_{1}^{2}}{a_{2}^{2}} =\frac{I_{1}}{I_{2}}=\frac{4}{1}$
$\frac{a_{1}}{a_{2}}=\frac{2}{1}$
$\therefore \frac{I_{\max }}{I_{\min }} =\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}} $
$=\left(\frac{2+1}{2-1}\right)^{2}$
$=\left(\frac{3}{1}\right)^{2}=\frac{9}{1}$
Therefore, difference of loudness is given by
$L_{1}-L_{2} =10 \log \frac{I_{\max }}{I_{\min }}=10 \log (9) $
$=10 \log 3^{2}=20 \log 3$