A) 10 log 2
B) 20 log 3
C) 10 log 3
D) 20 log 2
Correct Answer: B
Solution :
Given \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{4}{1}\] We know \[I\propto {{a}^{2}}\] \[\therefore \] \[\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{4}{1}\] or \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{1}\] \[\therefore \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[={{\left( \frac{2+1}{2-1} \right)}^{2}}\] \[={{\left( \frac{3}{1} \right)}^{2}}=\frac{9}{1}\] Therefore, difference of loudness is given by \[{{L}_{1}}-{{L}_{2}}=10\log \frac{{{I}_{\max }}}{{{I}_{\min }}}=10\log (9)\] \[=10\log {{3}^{2}}=20\log 3\]
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