A) \[49:1\]
B) \[225:81\]
C) \[3:1\]
D) \[9:1\]
Correct Answer: A
Solution :
Key Idea: \[Intensity\propto {{(amplitude)}^{2}}.\] We know that \[I=k{{a}^{2}}\] where \[a\] is amplitude and \[I\] the intensity \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{16}{9}=\frac{a_{1}^{2}}{a_{2}^{2}}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{4}{3}\] \[\therefore \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(4+3)}^{2}}}{{{(4-3)}^{2}}}=\frac{{{7}^{2}}}{1}=\frac{49}{1}\]
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