A) \[4:1\]
B) \[1:25\]
C) \[1:3\]
D) \[49:1\]
Correct Answer: D
Solution :
Key Idea: Intense of wave is proportional to square of amplitude of the wave. \[\text{Intensity}\,\,\,\propto \,\,\,{{\text{(Amplitude)}}^{\text{2}}}\] or \[I\propto {{a}^{2}}\] or \[\frac{{{I}_{2}}}{{{I}_{2}}}={{\left( \frac{{{a}_{1}}}{{{a}_{2}}} \right)}^{2}}\] Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{16}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}=\sqrt{\frac{9}{16}}=\frac{3}{4}\] Maximum intensity, \[{{I}_{\max }}={{({{a}_{1}}+{{a}_{2}})}^{2}}\] \[={{(3a+4a)}^{2}}\] \[={{(7a)}^{2}}\] \[{{I}_{\min }}={{({{a}_{1}}-{{a}_{2}})}^{2}}\] \[={{(3a-4a)}^{2}}\] \[={{(-a)}^{2}}\] Hence, the ratio, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(7a)}^{2}}}{{{(-a)}^{2}}}\] \[=\frac{49}{1}\]
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