A) 4 : 1
B) 1 : 25
C) 1 : 3
D) 49 : 1
Correct Answer: D
Solution :
Key Idea: Intensity of wave is proportional to square of amplitude of the wave. \[\text{Intensity}\,\propto {{\text{(Amplitude)}}^{\text{2}}}\] or \[I\propto {{a}^{2}}\] or \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{a}_{1}}}{{{a}_{2}}} \right)}^{2}}\] 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, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(7a)}^{2}}}{{{(-a)}^{2}}}\] \[=\frac{49}{1}\]You need to login to perform this action.
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