A) 49 : 1
B) 25 : 7
C) 10 : 9
D) 4 : 3
Correct Answer: A
Solution :
[a] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{16};\]\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{A}_{1}}^{2}}{{{A}_{2}}^{2}}\] [\[{{A}_{1}}\]& \[{{A}_{2}}\] are amplitudes of waves] \[\frac{{{A}_{1}}}{{{A}_{2}}}=\sqrt{\frac{9}{16}}=\frac{3}{4}\] \[\frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{2}}-{{A}_{1}}}=\frac{7}{1}\] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{2}}-{{A}_{1}})}^{2}}}={{\left( \frac{7}{1} \right)}^{2}}=\frac{49}{1}\] \[{{\operatorname{I}}_{\max }}:{{I}_{\min }}=49:1\]
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