A) 5 : 7
B) 7 : 4
C) 4 : 7
D) 7 : 5
Correct Answer: D
Solution :
Let \[{{a}_{1}},{{a}_{2}}\] be amplitudes for intensities \[{{I}_{1}},{{I}_{2}}\]respectively. Then \[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\] Given, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{36}{1}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\] \[\Rightarrow \] \[{{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}=6\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{7}{5}\]
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