A) \[5I\]and\[I\]
B) \[5I\]and\[3I\]
C) \[9I\]and\[I\]
D) \[9I\] and \[3I\]
Correct Answer: C
Solution :
Intensity \[\propto \](Amplitude)2 \[\Rightarrow \] \[I\propto {{A}^{2}}\] When two waves (beams) of amplitude A1 and A2 superimose, at maxima and minima, the amplitude of the resulting wave are \[({{A}_{1}}+{{A}_{2}})\] and \[({{A}_{1}}-{{A}_{2}})\], respectively. If the maximum and minimum possible intensities are \[{{I}_{\max }}\] and \[{{I}_{\min ,}}\] respectively. Then, \[{{I}_{\max }}\,\propto \,{{({{A}_{1}}+{{A}_{2}})}^{2}}\] and \[{{I}_{\min }}\propto \,{{({{A}_{1}}-{{A}_{2}})}^{2}}\] \[\Rightarrow \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\,{{\left( \frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{1}}-{{A}_{2}}} \right)}^{2}}=\,{{\left\{ \frac{\frac{{{A}_{1}}}{{{A}_{2}}}+1}{\frac{{{A}_{1}}}{{{A}_{2}}}-1} \right\}}^{2}}\] where, \[\frac{{{A}_{1}}}{{{A}_{2}}}=\,\sqrt{\frac{I}{4I}}=\frac{1}{2}\] \[\Rightarrow \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{9}{1}\] \[\Rightarrow \] \[{{I}_{\max }}=9I\] and \[{{I}_{\min }}=I\]
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