A) 5I and I
B) 5I and 3I
C) 9I and I
D) 9I and 3I
Correct Answer: C
Solution :
[c] : Intensity\[\propto \](Amplitude)\[^{2}\]\[\Rightarrow I\propto {{A}^{2}}\] When two waves (beams) of amplitude \[{{A}_{1}}\]and \[{{A}_{2}}\]superimpose, 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 \[{{I}_{\min }}\propto {{({{A}_{1}}+{{A}_{2}})}^{2}}\]and\[{{I}_{\min }}\propto {{({{A}_{1}}-{{A}_{2}})}^{2}}\] \[\Rightarrow \]\[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{1}}-{{A}_{2}}} \right)}^{2}}={{\left\{ \frac{\frac{{{A}_{1}}}{{{A}_{2}}}+1}{\frac{{{A}_{1}}}{{{A}_{2}}}-1} \right\}}^{2}}\] Given,\[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{\sqrt{I}}{\sqrt{4I}}=\frac{1}{2}\]\[\Rightarrow \]\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{9}{1}\] \[\therefore \]\[{{I}_{\max }}=9I\]and\[{{I}_{\min }}=I\]
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