A) \[\frac{{{I}_{m}}}{3}\left( 1-2{{\cos }^{2}}\frac{\phi }{2} \right)\]
B) \[\frac{{{I}_{m}}}{5}\left( 1+4{{\cos }^{2}}\frac{\phi }{2} \right)\]
C) \[\frac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\frac{\phi }{2} \right)\]
D) \[\frac{{{I}_{m}}}{9}\left( 8+{{\cos }^{2}}\frac{\phi }{2} \right)\]
Correct Answer: C
Solution :
[c] : Here,\[A{{ & }_{2}}=2A{{ & }_{1}}\] \[\because \]Intensity\[\propto \](Amplitude)2 \[\therefore \]\[\frac{{{I}_{2}}}{{{I}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}={{\left( \frac{2{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}=4\]\[{{I}_{2}}=4{{I}_{1}}\] Maximum intensity, \[{{I}_{m}}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}\] \[={{\left( \sqrt{{{I}_{1}}}+\sqrt{4{{I}_{2}}} \right)}^{2}}={{\left( 3\sqrt{{{I}_{1}}} \right)}^{2}}=9{{I}_{1}}\]or\[{{I}_{1}}=\frac{{{I}_{m}}}{9}\]?(i) Resultant intensity,\[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[={{I}_{1}}+4{{I}_{2}}+2\sqrt{{{I}_{1}}(4{{I}_{1}})}\cos \phi \] \[=5{{I}_{1}}+4{{I}_{2}}\cos \phi ={{I}_{1}}+4{{I}_{1}}+4{{I}_{1}}\cos \phi \] \[={{I}_{1}}+4{{I}_{1}}(1+\cos \phi )\] \[={{I}_{1}}+8{{I}_{1}}{{\cos }^{2}}\frac{\phi }{2}\]\[\left( \because 1+\cos \phi =2co{{s}^{2}}\frac{\phi }{2} \right)\] \[={{I}_{1}}\left( 1+8co{{s}^{2}}\frac{\phi }{2} \right)\] Putting the value of \[{{I}_{1}}\]from eq. (i), we get \[I=\frac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\frac{\phi }{2} \right)\]
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