A) \[\frac{{{I}_{m}}}{9}(4+5cos\phi )\]
B) \[\frac{{{I}_{m}}}{3}\left( 1+2co{{s}^{2}}\frac{\phi }{2} \right)\]
C) \[\frac{{{I}_{m}}}{5}\left( 1+4co{{s}^{2}}\frac{\phi }{2} \right)\]
D) \[\frac{{{I}_{m}}}{9}\left( 1+8co{{s}^{2}}\frac{\phi }{2} \right)\]
Correct Answer: D
Solution :
[d] It is given, \[{{A}_{2}}=2{{A}_{1}}\] We know, intensity \[\propto {{\left( Amplitude \right)}^{2}}\] Hence\[\frac{{{I}_{2}}}{{{I}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}={{\left( \frac{2{{A}_{1}}}{{{A}_{1}}} \right)}^{2}}=4\] \[\Rightarrow {{I}_{2}}=4{{I}_{1}}\] Maximum intensity, \[{{I}_{m}}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}\] \[={{\left( \sqrt{{{I}_{1}}}+\sqrt{4{{I}_{1}}} \right)}^{2}}={{\left( 3\sqrt{{{I}_{1}}} \right)}^{2}}=9{{I}_{1}}\] Hence \[{{I}_{1}}=\frac{{{I}_{m}}}{9}\] Resultant intensity, \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[={{I}_{1}}+4{{I}_{1}}+2\sqrt{{{I}_{1}}(4{{I}_{1}})}\cos \phi \] \[=5{{I}_{1}}+4{{I}_{1}}\cos \phi ={{I}_{1}}+4{{I}_{1}}+4{{I}_{1}}\cos \phi \] \[={{I}_{1}}+4{{I}_{1}}(1+cos\phi )\] \[={{I}_{1}}+8{{I}_{1}}{{\cos }^{2}}\phi \]\[\left( \therefore 1+\cos \phi =2{{\cos }^{2}}\frac{\phi }{2} \right)\] \[I=\frac{{{I}_{m}}}{9}\left( 1+8cs{{o}^{2}}\frac{\phi }{2} \right)\] Putting the value of \[{{I}_{1}}\]from eqn. (i), we get \[I=\frac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\frac{\phi }{2} \right)\]
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