A) \[4{{I}_{0}}{{\cos }^{2}}\phi /2\]
B) \[\frac{4{{I}_{0}}}{3}{{\sin }^{2}}\phi \]
C) \[\frac{4{{I}_{0}}}{9}[5+4\cos 2\phi ]\]
D) \[\frac{4{{I}_{0}}}{9}[5+8\cos 2\phi ]\]
Correct Answer: C
Solution :
As amplitudes are \[2A\] and\[4A\], so intensities would be in the ratio\[1:4\], let us say \[4I\] and\[16I\]. \[{{I}_{\max }}=4I+16I+2\sqrt{4I\times 16I}\] \[=20I+2\times 2\times 4I=36I\] According to the question, \[{{I}_{\max }}=4{{I}_{0}}\Rightarrow 4{{I}_{0}}=36I\]or,\[I=\frac{{{I}_{0}}}{9}\] Thus, intensity of general point \[I'=4I+16+2\sqrt{64{{I}^{2}}}\cos (2\phi )\] \[=20I+16I\cos 2\phi \] \[=\frac{20{{I}_{0}}}{9}+16\times \frac{{{I}_{0}}}{9}\cos 1\phi \] \[=\frac{4{{I}_{0}}}{9}(5+4\cos 2\phi )\]
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