A) \[4{{l}_{0}}{{\cos }^{2}}\text{o}|/2\]
B) \[\frac{4{{l}_{0}}}{3}{{\sin }^{2}}\text{o }\!\!|\!\!\text{ }\]
C) \[\frac{4{{l}_{0}}}{9}[5+4cos2o|]\]
D) \[\frac{4{{l}_{0}}}{9}[5+8cos2o|]\]
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 \[I=4I+16I+2\sqrt{64{{I}^{2}}}\cos (2o|)\] \[=20I+16I\cos 2\text{o }\!\!|\!\!\text{ }\] \[=\frac{20I}{9}+16\times \frac{{{I}_{0}}}{9}\cos 2o|\] \[=\frac{4{{I}_{0}}}{9}(5+4cos2o|)\]
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