A) \[I(\theta )={{I}_{0}}\]for \[\theta ={{0}^{o}}\]
B) \[I(\theta )={{I}_{0}}/2\]for \[\theta ={{30}^{o}}\]
C) \[I(\theta )={{I}_{0}}/4\]for \[\theta ={{90}^{o}}\]
D) \[I(\theta )\] is constant for all values of q
Correct Answer: B
Solution :
For microwave \[\lambda =\frac{c}{f}=\frac{3\times {{10}^{8}}}{{{10}^{6}}}=300\ m\] As \[\Delta x=d\sin \theta \] Phase difference \[\varphi =\frac{2\pi }{\lambda }\](Path difference) \[=\frac{2\pi }{\lambda }(d\sin \theta )=\frac{2\pi }{300}(150\sin \theta )=\pi \sin \theta \] \[{{I}_{R}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \varphi \] Here \[{{I}_{1}}={{I}_{2}}\]and \[\varphi =\pi \sin \theta \] \[\therefore {{I}_{R}}=2{{I}_{1}}[1+\cos (\pi \sin \theta )]=4{{I}_{1}}{{\cos }^{2}}\left( \frac{\pi \sin \theta }{2} \right)\] IR will be maximum when \[{{\cos }^{2}}\left( \frac{\pi \sin \theta }{2} \right)=1\] \[\therefore {{({{I}_{R}})}_{\max }}=4{{I}_{1}}={{I}_{o}}\] Hence \[I={{I}_{o}}{{\cos }^{2}}\left( \frac{\pi \sin \theta }{2} \right)\] If \[\theta =0,\]then \[I={{I}_{o}}\cos \theta ={{I}_{o}}\] If \[\theta =30{}^\circ ,\] then \[I={{I}_{o}}{{\cos }^{2}}(\pi /4)={{I}_{o}}/2\] If \[\theta ={{90}^{o}},\]then \[I={{I}_{o}}{{\cos }^{2}}(\pi /2)=0\]You need to login to perform this action.
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