A) \[4{{I}_{0}}\]
B) \[{{I}_{0}}\]
C) \[2{{I}_{0}}\]
D) 0
Correct Answer: C
Solution :
Path difference \[=\Delta x={{s}_{2}}D-{{s}_{1}}D=5-4=1\,m\] \[\therefore \]The corresponding phase difference will be \[\phi =\frac{2\pi }{\lambda }.\Delta x=\left( \frac{2\pi }{4} \right)(1)=\frac{\pi }{2}\] Using \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[I={{I}_{0}}+{{I}_{0}}+2\sqrt{{{I}_{0}}{{I}_{0}}}\cos \frac{\pi }{2}=2{{I}_{0}}\] Hence, the correction option is (c).You need to login to perform this action.
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