A) \[(2n+1)\frac{\lambda }{2}\]
B) \[(2n+1)\frac{\lambda }{4}\]
C) \[(2n+1)\frac{\lambda }{8}\]
D) \[(2n+1)\frac{\lambda }{16}\]
Correct Answer: B
Solution :
[b] \[{{I}_{\max }}=4I\] \[2I=4I{{\cos }^{2}}\frac{\phi }{2}\] \[\cos \frac{\phi }{2}=\frac{1}{\sqrt{2}}\] \[\frac{\phi }{2}=(2n+1)\frac{\pi }{4}\] \[\frac{1}{2}.\frac{2\pi }{\lambda }.\Delta x=(2n+1)\frac{\pi }{4}\,\,\,\,\Rightarrow \,\,\,\,\Delta x=(2n+1)\frac{\lambda }{4}\]You need to login to perform this action.
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