A) O is \[3\,{{I}_{0}}\]
B) O is zero
C) a point 1 m below O is \[4\,{{I}_{0}}\]
D) a point on the screen 1 m below O is zero
Correct Answer: C
Solution :
[c] The path difference at O, \[\Delta x=d\,\sin \,\alpha =d\,\sin \,{{30}^{o}}=\frac{{{10}^{-3}}}{2}m\] Now \[\phi =\frac{2\pi }{\lambda }.\Delta x=\frac{2\pi }{5000\times {{10}^{-10}}}\times \frac{{{10}^{-3}}}{2}=2\pi \times {{10}^{3}}\] So \[I={{I}_{o}}+{{I}_{o}}+2\sqrt{{{I}_{o}}{{I}_{o}}}\cos (2\pi \times {{10}^{3}})=4{{I}_{o}}\] The angular position of P, \[\tan \theta =\frac{1}{\sqrt{3}}\]; or \[\theta ={{30}^{o}}\]. It means path \[dif{{f}^{n}}\] at P is also\[\Delta x\] and hence \[I=4{{I}_{o}}\]You need to login to perform this action.
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