A) \[\frac{{{I}_{0}}}{2}\]
B) \[\frac{3}{4}{{I}_{0}}\]
C) I0
D) \[\frac{{{I}_{0}}}{4}\]
Correct Answer: A
Solution :
Suppose P is a point infront of one slit at which intensity is to be calculated from figure it is clear that\[x=\frac{d}{2}\]. Path difference between the waves reaching at P \[\Delta =\frac{xd}{D}=\frac{\left( \frac{d}{2} \right)d}{10d}\] \[=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}\] Hence corresponding phase difference \[\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}\] Resultant intensity at P \[I={{I}_{\text{max}}}{{\cos }^{2}}\frac{\varphi }{2}\]\[={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi }{4} \right)=\frac{{{I}_{0}}}{2}\]You need to login to perform this action.
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