A) \[\frac{\beta }{2}\]
B) \[\frac{\beta }{4}\]
C) \[\frac{\beta }{3}\]
D) \[\frac{\beta }{6}\]
Correct Answer: B
Solution :
\[I=4{{I}_{0}}{{\cos }^{2}}\left( {}^{\phi }/{}_{2} \right)\] \[{{\operatorname{I}}_{max}}=4{{I}_{0}}\] Now\[\frac{{{\operatorname{I}}_{max}}}{2}=2{{I}_{0}}=4{{I}_{0}}{{\cos }^{2}}\left( {}^{\phi }/{}_{2} \right)\] \[\cos \left( {}^{\phi }/{}_{2} \right)=\frac{1}{\sqrt{2}}\] \[\therefore \]\[\frac{\phi }{2}=\frac{\pi }{2}\] \[\therefore \]\[\phi =\frac{\pi }{2}\] \[\frac{2\pi }{\lambda }\Delta x=\frac{\pi }{2}\] \[\therefore \]\[\Delta x=\frac{\pi }{4}\] \[y\frac{d}{D}=\frac{\lambda }{4}\] \[\therefore \]\[y=\frac{\lambda D}{4d}\] \[\therefore \]\[\]
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