A) 0.1mm
B) 0.2mm
C) 0.3mm
D) 0.4mm
Correct Answer: B
Solution :
given, \[\beta =1.22\operatorname{mm}\] |
For \[{{\operatorname{I}}_{R}}=\frac{3}{4}{{I}_{0}}\] |
\[\therefore \] \[\frac{3}{4}{{I}_{0}}=2I+21\cos \phi \] |
Or \[\frac{3}{4}(4I)=4I{{\cos }^{2}}\frac{\phi }{2}\] Or \[\cos \frac{\phi }{2}=\frac{\sqrt{3}}{2}\] Or \[\phi ={{60}^{0}}\] |
Now, \[\Delta x=\frac{\phi \times \lambda }{2\pi }=\frac{\left( \pi /3 \right)\times \lambda }{2\pi }=\frac{\lambda }{6}\] |
or \[d\sin \theta =\frac{\lambda }{6}\] or \[d\times \frac{y}{D}=\frac{\lambda }{6}\] or \[y=\left( \frac{D\lambda }{d} \right)\times \frac{1}{6}\]\[=\frac{1.2}{6}=0.2\operatorname{mm}.\] |
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