A) 2 mm
B) 1 mm
C) 0.5 mm
D) 4 mm
Correct Answer: A
Solution :
| \[l=4{{l}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right);\]\[{{l}_{0}}=4{{l}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right)\] |
| \[\therefore \,\,\,\,\,\cos \left( \frac{\phi }{2} \right)=\frac{1}{2}\]or \[\frac{\phi }{2}=\frac{\pi }{2}\] |
| or \[\phi =\frac{2\pi }{3}=\left( \frac{2\pi }{\lambda } \right).\Delta x\] or \[\frac{1}{3}=\left( \frac{1}{\lambda } \right)y.\frac{d}{D}\] \[\left( \Delta x=\frac{yd}{D} \right)\] |
| \[\therefore \,\,\,\,\,y=\frac{\lambda }{3\times \frac{d}{D}}=\frac{6\times {{10}^{-7}}}{3\times {{10}^{-4}}}=2\times {{10}^{-3}}m=2mm\] |
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