A) 0.01 mm
B) 0.03 mm
C) 0.05 mm
D) 0.1 mm
Correct Answer: C
Solution :
Resultant intensity \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[I=\frac{3{{I}_{0}}}{4}={{I}_{1}}+{{I}_{2}}+2{{I}_{1}}\cos \phi \] or \[\frac{3}{4}\times 4{{I}_{1}}=2{{I}_{1}}(1+\cos \phi )\] or \[\cos \phi =\frac{1}{2}=\cos \frac{\pi }{3}\] \[\therefore \]\[\phi =\frac{\pi }{3}\]and path differenced\[\frac{\lambda }{6}=\frac{xd}{D}\] \[\therefore \] \[x=\frac{D\lambda }{6d}=\frac{1\times 6\times {{10}^{-7}}}{6\times 2\times {{10}^{-3}}}=0.05\,mm\]
You need to login to perform this action.
You will be redirected in
3 sec
