A) \[0.2\,\,mm\]
B) \[0.4\,\,mm\]
C) \[0.5\,\,mm\]
D) \[0.6\,\,mm\]
Correct Answer: C
Solution :
Key Idea: Refractive index\[\mu =\frac{{{\lambda }_{a}}}{{{\lambda }_{w}}}\] The fringe width \[(W)\] is given by \[W=\frac{D\lambda }{d}\] where, \[\lambda \] is wavelength, \[D\] is distance between source and screen and \[d\] is distance between coherent sources. \[\frac{{{W}_{a}}}{{{W}_{b}}}=\frac{{{\lambda }_{a}}}{{{\lambda }_{w}}}\] Also, \[\frac{0.8}{{{W}_{w}}}=\mu =1.6\] \[\Rightarrow \] \[{{W}_{w}}=\frac{0.8}{1.6}=\frac{1}{2}=0.5\,\,mm\]
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