A) \[1\]
B) \[0.8\]
C) \[0.4\]
D) \[0.6\]
Correct Answer: B
Solution :
Fringe visibility \[V=\frac{2\sqrt{{{I}_{1}}{{I}_{2}}}}{{{I}_{1}}+{{I}_{2}}}\] where \[{{I}_{1}}\] and \[{{I}_{2}}\] are intensities of coherent sources. Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{4}\] \[\therefore \] \[{{I}_{2}}=4{{I}_{1}}\] \[\therefore \] Fringe visibility \[=\frac{2\sqrt{{{I}_{1}}\times 4{{I}_{1}}}}{({{I}_{1}}\times 4{{I}_{1}})}\] \[=\frac{2\times 2{{I}_{1}}}{5{{I}_{1}}}=\frac{4}{5}\] \[\Rightarrow \] \[V=0.8\]
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