A) 1
B) 0.8
C) 0.4
D) 0.6
Correct Answer: B
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{4}\Rightarrow {{I}_{1}}=k\] and \[{{I}_{2}}=4k\] \[\therefore \] Fringe visibility \[V=\frac{2\sqrt{{{I}_{1}}{{I}_{2}}}}{({{I}_{1}}+{{I}_{2}})}=\frac{2\sqrt{k\times 4k}}{(k+4k)}=0.8\]
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