A) \[\frac{1+p}{2\sqrt{p}}\]
B) \[\frac{1\sqrt{p}}{1+p}\]
C) \[\frac{p}{1+p}\]
D) \[\frac{2p}{1+p}\]
Correct Answer: B
Solution :
Visibility, \[V=\frac{{{I}_{\max }}-{{I}_{\min }}}{{{I}_{\max }}+{{I}_{\min }}}=\frac{2\sqrt{{{I}_{1}}{{I}_{2}}}}{({{I}_{1}}+{{I}_{2}})}\] \[=\frac{2\sqrt{{{I}_{1}}/{{I}_{2}}}}{\frac{{{I}_{1}}}{{{I}_{2}}}+1}=\frac{2\sqrt{P}}{P+1}\]
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