Explanation:
Intensity of two interfering waves on the screen is given by,
\({I=4 I_{0} \cos ^{2} \dfrac{\phi}{2} \quad}\) where \({\phi=}\) phase difference
\({\therefore}\) For constructive interference
\(\phi=0, \pm 2 \pi, \pm 4 \pi, \ldots \ldots\)
We get only even multiples of \({\pi}\).
So, maximum intensity corresponds to constructive interference, \({I=4 I_{0}}\)
So, \({\phi}\) cannot be \({5 \pi}\).
So correct option is (3)