Correct Answer - D
As `Delta x = 1.5 m = (3)/(2)m = (lambda)/(2) :. Delta phi = pi` ,br. Reflection from water surface introduces a phase diff. of `pi`. Therefore, total phase diff. `= pi + pi = 2pi`.
`:.` Intensity obtained is maximum.
Now `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 2pi`
`= I + (I)/(4) + 2sqrt(I xx I//4) = (9 I)/(4)`