Is the resultant intensity in case of constructive interference always 4 times? I get that intensity is proportional to square of ampltude and hence if two waves of same amplitude meet in phase, it results in 4 times intensity. But they need not have the same ampltude. One can have twice the ampltudue but still be in phase. In that case will they not have intensity different than 4 times.? 
 A: You are correct.  Strictly speaking, "destructive interference" occur whenever the intensity produced by two (or more) intersecting beams is less than the sum of the individual beam intensities.  And "constructive interference" is the reverse case, when the local intensity is greater than the sum of the individual beam intensities.
The simplest cases to consider are two beams with equal intensity, which are either exactly in phase or exactly $180^{\circ}$ out of phase.  That produces what is sometimes called "perfect" constructive or destructive interference.  With perfect interference, the net intensity is either zero or four times the intensity of a single beam.  However, you can still have either destructive or constructive interference when the beams are not of equal intensity, or when the phase relationships between them are different.  As long as the intensity of the superposition of the two beams is not the sum of the intensities, there is some interference, and whether it is greater or less decides whether the (imperfect) interference is constructive or destructive.
