# calculation of phase difference

Question:

Among two interfering sources , let $S_{1}$ be ahead of the phase by $\dfrac{\pi}{2}$ radians relative to $S_{2}$. If an observation point P is such that $PS_{1}-PS_{2}=1.5\lambda$,then the phase difference between the waves $S_{1}$ and $S_{2}$ is ........

my attempt :

initially , i assumed phasors for

$\ S_{1}=Ae^{-i ({kx+\pi/2})}$ and for $S_{2}=Be^{-i(kx)}$

also, for reaching point P if $S_{1}$ covers path length $x_{p}$ then $S_{2}$ has to cover $(x_{p}-1.5\lambda)$

and so, there phasors at point P will be

$S_{1}=Ae^{-i ({k(x-x_{P})+\pi/2})}$ and $S_{2}=Be^{-ik(x-x_{p}+1.5\lambda)}=Be^{-i{(k(x-x_{p})}+3\pi)}$

then subtracting there phases to get the phase difference as

$\delta \phi=3\pi-\dfrac{\pi}{2}=\dfrac{5\pi}{2}$

answer is also given as $\dfrac{5\pi}{2}$

But i am looking forward to intuitive solution (not purely mathematical like above) which totally don't rely on mathematical manipulation

$1.5 \lambda$ difference is coming already from the observing point, meaning that there is already a phase difference even if there is no phase difference between $S_1,S_2$. $1.5 \lambda$ corresponds to a phase difference of $2\pi+\pi=3\pi$. Then the waves themselves are out of phase, namely $\pi/2$ if there was no phase difference between the observing points i.e. $PS_1-PS_2=0$, so look two times at the same point. While in this situation we are dealing with both contributions. $S_1$ is ahead of $S_2$ meaning that $S_1-S_2=\pi/2$ without looking at the fact that there is a distance between the observing points. Combining those indeed boils down to the answer you gave us.