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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

please help me visualise the sitution which i'm unable to do so by myself and give short and nice answer. regards

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  • $\begingroup$ It might help if you could add a diagram, to explain S1, S2, P. $\endgroup$
    – Time4Tea
    Aug 15, 2018 at 17:14
  • $\begingroup$ @Time4Tea: sorry diagram is not given in the question ,i write it as it is written in book $\endgroup$
    – user182868
    Aug 15, 2018 at 17:16
  • $\begingroup$ this question is under the topic interference of waves $\endgroup$
    – user182868
    Aug 15, 2018 at 17:19

1 Answer 1

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$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.

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