If we had two generators producing waves, say $S$ and $T$, and the waves from $S$ were in phase with the waves from $T$, we know that at some point $P$, if the path difference is $n$$\lambda$, they will add constructively and if we have $(n+\frac{1}{2}$)$\lambda$, then they add destructively. Now say the waves from $S$ had a phase differnce of $\pi$ relative to the waves from $T$, now if the path difference was $n$$\lambda$, they would add destructively, and if they had a path difference of $(n+\frac{1}{2}$)$\lambda$, they would add constructively, now my question is, in my textbook it states If two sources are coherent, and the path difference is $(n+\frac{1}{2}$)$\lambda$, the waves will add destructively, but we know that if the waves maintain the phase difference of $\pi$, we can call them coherent, as its constant phase difference, yet the wrong type of interference happens (wrong as in not what the rule states). Can someone help my misunderstanding please? Or is the textbook wrong?


2 Answers 2


You are correct. When we talk about coherence, it really is about the ability for waves from those points to cause an interference pattern. For that to happen, any constant phase difference is all that you need.

What the textbook is doing, is NOT wrong. It is just that it is a simplification for the case of zero phase difference, which is quite usually the assumption needed for an argument in terms of path differences to make sense.

  • $\begingroup$ Alright, but I had another question where one wave was reflected and its phase shifted by $\pi$, here the condition for constructive interference was $(n +\frac{1}{2})$ $\lambda$ as the phase shift added $\lambda$$/2$ to its journey. $\endgroup$ May 27 at 15:46
  • $\begingroup$ That is another question; if you want to ask it, ask it separately. Or you can just deduce the answer from knowing the solution to this question. $\endgroup$ May 27 at 15:47
  • $\begingroup$ Yeah I know why, but is the textbook just assuming $0$ or phase difference of the form $2n\pi$? $\endgroup$ May 27 at 15:48
  • $\begingroup$ it goes by modulo $2\pi$ so that is the same. In practice, wave sources are only coherent up to a certain spatial length and temporal time, so when $n$ is large the $\Delta n$ correction to the phase difference becomes random enough to wash away interference. $\endgroup$ May 28 at 4:04

The relation between path difference and phase difference is as follows: $$\Delta{\theta}=\frac{2\pi}{\lambda}\Delta{x}$$ If $$\Delta{\theta}=\pi$$,then $$\Delta{x}=\frac{\lambda}{2}$$ which is in the form of $$\lambda({n+\frac{1}{2}})$$ Which leads to destructive interference. 1st source is in the form of $$y_1=Asin(x)$$ 2nd source $$y_2=Asin(x-\pi)$$ A is same because both are coherent. $$sin(x-\pi)=-sin(x)$$ Therefore, $$y_1+y_2=0$$ $$(y_1)^2=I_1$$ $$(y_2)^2=I_2$$ Where I is the intensity of light. $$I_1=-I_2$$ $$I_1+I_2=0$$ Phase difference doesn't corresponds to coherency it is determined by the source.

  • $\begingroup$ This makes little sense to me, I will mark the other answer as the solution, could you explain yours a bit more? Thanks tho! $\endgroup$ May 27 at 15:49
  • $\begingroup$ @NavBhatthal you can of this as two wave trying to combine and form a new wave but due to the phase difference of pi, the second wave is having the exact opposite values wrt first wave. Which leads to complete destruction interference. $\endgroup$
    – Alv
    May 27 at 15:56

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