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If two light waves of same wavelength reaches the point $ P$, then what is the relationship between path difference and phase difference? In many books I have seen that they assume two 1-D equation for each wave. Such as,

\begin{align} y_{1}= a \sin\frac{2π}{\lambda}(ct-x_1) \end{align}

\begin{align} y_{2}= a \sin\frac{2π}{\lambda}(ct-x_2) \end{align}

Now,phase difference, \begin{align} \delta = \frac{2π}{\lambda}(ct-x_1)-\frac{2π}{\lambda}(ct-x_2) \end{align} \begin{align} \Rightarrow \delta = \frac{2π}{\lambda}(x_2 - x_1) \end{align} \begin{align} \Rightarrow \delta = \frac{2π}{\lambda} ∆x \end{align}

But the problem is that both wave sources can be different and the waves may not be parallel or along same axis. So how can we use this formula then?

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  • $\begingroup$ Your equations are only in one spatial dimension. How can you have different axes? $\endgroup$ Commented Jun 16, 2022 at 12:28
  • $\begingroup$ @BioPhysicist in case of Interference this formula is actually used . Maybe they assume that difference between two slits are too small thus two light waves are almost parallel . I am not sure though $\endgroup$ Commented Jun 16, 2022 at 12:39
  • $\begingroup$ It is assumed that these are plane waves, so the path is measured along the direction of propagation - as if they were straight rays. Thus, $x_2-x_2$ is a difference in path lengths, not in coordinates (it would be less ambiguous, if written as $L_2-L_1$). $\endgroup$
    – Roger V.
    Commented Jun 16, 2022 at 12:43

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$\delta$ is the phase-difference $\theta_2 - \theta_1$ at the meeting point, where $\Delta x$ is the path-difference (difference in path lengths) $\ell_2 - \ell_1$ between the two paths from the sources to the meeting point.

The paths need not be parallel. However, the paths must meet at a meeting point. (The tangent vectors to the paths need not be parallel at the meeting point.) At that meeting point, the difference of their phases is calculated.

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