# Phase difference and path difference relation confusion

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?

• Your equations are only in one spatial dimension. How can you have different axes? Jun 16, 2022 at 12:28
• @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 Jun 16, 2022 at 12:39
• 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 $L_2-L_1$). Jun 16, 2022 at 12:43

$$\delta$$ 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.