The path Difference between the two waves ${y_1} = {a_1}\,\sin \,\left( {\omega t - \frac{{2\pi x}}{\lambda }} \right)$ and ${y_2} = {a_2}\,\cos \,\left( {\omega t - \frac{{2\pi x}}{\lambda } + \phi } \right)$ is
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$y_{1}=a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$
$\mathrm{y}_{2}=\mathrm{a}_{2} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}+\phi\right)$
$\Delta \phi=\frac{\pi}{2}+\phi$
So, $\frac{\Delta \phi}{2 \pi}=\frac{\Delta x}{\lambda} \Rightarrow \Delta x=\frac{\lambda}{2 \pi}\left(\frac{\pi}{2}+\phi\right)$
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