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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(b) ${y_1} = {a_1}\sin \,\left( {\omega \,t - \frac{{2\pi x}}{\lambda }} \right)$ and
${y_2} = {a_2}\cos \,\left( {\omega \,t - \frac{{2\pi x}}{\lambda } + \phi } \right)$
$ = {a_2}\sin \,\left( {\omega \,t - \frac{{2\pi x}}{\lambda } + \phi + \frac{\pi }{2}} \right)$
So phase difference =$\phi + \frac{\pi }{2}$ and $\Delta$ =$\frac{\lambda }{{2\pi }}\left( {\phi + \frac{\pi }{2}} \right)$
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