The path difference between the two waves y_1 = a_1 (ω t - (2π...

The given wave equations are:

$y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \pi\right)$

$y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi\right)$

To compare the phases, we express $y_2$ in terms of the sine function:

$y_2 = a_2\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}\right)$

The phase of the first wave is $\Phi_1 = \omega t - \dfrac{2\pi x}{\lambda} + \pi$

The phase of the second wave is $\Phi_2 = \omega t - \dfrac{2\pi x}{\lambda} + \phi + \dfrac{\pi}{2}$

The phase difference $\Delta \Phi$ between the two waves is:

$\Delta \Phi = \Phi_2 - \Phi_1 = \left(\phi + \dfrac{\pi}{2}\right) - \pi = \phi - \dfrac{\pi}{2}$

The relation between path difference $\Delta x and phase difference$\Delta \Phi$is given by:<br><br>$\Delta x = \dfrac{\lambda}{2\pi} \Delta \Phi$<br><br>Substituting the value of$\Delta \Phi$:<br><br>$\Delta x = \dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)$<br><br>Answer:$\dfrac{\lambda}{2\pi}\left(\phi - \dfrac{\pi}{2}\right)