The path difference between two waves given b | vedclass

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(C) The phase difference between the two waves is given by the difference in their arguments.Let $\phi_1 = \omega t - \frac{2 \pi x}{\lambda}$ and $\p

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$\frac{\lambda}{2 \pi} \phi$

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(C) The phase difference between the two waves is given by the difference in their arguments.Let $\phi_1 = \omega t - \frac{2 \pi x}{\lambda}$ and $\phi_2 = \omega t - \frac{2 \pi x}{\lambda} + \phi$.The phase difference $\Delta \phi = \phi_2 - \phi_1 = \phi$.The relationship between path difference $(\Delta x)$ and phase difference $(\Delta \phi)$ is given by the formula:$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Rearranging this formula to solve for the path difference $\Delta x$:$\Delta x = \frac{\lambda}{2 \pi} \Delta \phi$.Substituting $\Delta \phi = \phi$ into the equation:$\Delta x = \frac{\lambda}{2 \pi} \phi$.

The path difference between two waves given by the equations $y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$ and $y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right)$ is

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