Physics - Wave Optics Question with Solution | TestHub

The given equations of the waves are:

$Y_1=a_1\sin \left(\omega t-\frac{2\pi x}{\lambda }\right)$

$Y_2=a_2\cos \left(\omega t-\frac{2\pi x}{\lambda }+\varphi \right)$

Expressing $Y_2$ in terms of sine function to compare the phases:

$Y_2=a_2\sin \left(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2}\right)$

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

$\Delta \Phi =\left(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2}\right)-\left(\omega t-\frac{2\pi x}{\lambda }\right)=\varphi +\frac{\pi }{2}$

The relation between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:

$\Delta x=\frac{\lambda }{2\pi }\Delta \Phi$

Substituting the value of $\Delta \Phi$ into the equation:

$\Delta x=\frac{\lambda }{2\pi }\left(\varphi +\frac{\pi }{2}\right)$

Answer: $\frac{\lambda }{2\pi }\left(\varphi +\frac{\pi }{2}\right)$