Physics - Wave Optics Question with Solution | TestHub

The given wave equations are:

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

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

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

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

The phase of the first wave is ${\Phi }_1=\omega t-\frac{2\pi x}{\lambda }+\pi$

The phase of the second wave is ${\Phi }_2=\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2}$

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

$\Delta \Phi ={\Phi }_2-{\Phi }_1=\left(\varphi +\frac{\pi }{2}\right)-\pi =\varphi -\frac{\pi }{2}$

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 $\Delta \Phi$:

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

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