Question

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

• A. $\frac{\lambda \varphi }{2\pi }$
• B. $\frac{\lambda }{2\pi }\left(\varphi +\frac{\pi }{2}\right)$
• C. $\frac{2\pi }{\lambda }\left(\varphi -\frac{\pi }{2}\right)$
• D. $\frac{2\pi \varphi }{\lambda }$

Answer 1

Answer: (B)

Given, $y_1=a_1\sin \left(\omega t-\frac{2\pi x}{\lambda }\right)$
and $y_2=a_2\cos \left(\omega t-\frac{2\pi x}{\lambda }+\varphi \right)=a_2\sin \left(\frac{\pi }{2}+\omega t-\frac{2\pi x}{\lambda }+\varphi \right)$
$\therefore$ Phase of first wave, ${\varphi }_1=\left(\omega t-\frac{2\pi x}{\lambda }\right)$
and phase of second wave, ${\varphi }_2=\frac{\pi }{2}+\omega t-\frac{2\pi x}{\lambda }+\varphi$
$\therefore$ Phase difference, $\Delta \varphi ={\varphi }_2-{\varphi }_1=\frac{\pi }{2}+\varphi$
$\therefore$ Path difference $=\frac{\lambda }{2\pi }\times$ phase difference $=\frac{\lambda }{2\pi }\left(\frac{\pi }{2}+\varphi \right)$