Question

The path difference between the two waves $y_{1}=a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2}=a_{2} \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$ is

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

Answer 1

Answer: (B)

$\frac{\lambda }{2\pi }\left( \phi +\frac{\pi }{2} \right)$

Answer 2

$y_{1} =a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2} =a_{2} \cdot \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$ $=a_{2} \sin \left(\omega t-\frac{2 \pi x}{\lambda}+\phi+\frac{\pi}{2}\right)$ So, phase difference $=\phi+\frac{\pi}{2}$ and path difference $\Delta=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$