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)

$y_{1} =a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ 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 $\Delta=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$