Question

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

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

Answer 1

Answer: (B)

Given, $y_{1}=a_{1} \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$
and $y_{2}=a_{2} \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right)=a_{2} \sin \left(\frac{\pi}{2}+\omega t-\frac{2 \pi x}{\lambda}+\phi\right)$
$\therefore $ Phase of first wave, $\phi_{1}=\left(\omega t-\frac{2 \pi x}{\lambda}\right)$
and phase of second wave, $\phi_{2}=\frac{\pi}{2}+\omega t-\frac{2 \pi x}{\lambda}+\phi$
$\therefore $ Phase difference, $\Delta \phi=\phi_{2}-\phi_{1}=\frac{\pi}{2}+\phi$
$\therefore $ Path difference $=\frac{\lambda}{2 \pi} \times$ phase difference $=\frac{\lambda}{2 \pi}\left(\frac{\pi}{2}+\phi\right)$