Question

The path difference between the two waves

$y_{1} = a_{1}\sin\left( \omega t - \frac{2\pi x}{\lambda} \right)$and$y_{2} = a_{2}\cos\left( \omega t - \frac{2\pi x}{\lambda} + \varphi \right)$ is

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

Answer 1

Answer: (B)

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

Answer 2

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

Phase difference =$\left( \omega t - \frac{2\pi x}{\lambda} + \varphi + \frac{\pi}{2} \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \left( \varphi + \frac{\pi}{2} \right)$

Path difference = $\frac{\lambda}{2\pi}$× Phase difference $= \frac { \lambda } { 2 \pi }$ $\left( \varphi + \frac{\pi}{2} \right)$