Q155 / 170 · Manipal (MET) 2018 · Physics

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

Solution

$\begin{gathered} \text { (b) } y_1=a_1 \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right) \text { and } \\ y_2=a_2 \cos \left(\omega t-\frac{2 \pi x}{\lambda}+\phi\right) \\ =a_2 \sin \left(\omega t-\frac{2 \pi x}{\lambda}+\phi+\frac{\pi}{2}\right) \end{gathered}$
So, phase difference $=\phi+\frac{\pi}{2}$ and path difference $\Delta=\frac{\lambda}{2 \pi}\left(\phi+\frac{\pi}{2}\right)$

Hence, the correct answer is option: 'B'.

Relevant Questions