Q163 / 170 · Manipal (MET) 2016 · Physics

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
    $\frac{\lambda \phi}{2 \pi}$
  • 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
    $\frac{2 \pi \phi}{\lambda}$

Solution

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, $\Delta \phi=\phi_2-\phi_1=\frac{\pi}{2}+\phi$ $\begin{aligned} \therefore \text { Path difference } & =\frac{\lambda}{2 \pi} \times \text { phase difference } \\ & =\frac{\lambda}{2 \pi}\left(\frac{\pi}{2}+\phi\right)\end{aligned}$

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

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