The path difference between two waves $y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right)$ and…
The path difference between two waves
$y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right)$ and
$y_2 = a_2 \cos\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right)$ [Manipal 2015]
$\frac{\lambda}{2\pi} (\phi + \pi/2)$
$\frac{\lambda}{2\pi} (\phi)$
$\frac{2\pi}{\lambda} (\phi - \pi/2)$
$\frac{2\pi}{\lambda} (\phi)$
Solution
Given, $y_2 = a_2 \cos \left(\omega t - \frac{2\pi x}{\lambda} + \phi\right)$
or $y_2 = a_2 \sin \left[\frac{\pi}{2} + \left(\omega t - \frac{2\pi x}{\lambda} + \phi\right)\right]$
and $y_1 = a_1 \sin \left(\omega t - \frac{2\pi x}{\lambda}\right)$
Phase difference between two waves $= \left(\frac{\pi}{2} + \phi\right)$
Path difference $= \frac{\lambda}{2\pi} \times \text{Phase difference}$
$\therefore \text{Path difference} = \frac{\lambda}{2\pi} \left(\frac{\pi}{2} + \phi\right)$