Question

The path difference between two waves $y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$ and $y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\varphi )$

• A. $\frac{\lambda }{2\pi }[\varphi +\pi /2]$
• B. $\frac{\lambda }{2\pi }\left[\varphi \right]$
• C. $\frac{\lambda }{2\pi }[\varphi -\pi /2]$
• D. $\frac{2\pi }{\lambda }\left[\varphi \right]$

Answer 1

The correct option is A $\frac{\lambda }{2\pi }[\varphi +\pi /2]$

Given, $y_2=a_2\cos \left[\begin{array}{c}\omega t-\frac{2\pi x}{\lambda }+\varphi \end{array}\right]$

or $y_2=a_2\sin \left[\begin{array}{c}\frac{\pi }{2}+(\omega t-\frac{2\pi x}{\lambda }+\varphi )\end{array}\right]$

and $y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda }+\varphi )$

We have phase difference between two waves $=\left[\begin{array}{c}\frac{\pi }{2}+\varphi \end{array}\right]$

Path difference $=\frac{\lambda }{2\pi }\times$ phase difference

$\therefore$ Path difference $=\frac{\lambda }{2\pi }\left[\begin{array}{c}\frac{\pi }{2}+\varphi \end{array}\right]$