Question

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

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

Answer 1

The correct option is B $\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$


Given, $y_1=a_1\sin \left(\omega t-\frac{2\pi x}{\lambda }\right)$
And $y_2=a_2\cos \left(\omega t-\frac{2\pi x}{\lambda }+\varphi \right)$
$=a_2\sin \left(\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2}\right)$
So, phase difference $=\varphi +\frac{\pi }{2}$
And path difference
$\Delta x=\frac{\lambda }{2\pi }$ (phase difference)
$\therefore \Delta x=\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$
Final answer: $\left(b\right)$