Question

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

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

Answer 1

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

Given ,
$y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })........\left(1\right)$
and
$y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\varphi )=a_2\sin (\omega t-\frac{2\pi x}{\lambda }+\varphi +\frac{\pi }{2})........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
Phase difference $\Delta \varphi =\varphi +\frac{\pi }{2}$
Since, Phase difference $\left(\Delta \varphi \right)=\frac{2\pi }{\lambda }\times$ Path difference $\left(\Delta x\right)$
$\Delta x=\frac{\lambda }{2\pi }\times \Delta \varphi \Rightarrow \Delta x=\frac{\lambda }{2\pi }(\varphi +\frac{\pi }{2})$

Thus, option (b) is the correct answer.