Question

The path difference between the two waves

$y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$ and
$y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\theta )$ is

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

Answer 1

Answer: (B)

$\frac{\lambda }{2\pi }(\theta +\frac{\pi }{2})$

$y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$
$y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\theta )$
$=a_2\sin (\omega t-\frac{2\pi x}{\lambda }+\theta +\frac{\pi }{2})$
Phase differece is $\theta +\frac{\pi }{2}$
Thus path difference is $\frac{\lambda }{2\pi }(\theta +\frac{\pi }{2})$