Question

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

• A. $\frac{\lambda \varphi }{2\pi }$
• B. $\frac{\lambda }{2\pi }(\frac{\pi }{2}+\varphi )$λϕ2π(ϕ+π2)
• C. $\frac{\lambda }{2\pi }(\frac{\pi }{2}-\varphi )$λϕ2π(ϕ−π2)
• D. $\frac{2\pi \varphi }{\lambda }$

Answer 1

The correct option is B $\frac{\lambda }{2\pi }(\frac{\pi }{2}+\varphi )$λϕ2π(ϕ+π2)

Given, $y_1=a_1\sin (\omega t-\frac{2\pi x}{\lambda })$ and
$y_2=a_2\cos (\omega t-\frac{2\pi x}{\lambda }+\varphi )$
$=a_2\sin (\frac{\pi }{2}+\omega t-\frac{2\pi x}{\lambda }+\varphi )$
$\therefore$ Phase of first wave, ${\varphi }_1=(\omega t-\frac{2\pi x}{\lambda })$
and phase of second wave, ${\varphi }_2=\frac{\pi }{2}+\omega t-\frac{2\pi x}{\lambda }+\varphi$
$\therefore$ Phase difference, $△\varphi ={\varphi }_2-{\varphi }_1=\frac{\pi }{2}+\varphi$
$\therefore$ Path difference $=\frac{\lambda }{2\pi }\times phasedifference$
$=\frac{\lambda }{2\pi }(\frac{\pi }{2}+\varphi )$.