Question

The path difference between the two waves \(y_{1}=a_{1}\sin \left(\omega t-\frac{2\pi x}{\lambda }+\pi \right)\) and \(y_{2}=a_{2}\cos \left(\omega t-\frac{2\pi x}{\lambda }+\phi \right)\) is

• A. \(\frac{\lambda }{2\pi }\phi\)
• B. \(\frac{\lambda }{2\pi }\left(\phi -\frac{\pi }{2}\right)\)
• C. \(\frac{\lambda }{2\pi }(\phi -\pi )\)
• D. \(\frac{\lambda }{2\pi }\left(2\phi -\frac{\pi }{2}\right)\)

Hint:

Convert both waves to the same trigonometric form, find the phase difference, then use path difference = (wavelength / 2 pi) times phase difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Path difference and phase difference are linked by \(\Delta x=\dfrac{\lambda}{2\pi}\Delta\phi\). First write both waves as sine functions.

Step 2: Rewrite the waves
\(y_{1}=a_{1}\sin\left(\omega t-\dfrac{2\pi x}{\lambda}+\pi\right)\)
\(y_{2}=a_{2}\cos(\theta)=a_{2}\sin\left(\theta+\dfrac{\pi}{2}\right)\), where \(\theta=\omega t-\dfrac{2\pi x}{\lambda}+\phi\).

Step 3: Phase difference
Phase of wave 2 = \(\phi+\dfrac{\pi}{2}\), phase of wave 1 = \(\pi\).
\(\Delta\phi=\phi+\dfrac{\pi}{2}-\pi=\phi-\dfrac{\pi}{2}\)

Step 4: Path difference
\(\Delta x=\dfrac{\lambda}{2\pi}\left(\phi-\dfrac{\pi}{2}\right)\)

Final Answer:
The path difference is lambda over 2 pi times (phi minus pi by 2). \[ \boxed{\text{(B) }\dfrac{\lambda}{2\pi}\left(\phi-\dfrac{\pi}{2}\right)} \]