Question

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

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

Hint:

Convert the cosine wave into a sine wave to read off the phase difference, then use path difference = lambda/2pi times phase difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Path difference \(\Delta x=\dfrac{\lambda}{2\pi}\times\) phase difference.

Step 2: Convert to Sine Form
\(Y_2=a_2\cos(\theta+\Phi)=a_2\sin\left(\theta+\Phi+\dfrac{\pi}{2}\right)\), where \(\theta=\omega t-\dfrac{2\pi x}{\lambda}\).

Step 3: Phase Difference
Comparing with \(Y_1=a_1\sin\theta\), the phase difference is \(\Phi+\dfrac{\pi}{2}\).

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 plus pi by 2. \[ \boxed{\text{(B) }\dfrac{\lambda}{2\pi}\left(\Phi+\dfrac{\pi}{2}\right)} \]