Phase difference from path length in waves

Steps:

• Locate points P and Q on the displacement-distance graph; Q is 1.5 wavelengths ahead of P.
• Identify wavelength λ as the graph distance for one full cycle.
• Compute phase difference: (path difference / λ) × 360° = (1.5λ / λ) × 360° = 540°.
• Confirm no aliasing; 540° is the direct calculation.

Why C is correct:

• Phase difference δ = (2π Δx / λ) radians or (360° Δx / λ), yielding 540° for Δx = 1.5λ per wave equation.

Why the others are wrong:

• A: 90° implies Δx = 0.25λ, but graph shows 1.5λ.
• B: 270° implies Δx = 0.75λ, underestimating the separation.
• D: 630° implies Δx = 1.75λ, exceeding the observed distance.

Final answer: C