Question

The path difference between two waves given by the equations \( y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right) \) and \( y_2 = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right) \) is

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

Use \( \Delta x = \frac{\lambda}{2\pi} \cdot \Delta \phi \) to convert phase difference to path difference in wave equations.

Answer 1

The Correct Option is C

The phase difference between two waves determines their path difference. For the waves: \[ y_1 = a_1 \sin\left( \omega t - \frac{2\pi x}{\lambda} \right), \quad y_2 = a_2 \sin\left( \omega t - \frac{2\pi x}{\lambda} + \phi \right) \] The phase of \( y_1 \) is \( \omega t - \frac{2\pi x}{\lambda} \), and the phase of \( y_2 \) is \( \omega t - \frac{2\pi x}{\lambda} + \phi \). The phase difference is: \[ \Delta \phi = \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \phi \] The relationship between phase difference \( \Delta \phi \) and path difference \( \Delta x \) is: \[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \] Solve for \( \Delta x \): \[ \Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{\lambda}{2\pi} \phi \] Check options: - (1) \( \frac{\lambda}{2\pi} |\phi| \): Incorrect, as path difference is proportional to \( \phi \), not its absolute value. - (2) \( \frac{\lambda}{2\pi} \left( \frac{\pi}{2} - \phi \right) \): Incorrect, introduces an arbitrary phase shift. - (3) \( \frac{\lambda}{2\pi} \phi \): Correct. - (4) \( \frac{\lambda}{2\pi} \left( \frac{\pi}{2} - \frac{\phi}{2} \right) \): Incorrect, manipulates \( \phi \) unnecessarily. Thus, the path difference is \( \frac{\lambda}{2\pi} \phi \), confirming option (3).