AP EAMCET · PHYSICS · Wave Optics
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 \(\left(\frac{\lambda}{\pi} \phi\right)\)
- B \(\frac{\lambda}{\pi}\left(\phi-\frac{\pi}{2}\right)\)
- C \(\frac{\lambda}{2 \pi} \phi\)
- D \(\frac{\lambda}{2 \pi}\left(\phi-\frac{\pi}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda}{2 \pi} \phi\)
Step-by-step Solution
Detailed explanation
Phase difference \(\Delta \Phi = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi\right) - \left(\omega t - \frac{2 \pi x}{\lambda}\right) = \phi\). Path difference \(\Delta x = \frac{\lambda}{2\pi} \Delta \Phi = \frac{\lambda}{2\pi} \phi\).
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