A) \[\frac{\lambda }{2\pi }\phi \]
B) \[\frac{\lambda }{2\pi }\left( \phi +\frac{\pi }{2} \right)\]
C) \[\frac{2\pi }{\lambda }\left( \phi -\frac{\pi }{2} \right)\]
D) \[\frac{2\pi }{\lambda }\phi \]
Correct Answer: B
Solution :
\[{{y}_{1}}={{a}_{1}}\sin \left( \omega t-\frac{2\pi x}{\lambda } \right)\]and \[{{y}_{2}}={{a}_{2}}\cos \left( \omega t-\frac{2\pi x}{\lambda }+\phi \right)\] \[={{a}_{2}}\sin \left( \omega t-\frac{2\pi x}{\lambda }+\phi +\frac{\pi }{2} \right)\] So, phase difference \[=\phi +\frac{\pi }{2}\] and path difference \[\Delta =\frac{\lambda }{2\pi }\left( \phi +\frac{\pi }{2} \right)\]
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