A) \[\frac{\pi }{4}\]radian
B) \[\frac{\pi }{2}\] radian
C) \[\pi \] radian
D) \[\frac{\pi }{8}\] radian
Correct Answer: C
Solution :
For first minima at \[P,a\sin \theta =n\lambda \] where, \[N=1\Rightarrow a\sin \theta =\lambda ,\] So, phase difference, \[\Delta {{\phi }_{1}}=\frac{\Delta {{x}_{1}}}{\lambda }\times 2\pi =\frac{(a/2)sin\theta }{\lambda }\times 2\pi \] \[=-\frac{\lambda }{2\lambda }\times 2\pi =\pi \,rad\]You need to login to perform this action.
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