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, |
\[\Delta {{\phi }_{1}}=\frac{\Delta {{x}_{1}}}{\lambda }\times 2\pi =\frac{(a/2)\sin \theta }{\lambda }\times 2\pi \] |
\[=-\frac{\lambda }{2\lambda }\times 2\pi =\pi \,rad\] |
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