A) \[(2n+1)\,\lambda /2\]
B) \[\text{n}\]
C) \[\]
D) \[\text{/2}\]
Correct Answer: B
Solution :
We know that \[I\propto {{A}^{2}}\] \[A=\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \theta }\] \[\cos \theta =2\pi n\] \[\frac{2\pi }{\lambda }\times \] path difference = phase angle \[\therefore \] path difference\[=\frac{2\pi n\times \lambda }{2\pi }=n\lambda \]
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