A) \[\frac{K}{4}\]
B) \[\frac{K}{2}\]
C) K
D) 0
Correct Answer: B
Solution :
\[I=2{{I}_{0}}(1+cos\delta ).\]When path difference \[=\lambda ,\]then phase difference \[\delta =\frac{2\pi }{\lambda }\times \]path difference \[=2\pi {{I}_{1}}\] \[=2{{I}_{0}}(1+\cos 2\pi )=4{{I}_{0}}=K\] When the difference \[=\lambda /4.\]Then phase difference \[\delta =\frac{2\pi }{\lambda }=\frac{\lambda }{4}=\frac{\pi }{2}\] \[\therefore \]\[{{I}_{2}}=2{{I}_{0}}\left\{ 1+\cos \frac{\pi }{2} \right\}=2{{I}_{0}}=\frac{k}{2}\] Hence, the correction option is (b).
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