A) K
B) K/4
C) K/2
D) zero
Correct Answer: C
Solution :
For net intensity \[l'=4{{l}_{0}}{{\cos }^{2}}\frac{\phi }{2}\left( \phi =\frac{2\pi }{\lambda }\times \lambda \right)\] For the first case, \[K=4{{l}_{0}}{{\cos }^{2}}[\pi ]\] \[K=4{{l}_{0}}\,\,\,\,\,\,\,\,\,\,.....(i)\] For the second case \[K'=4{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi /2}{2} \right)\left( \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4} \right)\] \[4{{l}_{0}}{{\cos }^{2}}(\pi /2)\] \[K'=2{{l}_{0}}\,\,\,\,\,\,\,\,\,\,\,\,.....(ii)\] Comparing Eqs. (i) and (ii) \[K'=K/2\]
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