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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