A) \[\frac{K}{4}\]
B) \[\frac{K}{2}\]
C) K
D) Zero
Correct Answer: B
Solution :
By using phase difference \[\varphi =\frac{2\pi }{\lambda }(\Delta )\] For path differencel, phase difference \[{{\varphi }_{1}}=2\pi \]and for path difference l/4, phase difference f2 = p/2. Also by using \[I=4{{I}_{0}}{{\cos }^{2}}\frac{\varphi }{2}\] Þ \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{\cos }^{2}}({{\varphi }_{1}}/2)}{{{\cos }^{2}}({{\varphi }_{2}}/2)}\] Þ \[\frac{K}{{{I}_{2}}}=\frac{{{\cos }^{2}}(2\pi /2)}{{{\cos }^{2}}\left( \frac{\pi /2}{2} \right)}=\frac{1}{1/2}\] Þ \[{{I}_{2}}=\frac{K}{2}.\]
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