A) \[0.2\,\mu \,m\]
B) \[1.0\,\mu \,m\]
C) \[1.4\,\mu \,m\]
D) \[1.6\,\mu \,m\]
Correct Answer: D
Solution :
\[I={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\] |
\[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=4\,{{I}_{0}}\] |
\[I=0.75\,{{I}_{\max }}=3{{I}_{0}}\] |
So \[3{{I}_{0}}={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\] |
\[\cos f=\frac{1}{2}\] |
\[f=\frac{\pi }{3},\] \[2\pi -\frac{\pi }{3},\]\[2\pi +\frac{\pi }{3},\]\[4\pi -\frac{\pi }{3}\] |
\[f=\frac{\pi }{3},\] \[\frac{5\pi }{3},\]\[\frac{7\pi }{3},\] \[\frac{11\pi }{3},\] ??. |
Path difference \[Dx=\frac{\lambda }{2\pi }\phi =(m-1)\,t\] |
\[t=\frac{0.6}{\pi }\phi \,\mu m\]= 0.2 mm, 1.0 mm, 1.4 mm, 2.2 mm, |
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