A) \[0.2\text{ }\mu \text{m}\]
B) \[1.0\text{ }\mu \text{m}\]
C) \[1.4\text{ }\mu \text{m}\]
D) \[1.6\text{ }\mu \text{m}\]
Correct Answer: D
Solution :
\[I={{I}_{0}}+{{I}_{0}}+2{{I}_{0}}\cos f\] |
\[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=4\,\,{{I}_{0}}\] |
\[{{I}_{0}}=0.75\,\,{{I}_{\max }}=3\,\,{{I}_{0}}\] 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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