A) 2 mm
B) 1 mm
C) 3 mm
D) 4 mm
Correct Answer: A
Solution :
Angular fringe width \[\theta =\frac{\beta }{D}=\frac{\lambda }{d}\] \[\frac{\lambda }{{{d}_{0}}}=\frac{{{1}^{o}}}{60}=\frac{\pi }{180\times 60}\] \[{{d}_{0}}=\lambda \left( \frac{180\times 60}{\pi } \right)\]\[=2\times {{10}^{-3}}m=2mm.\]You need to login to perform this action.
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