A) 0.002 mm
B) 0.02 mm
C) 0.2mm
D) 2mm
Correct Answer: C
Solution :
Given, \[d=5\times {{10}^{-3}}m,\,\,D=2\,m\], \[\lambda =500\times {{10}^{-9}}m\] Now, fringe width \[\beta =\frac{\lambda D}{d}=\frac{500\times {{10}^{-9}}\times 2}{5\times {{10}^{-3}}}\] = 0.2 mmYou need to login to perform this action.
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