A) 1 mm
B) 2 mm
C) 4 mm
D) 5 mm
Correct Answer: C
Solution :
Width of central bright fringe. \[=\frac{2\lambda D}{d}=\frac{2\times 500\times {{10}^{-9}}\times 80\times {{10}^{-2}}}{0.20\times {{10}^{-3}}}\]\[=4\times {{10}^{-3}}m=4mm.\]You need to login to perform this action.
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