A) 1.2 cm
B) 1.2 mm
C) 2.4 cm
D) 2.4 mm
Correct Answer: D
Solution :
distance between the first dark fringes on either side of the centred bright fringe, \[\operatorname{x}=\frac{2\lambda D}{d}\] \[\operatorname{x}=\frac{2\times 600\times 1{{0}^{-9}}\times 2}{1{{0}^{-3}}}=24\times 1{{0}^{-4}}m\] \[\operatorname{x}=2.4mm\left( \therefore 1m=1{{0}^{3}}mm \right)\]You need to login to perform this action.
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