A) 1.2 cm
B) 1.2 mm
C) 2.4cm
D) 2.4mm
Correct Answer: D
Solution :
\[\beta =\frac{\lambda D}{d}=\frac{600\times {{10}^{-9}}\times 2}{1\times {{10}^{-3}}}=12\times {{10}^{-4}}\,m\] So, distance between the first dark fringes on either side of the central bright fringe \[X=2\beta \] \[=2\times 12\times {{10}^{-4}}m\] \[=24\times {{10}^{-4}}m\] \[=2.4\,mm\]
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