A) 1.2 cm
B) 1.2 mm
C) 2.4 cm
D) 2.4 mm
Correct Answer: D
Solution :
Given \[D\,\,=\,\,2\,\,m;\,\,d=1\,\,mm=1\times {{10}^{-}}^{3}m\] \[\lambda \,\,=\,\,600\,nm=600\times {{10}^{-\,6}}\,m\] Width of central bright fringe \[\left( =\,\,2\beta \right)\] \[=\,\,\frac{2\lambda D}{d}\,\,=\,\,\frac{2\times 600\times {{10}^{-6}}\,\times 2}{1\times {{10}^{-\,3}}}\,m\] \[=2.4\,\,\times \,\,{{10}^{-3}}m\,\,=\,\,2.4\,mm\]You need to login to perform this action.
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