A) \[3.44\times {{10}^{-4}}m\]
B) \[4.33\times {{10}^{-4}}m\]
C) \[5.44\times {{10}^{-4}}m\]
D) \[6.33\times {{10}^{-4}}m\]
Correct Answer: A
Solution :
Angular fringe width \[\alpha =0.1{}^\circ \] \[=\frac{0.1\times \pi }{180}\] \[=1.74\times {{10}^{-3}}rad\] \[\lambda =6000\overset{o}{\mathop{\text{A}}}\,\] \[=6000\times {{10}^{-10}}m\] \[=6\times {{10}^{-7}}m\] The angular fringe width is \[\alpha =\frac{\lambda }{d}\] (where d is the separation between the slits) \[\Rightarrow \] \[d=\frac{\lambda }{\alpha }\] \[=\frac{6\times {{10}^{-7}}}{1.746\times {{10}^{-3}}}\] \[=3.44\times {{10}^{-4}}m\]
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