A) 344 mm
B) 0.1344 mm
C) 0.0344 mm
D) 0.034 mm
Correct Answer: C
Solution :
[c] In Young's double slit fringe with \[\sin \theta =\theta =y/D\] So, \[\Delta y/D\]and hence angular fringe width \[{{\theta }_{0}}=\Delta \theta \,(with\Delta y=\beta )\] Will be \[{{\theta }_{0}}=\frac{\beta }{D}=\frac{D\lambda }{d}\times \frac{1}{D}=\frac{\lambda }{d}\] \[\Rightarrow {{\theta }_{0}}={{1}^{0}}=\left( \frac{\pi }{180} \right)\]rad, and \[\lambda =6\times {{10}^{-7}}m\] Or \[d=\frac{\lambda }{{{\theta }_{0}}}=\frac{180}{\pi }\times (6\times {{10}^{-7}})=3.44\times {{10}^{-5}}m\] Or \[d=0.0344mm\]
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