A) 1 mm
B) 0.0.5 mm
C) 0.03 mm
D) 0.01 mm
Correct Answer: C
Solution :
\[\sin \theta \simeq \theta \frac{y}{D}\] So, \[\Delta \theta =\frac{\Delta y}{D}\] Angular fringe width\[{{\theta }_{0}}=\Delta \theta \](width\[\Delta y=\beta \]) \[{{\theta }_{0}}=\frac{\beta }{D}=\frac{D\lambda }{d}\times \frac{1}{D}=\frac{\lambda }{d}\] \[{{\theta }_{0}}=1{}^\circ =\frac{\pi }{180}rad\] And \[\lambda =6\times {{10}^{-7}}m\] \[d=\frac{\lambda }{{{\theta }_{0}}}=\frac{180}{\pi }\times 6\times {{10}^{-7}}\] \[=3.44\times {{10}^{-5}}m\] \[=0.03\text{ }mm\]
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