A) 1.0 mm
B) 0.1 mm
C) 0.05 mm
D) 0.01 mm
Correct Answer: B
Solution :
Given, \[\lambda =5000\overset{\text{o}}{\mathop{\text{A}}}\,=5\times {{10}^{-7}}m.\] \[\beta =5\,mm=5\times {{10}^{-3}}m.\] \[D=1m.\] \[d=?\] Fringe width \[\beta =\frac{D\lambda }{d}\] \[\therefore \] \[d=\frac{D\lambda }{\beta }=\frac{1\times 5\times {{10}^{-7}}}{5\times {{10}^{-3}}}\] \[=1\times {{10}^{-4}}m=0.1mm\]
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