A) \[8\times {{10}^{-4}}\]rad
B) \[6\times {{10}^{-4}}\]rad
C) \[4\times {{10}^{-4}}\]rad
D) \[16\times {{10}^{-4}}\]rad
Correct Answer: A
Solution :
Destructive interference occurs when the path difference is an odd multiple of \[\lambda /2\]. i.e., \[\frac{xd}{D}=\frac{(2n-1)\,\lambda }{2}\] Angular width of first dark fringe is \[\frac{2x}{D}=\frac{2\,(2n-1)\,\lambda }{2\,d}\] Given, \[n=1,\,\lambda =4800\,\overset{o}{\mathop{A}}\,=4800\times {{10}^{-10}}m\], \[d=0.6\,mm=0.6\,\times {{10}^{-3}}m\] \[\therefore \] \[\frac{2\,x}{D}=\frac{2\,(2\times 1-1)\times 4800\times {{10}^{-10}}}{2\times 0.6\times {{10}^{-3}}}\] \[=8\times {{10}^{-4}}\]rad
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