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 :
Angular fringe width of first minima is \[\frac{2x}{D}=2(2n-1)\frac{A}{2d}\] \[=(2n-1)\frac{\lambda }{d}\] Given, \[d=0.6\text{ }mm=0.6\times {{10}^{-3}}m\] \[\lambda =4800{AA}=4.8\times {{10}^{-7}}m,\text{ }n=1\] \[\therefore \] \[\frac{2x}{D}=\frac{(2\times 1-1)\times 4.8\times {{10}^{-7}}}{0.6\times {{10}^{-3}}}\] \[=8\times {{10}^{-4}}rad.\]You need to login to perform this action.
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