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 :
\[\frac{2x}{D}=2(2n-1)\frac{A}{2d}\] \[=(2n-1)\frac{\lambda }{d}\] \[d=0.6mm\,=0.6\times {{10}^{-3}}m\] \[\lambda =4800\overset{\text{o}}{\mathop{\text{A}}}\,=4.8\times {{10}^{-7}}m,n=1\] \[\frac{2x}{D}=\frac{(2\times 1-1)\times 4.8\times {{10}^{-7}}}{0.6\times {{10}^{-3}}}\] \[=8\times {{10}^{-4}}\,\text{rad}\]
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