A) 0.072 mm
B) 0.063 mm
C) 0.037 mm
D) 0.019 mm
Correct Answer: A
Solution :
\[{{x}_{1}}=\frac{(2m-1)\,D{{\lambda }_{1}}}{2d}\] \[=\frac{(2\times 2-1)\times 1\times 480\times {{10}^{-9}}}{2\times 5\times {{10}^{-3}}}\] \[=3\times 48\times {{10}^{-6}}=0.144\,mm\] \[{{x}_{2}}=\frac{mD{{\lambda }_{1}}}{d}\] \[=\frac{3\times 1\times 480\times {{10}^{-9}}}{5\times {{10}^{-3}}}=288\times {{10}^{-6}}\] = 0.288 mm So, the third order bright fringes \[=\frac{{{x}_{2}}-{{x}_{1}}}{2}\] \[=\frac{0.288-0.144}{2}=\frac{0.144}{2}=0.072\,mm\]You need to login to perform this action.
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