A) 0.50 mm
B) 1.25 mm
C) 1.50 mm
D) 1.75 mm
Correct Answer: B
Solution :
Distance of nth minima from central bright fringe \[{{x}_{n}}=\frac{(2n-1)\lambda D}{2d}\] For n=3 i.e. 3rd minima \[{{x}_{3}}=\frac{(2\times 3-1)\times 500\times {{10}^{-9}}\times 1}{2\times 1\times {{10}^{-3}}}\] \[=\frac{5\times 500\times {{10}^{-6}}}{2}=1.25\times {{10}^{-3}}m=1.25\ mm.\]
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