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