A) 1.5 mm
B) 3 mm
C) 6 mm
D) 9 mm
Correct Answer: D
Solution :
Distance of the nth bright fringe from the centre\[{{x}_{n}}=\frac{n\lambda D}{d}\] Þ \[{{x}_{3}}=\frac{3\times 6000\times {{10}^{-10}}\times 2.5}{0.5\times {{10}^{-3}}}=9\times {{10}^{-3}}m=9\,mm.\]
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