A) 10 mm
B) 0.5 mm
C) 20 mm
D) 15 mm
Correct Answer: B
Solution :
Given\[d=1\times {{10}^{-3}}m,\lambda =500\times {{10}^{-9}}m\]and\[D=1m.\] \[Fringe\text{ }separation\text{ }\beta =\frac{\lambda D}{d}\] \[=\frac{500\times {{10}^{-3}}\times 1}{1\times {{10}^{-3}}}\] \[=500\times {{10}^{-6}}\] \[=0.5\times {{10}^{-3}}m\] \[=0.5\text{ }mm\]
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