A) 0.10 mm
B) 0.15 mm
C) 0.20 mm
D) 0.25 mm
Correct Answer: D
Solution :
\[\beta =0.10\] mm \[=0.10\times {{10}^{-3}}m\] \[\beta =\frac{\lambda D}{d}\Rightarrow \frac{\beta }{\lambda }=\frac{D}{d}\] Again \[\beta =\frac{4.0\times {{10}^{7}}\times D}{\frac{d}{2}}\] \[\beta =\frac{4.0\times {{10}^{-7}}\times 4D}{d}\] \[\beta =\frac{4.0\times {{10}^{-7}}\times 4\times 0.10\times {{10}^{-3}}}{6.4\times {{10}^{-7}}}\] \[\beta =0.25\,mm\]
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