A) 18 mm
B) 12 mm
C) 6 mm
D) 9 mm
Correct Answer: B
Solution :
\[I'=I\,{{e}^{-\mu x}}\] Þ \[x=\frac{1}{\mu }{{\log }_{e}}\frac{I}{I\,'}\] (where I = original intensity, I' = changed intensity) \[36=\frac{1}{\mu }{{\log }_{e}}\frac{I}{I/8}\] = \[\frac{3}{\mu }{{\log }_{e}}2\] ....(i) \[x=\frac{1}{\mu }{{\log }_{e}}\frac{I}{I/2}\] \[=\frac{1}{\mu }{{\log }_{e}}2\] .....(ii) From equation (i) and (ii), \[x=12\,mm\].
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