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