Optics Question 326
Question: The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, it is reduced to $ \frac{I}{8} $ . The thickness of lead which will reduce the intensity to $ \frac{I}{2} $ will be
[AIEEE 2005]
Options:
A) 18 mm
B) 12 mm
C) 6 mm
D) 9 mm
Show Answer
Answer:
Correct Answer: B
Solution:
$ I’=I{{e}^{-\mu x}} $
Therefore $ 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=12mm $ .
BETA
