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 $ .



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