JEE Main On 16 April 2018 Question 12
Question: Unpolarized light of intensity I is incident on a system of two polarizes, A followed by B. The intensity of emergent light is $ I/2. $ If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $ \theta . $ Then [JEE Main 16-4-2018]
Options:
A) $ \cos \theta ={{( \frac{2}{3} )}^{1/4}} $
B) $ \cos \theta ={{( \frac{1}{3} )}^{1/4}} $
C) $ \cos \theta ={{( \frac{1}{3} )}^{1/2}} $
D) $ \cos \theta ={{( \frac{2}{3} )}^{1/2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let initial intensity be I.
intensity of the beam after passing through A is $ I _1=\frac{I}{2} $
given that intensity after B is $ \frac{I}{2}. $
then angle between A and B is zero.
A polariser C is introduced between A and C then by Molus law after B
$ I _{b}=\frac{I}{2}{{\cos }^{2}}\theta $ and after $ C,I _{c}=I _{b}{{\cos }^{2}}\theta $
so given $ I _{c}=\frac{I}{3} $ from here solving all the three equations $ \frac{I}{3}=\frac{I}{2}{{\cos }^{4}}\theta $
$ \cos \theta ={{(\frac{2}{3})}^{\frac{1}{4}}} $
