### 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}}} $