Unpolarized light of intensity I is incident on a system of two polarizers, 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 :

cos \theta = ( 2 \over 3 )^\frac14

cos \theta = ( 2 \over 3 )^\frac12

cos \theta = ( 1 \over 3 )^\frac12

cos \theta = ( 1 \over 3 )^\frac14

Wave Optics PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q51 with Solution

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Correct answer: B

Unpolarized light of intensity $I$ falls on polarizer $A$ . After passing through the first polarizer, intensity becomes

$I_A=\frac{I}{2}$

Now this light passes through polarizer $B$ . Let the angle between axes of $A$ and $B$ be $\phi$ . By Malus' law,

$I_{AB}=\frac{I}{2}\cos^2\phi$

Given that emergent intensity is $\frac{I}{2}$ , so

$\frac{I}{2}\cos^2\phi=\frac{I}{2}$

$\cos^2\phi=1$

Hence,

$\phi=0^\circ$

So polarizers $A$ and $B$ are parallel.

Let the angle between $A$ and $C$ be $\theta$ . Since $B$ is parallel to $A$ , the angle between $C$ and $B$ is also $\theta$ .

After $A$ , intensity is

$I_1=\frac{I}{2}$

After $C$ ,

$I_2=\frac{I}{2}\cos^2\theta$

After $B$ ,

=\frac{I}{2}\cos^4\theta$$ Given that this final intensity is reduced to $\frac{I}{3}$: $$\frac{I}{2}\cos^4\theta=\frac{I}{3}$$ Cancel $I$: $$\frac{1}{2}\cos^4\theta=\frac{1}{3}$$ $$\cos^4\theta=\frac{2}{3}$$ Therefore, $$\cos\theta=\left(\frac{2}{3}\right)^{1/4}$$ --- 3. **Match with options** This corresponds to **Option B**. $$\boxed{\cos\theta=\left(\frac{2}{3}\right)^{1/4}}$$