Question

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 :

• A. $\cos \theta = \left( \frac{2}{3} \right)^{1/2}$
• B. $\cos \theta = \left( \frac{2}{3} \right)^{1/4}$
• C. $\cos \theta = \left( \frac{1}{3} \right)^{1/2}$
• D. $\cos \theta = \left( \frac{1}{3} \right)^{1/4}$

Answer 1

Answer: (B)

As intensity of emergent beam is reduced to half after passing through two polarisers.
It means angle between A and B is $0^{\circ}$

Now, on placing between $A$ and $B$.

Intensity after passing through $A$ is
Let $\theta$ is angle between $A$ and $C$. Intensity of light after passing through $C$ is given by
$I_{ C }=\frac{I}{2} \cos ^{2} \theta$
Intensity after passing through polariser $B$ is $\frac{I}{3}$.
Angle between $C$ and $B$ is also $\theta$ as $A$ is parallel to $B$.
So, $\frac{I}{2}=I_{c} \cos ^{2} \theta=\frac{I}{2} \cos ^{2} \theta \cdot \cos ^{2} \theta$
$\cos ^{4} \theta=\frac{2}{3}$
$\Rightarrow \cos \theta=\left(\frac{2}{3}\right)^{\frac{1}{4}}$