Unpolarized light of intensity I passes through an ideal polariser A . Another identical polariser B is placed behind A . The intensity of light beyond B is found to be I 2 . Now another identical polariser C is placed between A and B . The intensity beyond B is now found to be I 8 . The angle between polariser A and C is

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsPolarizationMedium2 minPYQ_2018

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Unpolarized light of intensity $I$ passes through an ideal polariser $A$ . Another identical polariser $B$ is placed behind $A$ . The intensity of light beyond B is found to be $\frac{I}{2}$ . Now another identical polariser $C$ is placed between $A$ and $B$ . The intensity beyond B is now found to be $\frac{I}{8}$ . The angle between polariser $A$ and $C$ is

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Answer:

Solution:

Transformation axis $A$ and $B$ are parallel. If the transformation axis of $C$ is making angle $θ$ with $A$ , it will make angle $(90 - θ)$ with $B$ . Therefore,

$\Rightarrow \frac{I}{2} \cos^{2} {θcos}^{2} (90 - θ) = \frac{I}{8}$ $\Rightarrow \frac{1}{4} \sin^{2} 2 θ = \frac{1}{4}$
$\Rightarrow 2 θ = \frac{π}{2}$ $\Rightarrow θ = \frac{π}{4}$

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:Polarization

⏱ 2mℹ️ Source: PYQ_2018

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