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

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsPolarizationEasy2 minPYQ_2020

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

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Solution:

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

$\frac{I}{2} \cos^{2} {θcos}^{2} (90 - θ) = \frac{I}{8}$

$\frac{1}{4} \sin^{2} 2 θ = \frac{1}{4}$

$2 θ = \frac{π}{2}$

$θ = \frac{π}{4}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Wave OpticsSubtopic:Polarization

⏱ 2mℹ️ Source: PYQ_2020

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