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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Wave Optics

JEE Main 2018 · 8 April · Physics · Wave Optics

Unpolarized light of intensity I passes through an ideal polariser...

Unpolarized light of intensity II passes through an ideal polariser AA. Another identical polariser BB is placed behind AA. The intensity of light beyond B is found to be I2\dfrac{I}{2}. Now another identical polariser CC is placed between AA and BB. The intensity beyond B is now found to be I8\dfrac{I}{8} . The angle between polariser AA and CC isUnpolarized light of intensity I passes through an ideal polariser A . Another identical polariser B is placed behind AUnpolarized light of intensity I passes through an ideal polariser A . Another identical polariser B is placed behind A

Options

  1. A60°60 °
  2. B0°0 °
  3. C30°30 °
  4. D45°45 °

Correct answer

D. 45°45 °

Step-by-step solution

Unpolarized light of intensity I passes through an ideal polariser A . Another identical polariser B is placed behind A

Transformation axis AandBA and B are parallel. If~the transformation& axis of CC is making angle θ\theta with AA, it will make angle (90−θ)\left(90 - \theta\right) with BB. Therefore,

⇒I2cos⁡2θcos2(90−θ)=I8⇒14sin⁡22θ=14⇒ \dfrac{I}{2}\cos^{2}{θcos}^{2}( 90 - \theta ) = \dfrac{I}{8}⇒ \dfrac{1}{4}\sin^{2}2\theta = \dfrac{1}{4}
⇒2θ=π2⇒θ=π4⇒ 2\theta = \dfrac{π}{2}⇒ \theta = \dfrac{π}{4}
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