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Three circularly shaped linear polarisers are placed coaxially. The transmission axis of the first polariser is at $30^{\circ}$, the second one is at $60^{\circ}$ and the third at $90^{\circ}$ to the vertical all in the clockwise sense. Each polariser additionally absorbs $10 \%$ of the light. If a vertically polarised beam of light of intensity $\mathrm{I}=100$ $\mathrm{W} / \mathrm{m}^{2}$ is incident on this assembly of polarisers, then the final intensity of the transmitted light will be close to
PhysicsWave OpticsKVPYKVPY 2017 (19 Nov SB/SX)
Options:
  • A $10 \mathrm{~W} / \mathrm{m}^{2}$
  • B $20 \mathrm{~W} / \mathrm{m}^{2}$
  • C $30 \mathrm{~W} / \mathrm{m}^{2}$
  • D $50 \mathrm{~W} / \mathrm{m}^{2}$
Solution:
2417 Upvotes Verified Answer
The correct answer is: $30 \mathrm{~W} / \mathrm{m}^{2}$


$\mathrm{I}_{1}=\mathrm{I}_{0} \times 0.9 \cos ^{2} 30^{\circ}=\mathrm{I}_{0} \times 0.9 \times \frac{3}{4}$
$\mathrm{I}_{2}=\mathrm{I}_{1} \times 0.9 \cos ^{2} 30^{\circ}=\mathrm{I}_{1} \times 0.9 \times \frac{3}{4}$
$\mathrm{I}_{3}=\mathrm{I}_{2} \times 0.9 \cos ^{2} 30^{\circ}=\mathrm{I}_{2} \times 0.9 \times \frac{3}{4}$
$\Rightarrow \mathrm{I}_{3}=\mathrm{I}_{0}(0.9)^{3}\left(\frac{3}{4}\right)^{3}$
$\mathrm{I}_{3}=30.75 \mathrm{w} / \mathrm{m}^{2}$

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