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

• A

  $10 \mathrm{~W} / \mathrm{m}^{2}$
• B

  $20 \mathrm{~W} / \mathrm{m}^{2}$
• Correct answer
  C

  $30 \mathrm{~W} / \mathrm{m}^{2}$
• D

  $50 \mathrm{~W} / \mathrm{m}^{2}$

Solution

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

Hence, the correct answer is option: 'C'.