Q108 / 150 · MHT CET 2025 (20 Apr Shift 2) · Physics

Four polaroids are placed such that the optic axis of each is inclined at an angle of $30^{\circ}$ with the optic axis of the preceding one. If unpolarised light of intensity ' $\mathrm{I}_0$ ' falls on the first polaroid, the intensity of light transmitted from the fourth polaroid is $[\cos 30=\sqrt{3} / 2]$

  • A
    $\frac{9 \mathrm{I}_0}{32}$
  • Correct answer
    B
    $\frac{27 \mathrm{I}_0}{128}$
  • C
    $\frac{35 \mathrm{I}_0}{128}$
  • D
    $\frac{27 \mathrm{I}_0}{32}$

Solution

Unpolarized light of intensity $I_0$ passes through the first polarizer, reducing the intensity by half: $I_1 = \frac{I_0}{2}$.

The second polarizer is oriented at $30^\circ$ to the first. Applying Malus's Law: $I_2 = I_1 \cos^2 30^\circ = \frac{I_0}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3I_0}{8}$.

Each subsequent polarizer is rotated an additional $30^\circ$ from the previous one, multiplying by $\cos^2 30^\circ = \frac{3}{4}$ each time.

After three such rotations: $I_4 = \frac{I_0}{2} \cdot \left(\frac{3}{4}\right)^3 = \frac{27I_0}{128}$.

The final intensity is $\boxed{\text{B}}$.

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

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