Ratio of Output Intensities With and Without Intermediate Polarizer

Let the initial intensity of the unpolarized light be $I_0$.

Case 1: Without the third polarizer

1. When unpolarized light of intensity $I_0$ passes through the first polarizer (P1) oriented at $\theta_1 = 30^\circ$ with respect to the horizontal axis, the intensity of light transmitted through P1 is: $I_1 = \frac{I_0}{2}$

2. The transmitted light from P1 is linearly polarized at $30^\circ$. It then passes through the second polarizer (P2) oriented at $\theta_2 = 90^\circ$ with respect to the horizontal axis. The angle between the transmission axes of P1 and P2 is: $\Delta\theta = 90^\circ - 30^\circ = 60^\circ$

3. Using Malus's Law, the final transmitted intensity without the third polarizer ($I_{\text{without}}$) is: $I_{\text{without}} = I_1 \cos^2(60^\circ) = \left(\frac{I_0}{2}\right) \left(\frac{1}{2}\right)^2 = \frac{I_0}{8}$

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Case 2: With the third polarizer inserted in between

1. A third polarizer (P3) is placed between P1 and P2 with its transmission axis at $\theta_3 = 60^\circ$ to the horizontal axis.

2. The angle between P1 ($\theta_1 = 30^\circ$) and P3 ($\theta_3 = 60^\circ$) is: $\Delta\theta_{13} = 60^\circ - 30^\circ = 30^\circ$ The intensity of light emerging from P3 is: $I_3 = I_1 \cos^2(30^\circ) = \left(\frac{I_0}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3 I_0}{8}$

3. The angle between P3 ($\theta_3 = 60^\circ$) and P2 ($\theta_2 = 90^\circ$) is: $\Delta\theta_{32} = 90^\circ - 60^\circ = 30^\circ$ The final output intensity with the third polarizer ($I_{\text{with}}$) is: $I_{\text{with}} = I_3 \cos^2(30^\circ) = \left(\frac{3 I_0}{8}\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{9 I_0}{32}$

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Ratio of Output Intensities

The required ratio of the output intensities with and without the third polarizer is: $\text{Ratio} = \frac{I_{\text{with}}}{I_{\text{without}}} = \frac{\frac{9 I_0}{32}}{\frac{I_0}{8}} = \frac{9}{32} \times 8 = \frac{9}{4}$

Thus, the correct answer is Option C.