An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at 30° and 90°, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at 60° with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is ______.

Let the intensity of the incident unpolarised light be $$I_0$$. When unpolarised light passes through a polariser, its intensity is reduced to half.

Case 1: Only the original two polarisers (axes at $$30^{\circ}$$ and $$90^{\circ}$$)
After the first polariser (axis $$30^{\circ}$$):
$$I_1 = \frac{I_0}{2}$$

The second polariser’s axis is at $$90^{\circ}$$, i.e. it is at an angle $$\theta = 90^{\circ}-30^{\circ}=60^{\circ}$$ to the first axis.
By Malus’ law, transmitted intensity is
$$I_{\text{without}} = I_1 \cos^{2} \theta = \frac{I_0}{2}\cos^{2}60^{\circ} = \frac{I_0}{2}\left(\frac{1}{2}\right)^{2} = \frac{I_0}{8}$$

Case 2: A third polariser (axis $$60^{\circ}$$) inserted between the first two
Order of axes: $$30^{\circ} \rightarrow 60^{\circ} \rightarrow 90^{\circ}$$.

Step-1 First polariser (axis $$30^{\circ}$$):
$$I_1 = \frac{I_0}{2}$$

Step-2 Second polariser (axis $$60^{\circ}$$) is at $$30^{\circ}$$ to the first.
$$I_2 = I_1 \cos^{2}30^{\circ} = \frac{I_0}{2}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{I_0}{2}\left(\frac{3}{4}\right) = \frac{3I_0}{8}$$

Step-3 Third polariser (axis $$90^{\circ}$$) is at $$30^{\circ}$$ to the second.
$$I_{\text{with}} = I_2 \cos^{2}30^{\circ} = \frac{3I_0}{8}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3I_0}{8}\left(\frac{3}{4}\right) = \frac{9I_0}{32}$$

Required ratio
$$\frac{I_{\text{with}}}{I_{\text{without}}} = \frac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \frac{9}{32}\times\frac{8}{1} = \frac{9}{4}$$

Hence, the intensity becomes $$\dfrac{9}{4}$$ times larger when the third polariser at $$60^{\circ}$$ is inserted.

Option C which is: $$\frac{9}{4}$$