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 unpolarized light be \(I_0\).

After passing through the first polarizer, the intensity becomes \(I_1 = \dfrac{I_0}{2}\).

Without the third polarizer, the light passes directly from the first polarizer (at \(30^{\circ}\)) to the second polarizer (at \(90^{\circ}\)). The angle between their transmission axes is \(90^{\circ} - 30^{\circ} = 60^{\circ}\).

Using Malus's law, the output intensity without the third polarizer is:
\(I_{without} = I_1 \cos^2(60^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}\)

When the third polarizer (at \(60^{\circ}\)) is placed between them, the light first passes from the first polarizer to the third polarizer. The angle between their axes is \(60^{\circ} - 30^{\circ} = 30^{\circ}\).

The intensity after the third polarizer is:
\(I_3 = I_1 \cos^2(30^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{8}\)

The light then passes from the third polarizer to the second polarizer. The angle between their axes is \(90^{\circ} - 60^{\circ} = 30^{\circ}\).

The final output intensity with the third polarizer is:
\(I_{with} = I_3 \cos^2(30^{\circ}) = \dfrac{3I_0}{8} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{9I_0}{32}\)

The ratio of the output intensities with and without the third polarizer is:
\(\text{Ratio} = \dfrac{I_{with}}{I_{without}} = \dfrac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \dfrac{9}{4}\)

Answer: \(9/4\)