To solve this problem, we will use the concept of polarization of light.$\\$When unpolarized light of intensity $I_0$ is incident on a polaroid $P_1,$ the intensity of$\\$light transmitted through $P_1$ is given by:$\\ I_1 = \frac{I_0}{2} \\$This is because a polaroid transmits only the component of light polarized along its axis,$\\$which is half of the unpolarized light intensity.$\\$The third polaroid $P_3$ is placed between $P_1$ and $P_2$ with its axis making an angle $45^\circ$ with $\\$the axis of $P_1.$ The intensity of light transmitted through $P_3$ is given by Malus's Law:$\\ I_2 = I_1 \cos^2(45^\circ) \\$Substitute $I_1 = \frac{I_0}{2}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}: \\ I_2 = \frac{I_0}{2} \left(\frac{1}{\sqrt{2}}\right)^2 \\ I_2 = \frac{I_0}{2} \cdot \frac{1}{2} \\ I_2 = \frac{I_0}{4} \\$Finally, the light is incident on $P_2,$ which is perpendicular to $P_1.$ The intensity of light$\\$transmitted through $P_2$ is again given by Malus's Law:$\\ I_3 = I_2 \cos^2(45^\circ) \\$Substitute $I_2 = \frac{I_0}{4}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}: \\ I_3 = \frac{I_0}{4} \left(\frac{1}{\sqrt{2}}\right)^2 \\ I_3 = \frac{I_0}{4} \cdot \frac{1}{2} \\ I_3 = \frac{I_0}{8} \\$Therefore, the intensity of transmitted light through $P_2$ is $\frac{I_0}{8}. \\$This corresponds to Option 2.