Let the intensity of unpolarised light be $I_o$. When unpolarised light passes through the first polaroid $P_1$, its intensity becomes half: $I_1 = \frac{I_o}{2}$ The transmitted light is now plane-polarised. The third polaroid $P_3$ is placed between $P_1$ and $P_2$ with its pass axis at an angle of $45^\circ$ with respect to $P_1$. According to Malus's Law, the intensity of light transmitted through $P_3$ is: $I_3 = I_1 \cos^2(45^\circ) = \frac{I_o}{2} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{2} \times \frac{1}{2} = \frac{I_o}{4}$ The polaroids $P_1$ and $P_2$ are crossed (their axes are perpendicular to each other). Since $P_3$ makes an angle of $45^\circ$ with $P_1$, the angle between the transmission axes of $P_3$ and $P_2$ is $90^\circ - 45^\circ = 45^\circ$. Again, applying Malus's Law, the intensity of light transmitted through $P_2$ is: $I_2 = I_3 \cos^2(45^\circ) = \frac{I_o}{4} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{4} \times \frac{1}{2} = \frac{I_o}{8}$