Unpolarised light of intensity I0 falls on a polariser P1. A second polariser P3 with its axis perpendicular to that of P1 is placed behind it, and…

The first polariser transmits half of the unpolarised intensity: $I_1=\dfrac{I_0}{2}$. By Malus's law, the intensity after $P_2$ (axis at $30^\circ$ to $P_1$) is $I_2=I_1\cos^230^\circ=\dfrac{I_0}{2}\cdot\dfrac34=\dfrac{3I_0}{8}$. The axis of $P_3$ is at $90^\circ$ to $P_1$, so it makes $60^\circ$ with $P_2$, and $I_3=I_2\cos^260^\circ=\dfrac{3I_0}{8}\times\dfrac14=\dfrac{3I_0}{32}$. Without $P_2$ the crossed pair would block all light; the middle polariser rotates the plane of polarisation in two steps. The value $\dfrac{I_0}{8}$ is the result for the $45^\circ$ arrangement.