Home exercise 6.1.3: Three polarisers

Unpolarised light loses half: $I_1=I_0/2=400W/{\mathrm{m}}^2$, now vertically polarised.

The horizontal sheet is at $90^{\circ }$ to the field: $I_2=400{\cos }^{2}90^{\circ }=0$ (crossed polarisers).

Sheet 3 at $\theta$ passes $400{\cos }^2\theta$ and leaves the field along its own axis. That field is at $90^{\circ }-\theta$ to sheet 2:

At $\theta =0$ and $\theta =90^{\circ }$ the third sheet is parallel to one of the others and the pair is crossed again: $I=0$, as the formula gives.