Usefulness of half-shade in Laurent half-shade polarimeter
The principles of the Laurent half-shade polarizer are well explained in this tutorial. In essence, the polarimeter consists of a collimated linearly polarized light source (S, L and P) that is passed through the sample tube T (which will rotate the polarization if the sample is optically active), and then polarized (A) and seen through a telescope E.
In this basic setup (without the half-shade A) you are looking for the maximum and minimum brightness, which then tells you that the analyzer A is precisely aligned with the output rotation.
The half-shade H goes between the polarized light source and the sample, and it consists of two half-disks of equally absorptive material. One half, ACB, is glass, and it lets the polarized light pass through unchanged. The other half, ADB, is made of quartz, with the optical axes along AOB and DOC, and it is cut to half-wave-plate thickness; in essence, this means that if the original polarization was along SOP, in the quartz half it gets reflected over into ROQ, i.e. orthogonal to SOP.
In this version of the polarimeter, you're looking for the point where both halves are exactly equal in brightness, i.e. where the analyzer is at exactly $45°$ from the polarization of both halves of the image.
The usefulness of the half-shade is on the sensitivity of the apparatus. The normal polarimeter requires the experimenter to look for the maximum or minimum brightness, and this is a tricky thing to do: on one side, you don't have an objective standard (you're comparing to other angles right next to the one you're trying), but more importantly you're taking variations around a point where the signal looks quadratic, so even large changes in the angle won't change the output by that much.
To make this more evident, suppose that your eyes can detect variations in brightness of $\delta=10\%$ and bigger. In the non-shaded polarimeter, this means that all points in the region $\cos^2(\theta)>1-\delta$ will look exactly like the maximum, so that is your range of uncertainty, which looks like this:
With the half-shade, on the other hand, you're comparing two different signals at the point where they match, which means that you have an immediate reference (the signals are equal or they're not, and you don't have to fiddle with the analyzer angle to check whether you're at the correct spot or not.
More importantly, though, you're comparing two signals that have positive slope, and this does wonders for the sensitivity. If you keep that same $\delta=10\%$ deviation of either signal with respect to the baseline, what you get looks like this:
It is immediately obvious that the larger slope makes the 'shadow' on the angle axis, which gives the range of uncertainty is much smaller. In fact, this is really over-counting the uncertainty, since you're probably able to detect $\delta=10\%$ differences in the brightness of the two halves, which means that you should slice in two the vertical range and therefore (because of linearity) the horizontal range should also be half as big.
It's important to emphasize that this is not just a one-off sort of difference: in fact, the difference in performance increases more and more as the sensitivity in the detector (i.e. your eyes, and eventually a pair of photodiodes on the two halves of the half-shade) improves. This is because the relationship between the angle uncertainty $\delta\theta$ and the signal uncertainty $\delta$ is quadratic for the non-shaded apparatus, since $$ \delta\sim\frac12 \delta\theta^2, $$ from the Taylor expansion of the cosine about its maximum, whereas the relationship is linear, $$ \delta\sim\delta\theta, $$ for situations with linear slope. This is the core of the method, and it is replayed over and over again in metrology - you get better measurements with differential measurements of quantities that have a sensitive change with respect to the thing you care about, rather than looking at the extrema where the change is more shallow.