Worked example 6.1.2: Glare, sunglasses and an LCD pixel

Without glasses $I=60+200=260W/{\mathrm{m}}^2$, of which $60/260=23\%$ is glare. With the glasses the glare field is at $\theta =90^{\circ }$ to the axis:

With the head tilted by $20^{\circ }$ the axis is $70^{\circ }$ from the glare field:

The diffuse part is still $100W/{\mathrm{m}}^2$, so $I=107W/{\mathrm{m}}^2$ with $7.02/107=6.6\%$ glare.

After the rear polariser the intensity is $I_0/2$. The cell turns the field by $\phi$, so it arrives at $90^{\circ }-\phi$ to the front axis:

The maximum, $50\%$, occurs at $\phi =90^{\circ }$ (no voltage, white), zero at $\phi =0$ (fully driven, black). Half of the maximum: ${\sin }^2\phi =\frac{1}{2}$, $\phi =45^{\circ }$.

With the head tilted by $90^{\circ }$ the glasses would pass all the glare and the result would be $60+100=160W/{\mathrm{m}}^2$, the formula $60{\cos }^2(90^{\circ }-90^{\circ })+100$.