# How to better view LCD through polarized glasses

When I put on my polarized sunglasses and look at my LCD monitor, I found the best viewing angle to be at $45^\text{o}$ to the left. Obviously the ergonomics of tilting my head to view my screen would be horrible.

I know why polarized sunglasses are polarized vertically (to reduce reflections off horizontal surfaces), and I have asked elsewhere why LCD monitors are polarized at $45^\text{o}$ (I suspect it's an engineering issue).

What I'd like to know is: Is there a way to rotate the polarization of the light from my monitor so that it's polarized vertically, without losing any intensity? Some combination of quarter-wave plates or something?

Why I'm asking: my workspace has very bright overhead lights which I'm not allowed to turn off. My eyes adjust for that, and it makes it hard to see the text on my monitor. If I wore my polarized glasses, that would block half the overhead light, but due to the polarization, it also blocks half the light from the monitor. If the light from the monitor were vertically polarized, then the glasses would be an effective filter, allowing me to see my screen more clearly.

From a theoretical point of view, you can get arbitrarily close to 100% transmission by using more and more linear polarizers. As the angle between each gets closer to zero, you get less polarization loss even as the number of filters increases.

Unfortunately for you, while the theoretical loss goes down, the efficiency of real filters are not 100% and the other losses would go up.

But you could try 2 or 3 linear polarizers. It could be that the monitor is bright enough that you would still be able to view it.

So for your normal case where $\theta$ is $\pi/4$, intensity becomes $$I = I_0 \cos^2(\pi/4)$$ $$I = 0.5 I_0$$

But if you had 3 more, each with an angle of $(\pi/16)$, you'd have a theoretical intensity of

$$I = I_0 (\cos^2(\pi/16))^4$$ $$I = 0.86 I_0$$

It would be whether or not the loss through having additional film removes any of that intensity gain.