Polarizing Puzzler
Logic, Trigonometry ; Difficulty: Easy
A Polarizing Puzzler from The Riddler
This week’s Riddler Classic about polarized light is a math+physics puzzle. The question is reproduced below from the Riddler site.
I have a light source that’s polarized in the vertical direction, but I want it to be polarized in the horizontal direction. To make that happen I need … polarizers! When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they aren’t perfectly aligned, only the component of the light that’s in the direction of the polarizer passes through.
Unfortunately, that means I can’t turn vertically polarized light into horizontally polarized light with a single polarizer. But I can do it with two polarizers, as shown below.
If the first polarizer is positioned at an acute angle with respect to the light, and then the second polarizer is positioned at a complementary angle with respect to the first polarizer, some light will be horizontally polarized in the end.
Now, I have tons of polarizers, and each one also reflects 1 percent of any light that hits it — no matter its polarization or orientation — while polarizing the remaining 99 percent of the light.
I’m interested in horizontally polarizing as much of the incoming light as possible. How many polarizers should I use?
*spoiler alert* My solution below
Understanding the question: Malus' law
The physics of light is endlessly fascinating and if nothing else, this reasonably easy puzzle should provoke some more curiosity (among the uninitiated) about the electromagnetic field, polarization of light and it’s numerous applications (sunglasses, photography, LCDs…)
This video illustrates the central principle of the puzzle: how the use of two polarizers allows rotation of polarization by \(90^{\large\circ} \)
What’s most relevant to solving the puzzle though is the relationship between amplitude and intensity of transverse waves: intensity is proportional to the square of the amplitude. What this implies is that the first polarizer placed at an angle of \( \theta \) changes the amplitude of the wave by \(cos\theta \) and the intensity is now \(cos ^2\theta \) times the original. This is known as Malus’s law. You can find a good explanation here. Or this video if you prefer that format.
Accounting for the 1% loss due to reflection, the intensity transmitted after one polarizer at an angle \( \theta \) is \(0.99 \ cos^2\theta \) .
Now if there was no loss to reflection, then you could achieve as near 100% conversion as desired by using an appropriately large number of polarizers, varying the angle of each infinitesimally from the previous. But with reflection loss there is a tax to be paid with the use of each additional polarizer resulting in a limit to the number of polarizers when the reflection tax becomes greater than the benefit of smaller cosines!
Solving the problem
Let’s say we use \(n\) polarizers set up at equal (acute) angles spanning the \(90^{\large\circ} \). I’m assuming we start with vertically polarized light. The total final Intensity \(I_{hor} \) achieved from the initial \(I_{ver}\) is
\[
I_{hor} = I_{ver} . 0.99^n .Cos^{2n}\left (\frac{90^{\large\circ}}{n}\right )
\]
It’s then relatively easy to calculate the \(n\) for which \(I_{hor}\) is maximized. Turns out 16 filters gives the maximum: 73% of the original intensity when 1% is reflected. The angle between adjacent filters is then \(5.625^{\large\circ} \)
As you would expect, as the reflection loss % increases, you are able to achieve a much smaller final intensity of rotation and forced to use fewer polarizers. In fact, when the reflection loss increases to ~41% or greater, 2 polarizers at \(45^{\large\circ} \) is optimal.
Now all that remains to be shown is that the equi angular spacing is indeed optimum and that we cannot do any better. The mere expectation of symmetry may not be sufficient for some 🙂
Why equal angles?
Consider using only 2 polarizers used to rotate the polarization of unit intensity polarized light by a total angle \(\theta\). Let the first one be placed at angle \(\phi\) as shown.
The final rotated intensity is:
\[
I = k^2\ Cos^2\phi \ Cos^2(\theta – \phi))
\]
Differentiating w.r.t. \(\phi \):
\[
\frac{dI}{d\phi} = k^2 \ Cos(\theta – \phi) \ Cos(\phi)\left [ Sin(\theta – \phi)Cos(\phi) – \ Cos(\theta – \phi) Sin(\phi) \right ]
\]
If you set this to zero, you can see that \(\displaystyle \phi\ = \frac{\theta}{2} \) is the solution. You can evaluate the second order differential to convince yourself that it is negative making the solution a maximum.
What this implies is that if any two of our \(n\) polarizers were not separated by equal angles we could always improve on the transmitted intensity by making it so. Extending this reasoning across all polarizers implies that the only way to have a solution that we cannot improve on is when the angles between all adjacent polarizers is the same!
What about all the re-reflections from the polarizers?
For the small 1% reflection value, it’s safe to ignore all the infinite reflections between polarizers. For the case of 2 polarizers at 45 degrees each, the difference in final intensity when accounting for the reflections is ~0.02%.