Half wave plate + polarizer vs rotatable polarizer + fixed polarizer

Question

I am learning about optical setups and am looking at how the intensity of the laser beam is attenuated for use. A lab in my institute uses a rotatable Half Wave plate and a polarizer fixed at the laser polarisation. Let's call this Setup 1.

Intensity attenuation can also be achieved by simply putting two polarizers: the first polarizer rotated at an angle and the second fixed at the laser polarisation. Let's call this setup 2.

Assume my laser polarisation is along the horizontal axis and the Half wave plate in setup 1 and the polarizer in setup 2 are rotated with respect to the laser polarisation by angle $\theta$ (in the same direction).

Calculating the final intensity of the beam after it passes through the fixed polarizers would give me:

For setup 1: $I = I_0 \cos^2(2\theta)$.

For setup 2: $I = I_0 \cos^4(\theta)$.

Here, $I$ is the final intensity and $I_0$ is the laser beam intensity before it interacts with any optics.

In an experimental setup I will vary $\theta$ to control the intensity of the beam. So in order to have a better control over the intensity, I would choose the setup whose intensity varies slower as $\theta$ is varied. Looking at the graphs of $cos^2(2x)$ and $cos^4(x)$ I can see that setup 2 would provide me with better control, but that is not being used at the lab.

I am trying to understand why the lab is using setup 1 over a much cheaper setup 2. Am I missing something? Are any of my assumptions incorrect in the previous paragraph? or are there any experimental considerations that favour the use of setup 1 over setup 2?

Answer 1

To see what is happening in the two setup scenarios, it is convenient to use the optical calculus simulation program previously used in answers here and here. The two setups are shown in the figure below, where the blocks are the program.

The rotation angle, for both the half wave plate (HWP) in setup 1 and the rotated polarizer in setup 2, is $\theta$, in degrees. In the simulation, $\theta$ runs from 0 to 360 degrees, in increments of 1 degree.

The linear polarizers are assumed to be ideal, though it is an easy change to accommodate non-zero polarizer extinction ratios. The laser is modeled by two optical elements: an unpolarized Stokes vector with intensity 2 followed by an x oriented linear polarizer. Thus the output of the laser has unity intensity and is horizontally, i.e., x, polarized.

Running the simulation produces the following results, where the black trace is for setup 1 and red trace is for setup 2:

This clearly shows that setup 2 varies the laser intensity more gradually than does setup 1. As well, for a single full cycle of $\theta$, setup 2 cycles half as fast through maxima and minima.

The next two figures simply show the Mueller calculus calculations of the transmitted laser intensity for the two setups.

The results above assumed ideal polarizers. Among other things, the polarizers have 100% transmittance (T = 1) for linearly polarized light aligned with the polarizer. Thus, x polarized light is 100% transmitted through an ideal x linear polarizer and y polarized light is completely blocked by an ideal x linear polarizer.

Answer 2

A typical film polarizer will have a transmittance of roughly 70-90% so passing through two polarizers will leave you with 50-81% of the initial power assuming your input is perfectly polarized and parallel with the polarizers. Passing through one polarizer will give you $$I_1 = I_0\cos^2(\theta_1)\cdot T$$ Passing through the next will give you $$I_2 = I_1\cos^2(\theta_2)\cdot T = I_0\cos^2(\theta_1)\cos^2(\theta_2) \cdot T^2$$ $T$ is the transmittance of the polarizers.

A typical half-wave plate has a much higher transmission (Up to 99.9%) and you only have to pass through one polarizer.

If you hold one of the polarizers constant in setup 2, you see that you have the same level of control over the power of the beam by adjusting the other polarizer.

Setup 1 gives you flexibility over the polarization angle of the beam (something setup 2 cannot do) and has a higher maximum power throughput since you have a better transmittance.

Answer 3

One way the OP could answer their own question would be to perform the comparison, of their setups 1 and 2, in their own lab. For various reasons, this may be infeasible. But other options exist, e.g., optical calculus-based computer simulations and simple manual calculations, as shown here.

Alternatively, it is not difficult to perform an actual experimental comparison of setup 1 and 2. It should be noted that attenuating a high power laser beam is not the same as attenuating a low power laser beam. Aside from safety considerations, factors such as optical component damage thresholds must be considered.

With that important caveat, consider the breadboarded system below:

The laser, not yet powered up in the photo, is a USB-powered 10 mW diode laser with measured wavelength of 485 nm. It is already somewhat polarized, but not very well. The laser beam passes from left to right through a calcite polarizer, an adjustable iris aperture, the rotating optical component (either a half wave plate or a calcite polarizer), the calcite analyzer polarizer and finally is incident upon the photodiode. The two fixed orientation polarizers, on either side of the rotation mount, have their polarization axes aligned. As well, the diode laser was initially rotated so that its polarization axis aligned with those of the fixed pair. Consequently, transmission is maximum and attenuation is minimum if there is no optical component in the rotation mount.

In the photo, the calcite polarizer is mounted in the rotation mount, as per setup 2. The photocurrent from the photodiode is manually measured using a digital multimeter in ammeter mode.

The next photo shows the system when the diode laser is powered on and the calcite polarizer is in the rotation mount.

All that remains is to measure the photocurrent as a function of rotation angle of whichever optical component was in the rotation mount. This was done for both setup 1, with the 488 nm half wave plate (HWP) in the rotation mount, and for setup 2, with another calcite polarizer in the rotation mount.

Data were collected manually for every 10 degrees of rotation of the respective optical component. To facilitate comparison with the expected intensities versus rotation angle, the photocurrents were normalized by 1.57 mA, for setup 1, and by 1.52 mA, for setup 2. The raw (unnormalized) results are shown in the table:

Plotting the normalized photocurrents versus the rotation angle yields the following figure:

This confirms the OP’s intuition: setup 2 provides better control of the laser attenuation. Furthermore, polarizer transmittance is not a factor with the calcite polarizers used here. This may be an issue with lower quality polarizers, but those would not be used to attenuate a laser beam of, say, 1 W power.

Experiment notes:

1. Since I made the photocurrent measurements every 10 degrees of rotation, I did not make measurements at 45 or 135 degrees. It would not matter for setup 2, but it would be useful information for setup 1 because attenuation is maximum there. So I did the two measurements, with these results: 0.0353 mA photocurrent at 45 deg and 0.0348 mA photocurrent at 135 deg.

2. In setup 2, it was easy to attenuate the laser to the point where I could not see any transmitted light. With setup 1, this was not the case: even with maximum attenuation, a little light was transmitted. The data show this as well. My hypothesis is that this was due to the facts that the 485 nm laser center wavelength is not exactly the same as the 488 nm HWP wavelength and the diode laser’s bandwidth is significantly broader than that of the 488 nm argon ion laser line, which is what the HWP was intended for use with.