PHYS 158 Thin Film Interference

In thin film questions, we’re mostly concerning ourself with the general case — a singular light wave travels through a medium, and it enters a different medium which has a thin film of material on it.

Two common examples are:

1. Light travels from air into water, but the water has a thin layer of oil on top of it.

2. Light travels from air into glass, and the glass has a thin exterior coating.

Our goal is to understand what is seen as a result of these thin surface films. This guide will provide an idea on how to set up thin film problems for PHYS 158.

Below is the formula for the interference of two waves. Whenever we use this formula to solve a problem, it’s common to call this the “argument method”. We use the argument method to understand if two waves constructively or destructively interfere with each other given their properties.

• ‘Argument Method’ proof

  

The question is, what second wave are we comparing to? Aren’t we only considering a singular light wave? Let’s look at the setup for a thin film problem:

As light enters through a film, it reflects at two points, (1) at the interface between the film and the original medium, and (2) at the interface between the film and the background medium. The properties of these two reflected waves (wave 1 and wave 2) are what we will use for our argument expression.

Note: This diagram is not completely accurate — by Snell’s law the light wave path in the medium should not be at the same angle as in air.

Omitting the argument, let’s start to understand the terms in relation to our two reflected waves.

We know that $\Delta r$ is the path difference between wave 1 and wave 2. Zooming into our diagram:

The path difference is the extra distance that wave 2 (W2) travels within the film. Clearly, this distance is dependent on $\theta$, or the angle of entry that the light enters the medium from. If $\theta$ = 90°, the path difference would be equal to 2t (twice the thickness of the medium). Normally, this is what we assume for our thin film questions, since its a decent approximation at small scales. Of course, you can modify the path difference based on the angle it of entry if a question specifies. For explanation purposes, I’ll continue to draw the light entering the film at an angle.

Next, let’s take a look at $\Delta \phi$. As shown below, waves 1 and 2 have their own respective phase difference in respect to the original light wave. Though as of now, we might not know what exactly the phase differences are, since the light travelled through different mediums, we know that they may not be the same.

To obtain $\phi_1$ and $\phi_2$ requires a little background on boundary conditions. The rule is simple:

Here, $n_{a/b}$ refers to the refractive index of the medium in which the wave travels. $\phi$ of the reflected wave is based on the boundary condition, which is dependent on the two index of refractions.

In the context of thin film, we have two boundaries, (1) at the interface between the film and the original medium, and (2) at the interface between the film and the background medium. The phase of wave 1 is based on $n_1$ and $n_{film}$, and the phase of wave 2 is based on $n_{film}$and $n_2$.

Here’s an example to clear things up:

One important note — a common mistake is to say that the phase difference in this case is $\pi$. Recall that to calculate path difference, we calculated $\Delta r = r_2 - r_1 \approx 2t$. Keeping consistent, the phase difference is given by $\Delta \phi = \phi_2 - \phi_1 = -\pi$.

Lastly, as you may recall from 157, k is the wave number, which is equal to $2\pi$ / wavelength. In thin film, this wavelength is the wavelength in the film. Questions will normally give you the wavelength of light in air, so we convert accordingly using the following formula:

The last step is to return to $\Delta\text{arg}$, and what it means in the context of our problems. Here we introduce the - up to this point ignored - last piece of our model, the final transmitted wave.

Sometimes a question may ask if the reflected waves constructively or destructively interfere — in this case, we don’t care about the transmitted wave. More commonly however, a question will ask what happens to a transmitted wave as a result of thin film interference. For this, we just need to remember that:

1. MAXIMUM TRANSMISSION (highest visibility) of a transmitted wave means the two reflective waves will destructively interfere

2. MINIMUM TRANSMISSION (lowest visibility) of a transmitted wave means the two reflected waves will constructively interfere

Knowing this, we can link minimum/maximum transmission to constructive/destructive interference, and thus understand what our argument term should be.

Here is an example question going over everything we’ve learned:

The key steps (which apply for nearly all thin film questions in 158) are:

1. Understand what our argument should be equal to
   1. In this case, we want maximum visibility/transmission of the bug’s light, so we want a destructive interference of the reflected waves

2. Calculate unknown variables for problem setup
   1. Calculated the phase difference of the two waves (using refractive indexes)
   2. Calculated the wavelength of the blue light within the film

3. Rearrange for what we want to solve for
   1. In this case, we want to find film thickness, so we rearrange everything with our final unknown being n_odd

4. Solve
   1. Set n_odd = 1 to minimize film thickness
   2. Continuously increment n_odd until the acceptable film thickness is greater than 300 nm.