May 18, 2017

Soap Bubble Physics

This week, I rediscovered that you can estimate how heavy a bubble is by looking at its colour.


A boy adds a small amount of detergent to some water, and uses the resulting mixture to blow soap bubbles. The bubbles are mostly transparent but have a faint violet tint. Estimate the mass of a soap bubble. (The refractive index of water is 1.33. Assume spherical bubbles of diameter 3 cm.)

These questions in which seemingly impossible, yet simple conclusions are drawn highlight all that is beautiful about physics. In this instance, we’re talking about soap bubbles and colours; more specifically, the thin film diffraction of the light as it passes through the surface of the spherical liquid solution. In order to solve this problem from scratch, we must first be familiar with the physics behind waves, diffraction, interference, and refraction. However, scratch is a somewhat misleading term because there are always prior principles upon which models are built: one can ask why over and over again and never be satisfied. Thus, it should suffice for us to examine the main equations involved in thin film interference without understanding underlying mechanisms in too much detail, especially if we aren’t physicists by profession.

When waves enter a new medium at an angle, they are distorted because the speed of their propagation changes. Light is no different, and this phenomenon (Huygen’s Principle) is responsible for the bending of light in water and air. This bending of the light, in addition to a decrease in its speed (and some other geometric properties) causes a path length difference and phase shift (don’t worry about those terms if they are unfamiliar; you can Google them if you’re curious) which in turn cause some parts of the wave to cancel out and some to brighten. Imagine the peaks and troughs of a wave either canceling out because the peaks coincide with troughs or being exaggerated because the peaks coincide with peaks. This is the sort of phenomenon that causes the beautiful pattern you see in the photo above. In other words, the physics behind the following ‘interference pattern’ is similar to the soap bubble at hand:

Interference http://www.physics.louisville.edu/cldavis/phys299/notes

Essentially, the pattern in the soap bubble case is observed because of interference between the different wavelengths of light (that correspond to different colours) which means some wavelengths (colours) cancel and some brighten.

Thin Film Geometry http://emfsafetynetwork.org/wp-content/uploads/2011/11/interference.jpg

Returning to the problem, the value we’re looking for is the mass of the soap bubble. How the heck do we find that with light? Well, if the question gives us the details of physics pertaining to light, it’s safe to assume we will need this somewhere down the line. But for now let’s take it one step at a time. We don’t know how much the bubble weighs. But we do know how much water weighs for each cubic metre we’ve got: this is called the density, and for water it’s pretty easy to remember:

Density of water = $ 1 \ g / cm^3 = 1000 \ kg / m^3$

Okay, so now we know that water weighs 1000kg for every metre cubed; now what? The key to solving some physics problems is intuition from experience, and a hell of a lot of breaking things down into components. So now we KNOW one thing. Let’s look for another thing to know. Well, IF we just happened to know the volume, then we could calculate the mass of the water right now since the volume multiplied by the density (which we know) is equal to the mass. Great. So what’s the volume of a sphere? Well it’s 4 x pi x r^3. But the mass of the water in the soap bubble isn’t in the sphere. Ignoring the air inside (it’s relatively light), all the mass (water and soap) is in the thin layer of the surface of the bubble. So we need the volume of that. Well that’s just the sphere of the bubble minus the sphere of the air. Let’s call the bubble’s thickness L and its radius from the centre to the outer surface r.

Soap Sphere


$$ V = 4\pi r^3 - 4\pi (r-L)^3 $$ $$ V = 4\pi (r^3 - (r-L)^3)$$

Alright, sweet. So what’s L? Well, this is the point where the previous delving into thin film interference becomes useful to us. Those bright maxima for certain wavelengths we were discussing? They occur at regular intervals depending on the wavelength, according to:

$$ 2L = (m+\frac{1}{2}) \frac{\lambda}{n}$$

Where n is the refractive index (a measure of how much some substance refracts light), m is an integer (1, 2, 3, 4, …) and L is…exactly what we want! Rearranging gives us:

$$ L = \frac{\lambda}{2n}(m +\frac{1}{2}) $$
The brightest, most intense interference (the one we’re looking for because the question says the dominant colour is violet) occurs at m = 0. Noting that the density is 1000 kg / m^3 and letting M = mass:
$$ V = 4\pi (r^3 - (r-L)^3) $$ o$$ V = 4\pi (r^3 - (r-\frac{\lambda}{4n})^3) $$ $$ M = 4000\pi (r^3 - (r-\frac{\lambda}{4n})^3) $$

Substituting 1.33 for n (the value for the refractive index of water) and 0.03m for r and the value for the wavelength of violet light 400 nm = 400 x 10^(-9) m:

$$ M = 4000\pi (0.03^3 - (0.03-\frac{400 \times 10^{-9}}{4(1.33)})^3) $$ $$ M \approx 2.55 \times 10 ^ {-6} \ kg = 2.55 \times 10 ^ {-3} \ g = 2.55 \ mg $$
So if a bubble looks purple, 2.55 micrograms would be a good guess of its mass. And this is quite beautiful. Not only have we derived a very basic, observable quantity about the object: the question also opens the imagination to a host of other questions, like will this bubble float? At what temperature will it pop? Simple questions like these, which seem impossible to answer at first, often turn out to be rather elegant in their solutions.

© Angus Lowe 2018

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