2008-07-04

How thick is human hair? How on earth did I start wondering about this anyway? You know how very thin objects are always compared to human hair in the media? Well, today my order of copper magnet wire arrived for winding high-Q loading coils for my beacon projects, while looking up the properties of copper wire I found a gauge table that went down to #56 (about 10 um). Tanya and I were talking about how thin that gauge was and we wanted some common physical object to compare against, naturally human hair came up, and being a geek one thing lead to another; an impromptu experimental apparatus was assembled to measure hair diameter.

Obviously too small to measure with callipers, I decided on using diffraction. Guessing that hair was around 1/20 of a mm it seemed practical to use a common laser pointer as the source, as it has a sufficiently coherent light for reasonable measurements around this size. The tracking pointer on my radiometric thermometer was used as a source, according to the compliance label it is < 1mW at around 650 nm. The laser illuminated the hair, taped across the output lens (Tanya's idea, much better than my attempt at just holding it still which is nearly impossible). At a measured distance of several meters the diffraction minima are easily marked on a whiteboard and then measured with a rule. A quick calculation gave the answers.

Everyone in the family was plucked for the experiment, the results say Dad has the finest hair at only 48 um. Next is Mum with 56 um, then Tanya with 91 um and myself with 98 um. This is head hair of course; human terminal hair which generally taken to be about 40-120 um depending on age and a bunch of other factors. The results agree well with expected values and the experiment was amazingly easy to perform, anyone with a tape measure and a laser pointer could repeat this experiment and should get fairly accurate results.

To increase accuracy I measured across the higher-order minima where possible and divided down. For my hair this was pretty easy, for the parents much harder because of their finer gauge hair which produces more widely spaced minima and less intensity, it was only possible to see the first minima for my Mum's hair even in dimly lit conditions and slowly flattening batteries.

## The Maths

For single-slit diffraction (Fraunhofer diffraction - which can be applied here even though we have not a slit but an essentially opaque filament in the monochromatic beam an a screen a great distance from the slit relative to its size and wavelength - Fresnel number less than 1) the minima occur when the path length difference between each side of the slit (hair) is a multiple of half the wavelength of the radiation (phase cancellation). I had to derive this from scratch in Physics class at University, but because the projected image is so distant compared to the size of the slit the two rays can be considered parallel and the path length difference is simply the slit size times the sine of the subtended angle.

`d sin(theta_n) = n lambda`

There are an infinite number of minima, but the amplitude drops of rapidly with increasing angle (it is sinc shaped, why takes a lot of maths to explain but it is of little surprise and is related to why it crops up in DSP so much too; diffraction is essentially spatial sampling).

Computing `theta` is easy enough with some trigonometry, but some simplifications associated with the large projection distance compared to the minimal spacing can make the exercise completely trig-free so you won't even need a scientific calculator.

`tan(theta) = s/L`

But for very small `x` it can be show that `sin(x) ~~ x` and `tan(x) ~~ x`, (The Taylor expansion can prove this to yourself) so:

`d ~~ lambda L / s`

Where `d` is the hair diameter, `lambda` is the light wavelength 650 nm (or 0.65 um), `L` is the spacing from the hair to the projection screen/wall/whiteboard/paper etc and `s` is the spacing to the first minima from the central beam. I measured between corresponding high-order minima through the central spot and then divided the measurement by twice the minima order to get higher accuracy.

## A worked example

For my hair measurement the distance between the pairs of 3rd minima was 90 mm, so the spacing `s` is 90/6 = 15 mm. The distance to the screen `L` was 2230 mm for all tests. Plugging this into the equation above gives my hair diameter `d` = (0.65 * 2230) / 15, approximately 96 um (about one-tenth of a millimeter, or 3.8 thou/mil for the Imperial readers).

The difference when calculated using `sin(tan^-1(s/L))` is quite insignificant and well inside the experimental error of measurement and laser wavelength. You can measure `s` and `L` in any units as long as they are the same units, if you prefer imperial you can use it, the answer will still be in the units of wavelength of course (nano or micrometers, but you could use angstroms if you wanted). If your laser is a different colour you'll need to use its wavelength for `lambda` instead of 650 nm, green doubled-IR lasers are around 520 nm and Blu-ray diodes about 405 nm. For your average red laser pointer the key "point 65 times length on spacing" is a pretty easy thing to remember and calculate on any old calculator.

The technique works for any fine thread too. You could measure that 56# wire with it, or your pets hair, or really anything that is in the general 100 um region. It would be a simple classroom exercise for a physics/optics demonstration. A simple jig to hold the laser diode, hair and a screen could be rigged up, the screen could be calibrated directly in um if you wanted, but it would probably be more instructive to make the students do the maths themselves after tracing the minima onto paper in their own lab book. Alternatively you might like to dispense with the "instrument" completely to show just how easily you can "do science" with the most basic of apparatus rather than some expensive well crafted piece of "big science" gear.