When does the extinction coefficient, Qext reach 2.00?

by Alan Rawle, Thursday, 29th December 2016

In light scattering theories and approximations, we assume that for large particles the extinction coefficient, Qext, is exactly 2. This is the basis of the Fraunhofer approximation meaning that one unit of light is lost from the shadow of the particle and one unit of light by diffraction. In the far field this presents us with Babinet’s conjecture which we’ll not deal with at this time.

When we come to examine the actual light scattered by particles we find at smaller sizes that this assumption (Qext = 2) is not valid and the behavior is extremely complex but modeled correctly by Mie theory. We can look at the complex nature of the plot by using a very powerful Freeware program called MiePlot written by Philip Laven. Here’s TiO2 (rutile form; averaged RI = 2.73) in linseed oil (RI = 1.52) of various sizes which was described in a webinar on the optical properties listed in the references. Here we use an imaginary RI of 0 which gives rise to the wonderful maxima and minima in the plot, λ = 500 nm to correspond to an earlier paper by Loney:

Figure 1: Qext versus size (μm) for TiO2 (2.73/0/1.52 @ 500 nm)

Note the maximum (just over 5.0) in the vertical Qext axis at around 0.3 μm. This means that as well as the one unit of light lost directly from the particle’s shadow, we lose 4 units of light from diffraction. It’s this high scattering loss that makes TiO2 such a wonderful white opaque pigment when painted on walls.

Figure 2: Qext = 2.00; Fraunhofer approximation 
Figure 3: Diagrammatic representation of hiding power in relation to Qext

Note too how in the plot in Figure 1 the value of Qext appears to approach 2.00 as we have indicated as the assumption in the Fraunhofer approximation.

The reader may be aware that Mie’s solution to Maxwell’s equations of the interaction of light with matter deals with a complex sum of terms explaining the entire scattering behavior of a homogeneous sphere. What is probably less known is that the contributions of the individual terms can be split out and this is the basis of the Debye approach (apparently independently) one year after Mie’s classic Ann. Physik. paper. This is easily seen by means of a simple diagram:

Figure 4: The different rays corresponding to the different terms in the Debye analysis

If you have viewed the ‘Life of Mie’ webinar, then you’ll know that physicists tend to use a term called the size parameter (rather than size directly) to take care of wavelength simultaneously:

Size parameter, x = 2πr/λ (circumference divided by wavelength)
If we plot Qext as a function of the size parameter x for n = 1.33 (water droplets) calculated using Mie theory and also the sum of the Debye p = 0 and p = 1 terms, we see (you can do this yourself in MiePlot):

LavernPlot-H2O
Figure 5 Screen dump of MiePlot showing the Mie and Debye terms for water droplets of various sizes

Note how the first 2 Debye terms (ρ = 0 and ρ = 1) converge to the exact Mie solution at above a size parameter of x = 3 or so (red and blue plots). Indeed, van der Hulst stated that for gold colloids (the subject of Mie’s paper) the summation of three terms is usually sufficient – 98.5% of scattered energy (van der Hulst; page 231).

You will see that value of the p = 1 curve oscillates around zero, whereas the p = 0 curve declines from a high value when x < 1 towards 2 when x is very large. In other words, it is asymptotic towards the value of 2 in the geometric limit (i.e. when x = ∞). The problem about asymptotic curves is that they never quite reach the asymptotic value – even though they get very close to it. This behavior is illustrated in the following table showing Qext for the Debye p = 0 term when n = 1.33:

Qext reach 2.00 Table1

Note that x = 10,000 corresponds to a water drop with radius r = 1 mm when λ = 0.628 μm = 628 nm. So x = 1,000,000 corresponds to r = 100 mm = 10 cm = 0.1 m. Of course, this is an incredibly large water droplet! Water droplets form rainbows and glories, of course, but the important point here is that the assumption of 2.00 implicit in the Fraunhofer approximation is not even reached for 10 cm water droplets….. Fraunhofer simply doesn’t predict a rainbow:

Ext_coeff_blog_Fig5b

Acknowledgements

The author acknowledges Philip Laven for a very interesting discussion on the above subject and generating the enthusiasm to write this note. The author also acknowledges Charlie Fillingham of Lord Rayleigh’s Farms in Terling Essex for correcting my previous pronunciation misstatements: “Rayleigh is pronounced Ray-Lee; Sir Walter Rayleigh is more pronounced Rar-lee”

References