All cracked up (with no place to go)

To: Dick Hughes
From: John L. Emmett

Subject: Your Problem (but not your only one)

In this spreadsheet there are only four numbers you can enter: n₁ in two places, and n₂ in two places. The spreadsheet calculates the Relative Index of Refraction n = n₂/n₁, and the Critical Angle so don't try to enter these. From these numbers the spreadsheet calculates all of the reflectivities. Note you can change Emerald to any gemstone name you want, and likewise you can change the crack content to anything you want. Note that when you change the index of refraction of anything, the spreadsheet recalculates everything. In addition the reflectivity calculations are duplicated so you can directly compare an empty crack with a filled crack by seeing the numbers next to each other.

Now for some explanations. The geometry, nomenclature, and equations used are presented in Grant R. Fowles' Introduction to Modern Optics (1968, Holt, Rinehart and Winston, New York). The geometry is shown in Figure 2.9, page 48 of this reference. The incident light ray is always in the region with the index of refraction n₁, regardless of the relative index of refraction of the two materials. The reflected ray is always in region n₁ also. The transmitted ray is always in the region with index n₂. With these conventions, the spreadsheet always calculates the reflectivities correctly whether the incident ray is in the lower or higher refractive index region. Thus this spreadsheet is quite general, but you have to think carefully about what you are doing.

The equations programmed into the spreadsheet are 2.58 and 2.59 on page 51 of the reference.

Where E'/E is the amplitude reflectivity of the electric field of the light, and R is the light intensity reflectivity.

The Critical Angle is that angle where the reflectivity becomes 100%, and remains 100% for all larger angles. This occurs only for the situation where n<1.0. The equation for the critical angle is just:

Now what do TE and TM mean? As you know, there are two polarization directions of light which are perpendicular to each other and perpendicular to the direction of propagation of the light ray or beam. Unpolarized light can be described as equal amounts of the two polarizations. Because the reflectivities of interfaces are dependent on the polarization of the incident ray, we have to calculate it for each polarization. That is why the spreadsheet has two columns for the reflectivity. There are many ways to denote the two polarizations. This book has chosen TE (transverse electric) and TM (transverse magnetic). What does that mean? First, remember that the polarization direction and the electric field vector direction of light are the same thing. Second, transverse in this context means "perpendicular." Perpendicular to what? Perpendicular to the plane of incidence. This is an imaginary plane that contains both the incident and reflected ray. In my drawing Figure 1a, the plane of incidence is the sheet of paper. (Note: all my drawings are for the situation where n₁>n₂ which is why the transmitted ray bends the other way from the drawings in the book.) Note that in Figure 1a the little circles on the light rays are meant to denote the ends of the polarization direction arrows shown more clearly for the other polarization in Figure 1b. Now what does TM mean? Remember that light is made up of both an electric field and a magnetic field and that their vectors are mutually perpendicular to each other. So TM just means that the magnetic field vector is perpendicular to the Plane of Incidence. Since we normally think of polarization in terms of the electric vector direction, I have shown that as the arrows in the TM case in Figure 1b. Since light in most observational situations is made up of both polarizations we average the two single polarization reflectivities to calculate the real single surface reflectivity.

Now on to the explanation of the problem I have calculated for you. As I understand your current interest, it is to calculate the reflectivity of a crack in a gemstone filled with anything like air, oil, Opticon, kangaroo piss, or whatever. Thus I have set up the problem in that way, i.e., the incident ray is in the gemstone (index n₁) and is incident on the crack (index n₂). This is shown in Figure 2. Now comes the subtle part. Note that there are really TWO reflections that count towards what you see; the reflection off the top of the crack AND the reflection off the bottom of the crack. Thus the reflectivity from the first two reflections of the crack is R + R(1–R)², where R is the single surface reflectivity. For low reflectivities, this nearly DOUBLES the single surface reflectivity calculated in the spreadsheet! Don't forget this. Actually there are many more reflections but they contribute very little additional intensity (see figure 2).

The enclosed spreadsheet has a lot of #NUM! entries courtesy of Excel. What is that all about? If you look at Equation 2.58 or 2.59 for the case where n<1.0, you can see that for some angles it is possible for the quantity under the square root sign to be negative. As you know, you can't take the square root of a negative number. Well actually you can, and the result is called a complex number because it contains two parts, a real part and an imaginary part denoted by the multiplier "i" which is equal to square root(-1). (I know you don't want to hear about this [your're right, I don't] but I must persist a little longer.) When you square a complex number to get to get the intensity reflectivity, you multiply the complex number (like a + ib) by its complex conjugate (which is a – ib). The result for the equations we are using is 1.0, that is the reflectivity is 100% for any complex number reflectivity. Now this simple Excel spreadsheet does not know about complex numbers, so when faced with taking the square root of a negative number it shouts HELP! (#NUM! in Excelspeak). So you know what to do, OK? Whenever you see #NUM! In the reflectivity column, just read it as 100%. Now notice another thing. The reflectivity changes relatively slowly with increasing angle until you near the critical angle where it increases very rapidly. Thus you go from a few tens of percent reflectivity to 100% in less than one degree! Positively magical!

Now examine the reflectivity columns in the spreadsheet for the case where the indices of refraction of crack filler and emerald differ by only 0.02. Note that the normal incidence reflectivity is only .004%! If it is really this low, why can you see the crack at all? Its simple; either the surface is not emerald or it is not planar. Most cracks contain mineral alteration products or other impurities which have been deposited in the crack. Even boiling in acid does not fully remove these materials. These materials have a color contrast to the emerald which makes them more visible, or have indices of refraction which are different than emerald causing additional reflection. However, it is important to remember that even if it was totally an emerald surface, but that surface had a fine roughness, the oil, with its finite viscosity, might not completely wet the surface and thus not completely reduce the reflectivity.

Another point I might make is that the closeness of the index match is not too important. Just use the spreadsheet to do an example or two for yourself. Thus the practice of filling tourmaline fractures with spermaceti (n=1.43) works quite well as most tourmaline fractures are very clean.

I hope the foregoing answers your questions adequately.

Sincerely,
John Emmett