This document last updated on 01-Nov-2011EENS 2110 MineralogyTulane University Prof. Stephen A. NelsonInterference Phenomena, Compensation, and Optic SignAs you have probably noticed by now, viewing an anisotropic crystal under crossed polars (analyzer inserted) the crystal is extinct when either of the two privileged directions in the crystal are lined up parallel to the polarizing direction of the microscope. This is because when the privileged directions are parallel to the polarizer, the crystal does not change the polarization direction and the light will thus be vibrating perpendicular to the analyzer. When the privileged directions are not parallel to the polarizer some light is transmitted by the analyzer and this light shows a color, called the interference color. In this lecture we will discover what causes this interference color and how it can be used to determine some of the optical properties of the crystal. The Interference of LightWaves Polarized in the Same PlaneAs we discussed in the lecture on X-rays, when electromagnetic waves emerge from a substance, they can interfere with each other and either become enhanced, partially destroyed, or completely destroyed. The same is true for polarized light. In the upper diagram below, labeled A, two waves polarized in the vertical plane are emerging from the polarizer. These waves are in phase, and thus we only see one wave. The difference between points on the same wave is called the path difference, symbolized as '. If the path difference is an integral number of wavelengths, nO, the waves are in phase. For points that are completely out of phase, for example between crests and troughs of the wave, ' will be equal to 1/2O , 3/2O, 5/2O, etc. or [(2n+1)/2]O. For path differences where the waves are completely out of phase, complete destructive interference will occur and no wave will emerge from the polarizer. For path differences that are neither in phase or completely out of phase, the waves will be summed together to produce a resultant wave with crests and troughs at different positions than the original waves. This is shown in the part of the diagram labeled B, and the blow-up to the right. Interference of Light11/1/2011Page 1 of 20Furthermore, the vector sum of the two waves will produce a resultant wave with a different amplitude than the two original waves.Waves Polarized in Perpendicular PlanesIf the waves are traveling on the same path, but are polarized at 90o to each other, again the resultant wave becomes the vector sum of the two waves. To see this we need to look at a three dimensional view.In this case, the waves have emerged from a crystal polarized in the directions of the privileged directions in the crystal, but they are otherwise in phase. The resultant wave, shown here in dark gray, is the vector sum of the two waves, and is thus polarized at a 45o angle to the original waves.Path Difference or Retardation Resulting from Passage Through a Crystalz When polarized light enters an anisotropic crystal from below, and neither of the privileged directions in the crystal are parallel to the polarizer, the light is broken up into two mutually perpendicular polarized waves that travel through the crystal. z One of these waves will be vibrating in the direction of high refractive index, N, and the other will be vibrating in the direction of the low refractive index, n. z Because refractive index is inversely proportional to the velocity of the wave, the wave vibrating in the direction of the larger refractive index will travel more slowly in the crystal than the wave vibrating in the direction of the lower refractive index.z Thus, we refer to the wave vibrating in the higher refractive index direction, N, as the slow wave, and the wave vibrating in the lower refractive index direction, n, as the fast wave.z Since the wave velocity, V, is given by: V = OQ, if the velocity of the wave becomes lower, the wavelength, O, will also decrease, since the frequency, Q, remains constant.z Thus, since the slow wave has a lower velocity, it will have a shorter wavelength while passing through the crystal.z Further, when the two waves emerge from the top of the crystal, they may not be in the Interference of Light11/1/2011Page 2 of 20same position in their vibration as when they entered the crystal (i.e. they may be out of phase, depending on the path difference acquired while in the crystal). To see this, examine the figure shown below. Polarized light vibrating E-W enters the crystal from below. Upon entering the crystal this light is broken into a slow ray, vibrating parallel to the larger refractive index direction, N, and a fast ray, vibrating parallel to the lower refractive index direction, n. While in the crystal, the slow wave's wavelength decreases and it passes through 1½ wavelengths, while the fast wave passes through 1 wavelength. At the instant the slow wave emerges from the crystal, the fast wave has traveled a distance '. This is the path difference or retardation. z If the time required for the slow ray to pass through the crystal is TN and the time required for the fast ray to pass through the crystal is Tn, then the distance traveled by the fast wave as it waits for the slow wave to emerge from the crystal is C(TN-Tn), where C is the velocity of light in air. This is the path difference between the two waves. z Thus, ' = CTN-CTnz The actual velocities of the two waves while in the crystal are: CN = t/TN and Cn = t/Tn , where t is the thickness of the crystal.z Then TN = t/CN and Tn = t/Cnz Substituting this into our equation for ', gives the result: ' = t(C/CN-C/Cn)z But, C/CN is the refractive index, N, of the slow wave, and C/Cnis the refractive index of the fast wave, n.z So, ' = t(N - n)z Note that (N - n) is the birefringence, so the equation above tells us that the retardation is equal to the thickness of the crystal times the birefringence. (Note that this is not necessarily the absolute birefringence of the crystal [|H Z|] but only the difference between the high and low refractive index for this orientation of the crystal). Interference of Light11/1/2011Page 3 of 20Polarization of the Resultant WaveAs discussed above, when the waves emerge from the crystal they will be polarized at 90o to each other. But, they will then interfere with each other and the resultant wave will be polarized in a plane that is the vector sum of the two waves. Thus, the resultant wave will