Interference of Light

October 19, 21, and 23

Light behaves as a transverse wave, as discussed a few lectures ago. In the microscope in plane polarized light (PPL) waves of light can be:

In phase: particles in the same wave train are in phase if, at any and all times, they are always displaced from the rest position by the same amount and direction.

Out of phase: particles whose vibrational displacements are equal but opposite in direction are out of phase.

Path difference: the distance between two points on the same wave path is known as their path difference. In phase waves have a path difference (D) of nl

Out of phase waves have D = (2n+1)/2l

When waves are neither in or out of phase, they must interfere and combine to form a composite wave with amplitude R. Their vibrational vectors add vectorially where they coincide in time and space. The form of the resultant wave depends on the wavelengths of the two original waves, their amplitudes, and their path difference.

 

Anisotropic Crystals

The refractive index (and therefore velocity of light) in an anisotropic mineral depends on the direction of light vibration.

When PPL enters an anisotropic crystal, the light is resolved immediately and breaks into two rays - fast and slow.

Rays do not interfere while in the crystal, but when the light reaches the edge of the crystal, there are two resulting perpendicular vibration directions. The waves interfere immediately once they leave the crystal.

In the crystal: 1) the slow ray has an index of refraction N

2) the fast ray has an index of refraction n. N > n

3) the path difference D between fast and slow is zero, and the fast travels faster once in the crystal and comes out on top first.

If the crystal thickness is t and the time for passage of the slow ray is TN and the fast ray is Tn, TN < Tn

While waiting for the slow ray to emerge, the fast ray speeds onward, and the separation distance is c(TN-Tn) where c = speed of light. Velocities in the crystal are cN = t/TN, cn = t/Tn such that

D = ct/cN - ct/cn = t (N-n). Note that N- n = the birefringence for the mineral

 

The resultant ray at the top of the crystal is plane polarized, circularly polarized, or elliptically polarized. Depending on which type of polarization is present, the light may or may not get transmitted by the upper polarizer/analyzer.

1) Plane polarization of the resultant wave

1/frequency = period of a wave - the time required for the series of vibrations necessary to complete one full wavelength.

2 cases for the difference between the two waves:

    1. Slow wave requires two periods, fast wave requires 1 period. Thus, D = l. Whenever D = nl, the emergent waves are in phase and interact when they emerge from the crystal. The vibrations are vectorially added at all parts coincident in space and time. The resultant wave is Plane Polarized in the same plane as the incident wave. Since the analyzer (upper polarizer) is polarized perpendicular to the other polarizer, no light is transmitted to the ocular. Thus, in phase slow and fast rays act isotropically.
    2. Slow wave requires two periods, fast wave requires 1 1/2 periods. D = 1/2l upon emergence and the waves are out of phase. The resultant wave is Plane Polarized at 90º to the original incident light vibrational direction. This orientation allows the analyzer to fully transmit the resultant wave. The same holds for all D = (2n+1)/2 l

 

2) Circular polarization of the resultant wave

D = 1/4 l, 3/4 l, ... (2n+1)/4 l

In this case, the interference produces a wave with constant displacement vectors but varying azimuthals - i.e. like the thread of a screw.

 

3) Elliptical polarization of the resultant wave

 D = all above cases. The resultant wave spirals upwards and the vibrational vectors are of constant length.

Viewed down the beam, the wave will appear elliptical

 

Above the crystal, the fast and slow waves interfere by vector addition. In some cases, we will treat the resultant wave as two vector components the came out of the crystal.

Next, the resultant wave encounters the upper analyzer. The percentage of transmitted light can be calculated using the following variables:

  1. f – the angle between the privileged direction of the polarizer and analyzer – usually set at and assumed to be 90º.
  2. t – the angle between the privileged direction of the polarizer and the crystal’s closest privileged direction.
  3. l - the wavelength of the monochromatic light
  4. D - the path difference produced in the crystal

 

% transmission = [-sin2 *(180º*D)/l * sin2(t-90º)]*100 assuming no absorption of light and 90º polarizers

if t = 45º, % transmission = (sin2 *(180º*D)/l)*100

transmission % is at a maximum when D = 1/2l, 1 ½l, 2 ½l… incident light is rotated 90º and all passes through the analyzer.

transmission % is at a minimum when D = nl - no rotation of the incident beam and all light is killed at the analyzer

 

EXTINCTION

Isotropic minerals are always extinct in XN (crossed polarizers)

Anisotropic crystals:

Because of interference of fast and slow rays upon emergence, the plane of polarization generally won’t be the same as that of the incident light.

When there is no rotation by the crystal, the grain is extinct in XN. This occurs 4 times when the grain is rotated 360º (every 90º) and maximum brightness is at 45º to the extinction.

INTERFERENCE COLORS

If the illumination source for the microscope is white light, anisotropic minerals will be colored in XN. These colors are interference colors that result from the unequal transmission of light by the analyzer of the component wavelengths of white light. The colors depend upon the path difference in the crystal for the wavelength of light produced.

Suppose that D = 550 nm, t = 45º, and the illumination is white light. Consider only the wavelengths 400.0, 440.0, 488.9, 550.0, 628.6, and 733.3 nm

For l = 400, D = 550 -> 1 3/8 l 95% transmitted

For l = 440, D = 550 -> 1 1/4 l 50%

For l = 488.9, D = 550 -> 1 1/8 l 17%

For l = 550, D = 550 -> 1l Extinguished

For l = 628.6, D = 550 -> 7/8 l 17%

For l = 733.3, D = 550 -> ¾ l 50%

 

In this case, the red and violet are transmitted, and green and yellow are eliminated, resulting in a reddish violet color.

 

See the chart in K&H or Nesse of the colors that are associated with the various path differences and thin section thickness.