The title could be: How Do Colors Appear on a Birefringence Chart?

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    Birefringence Color
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When polarized light passes through an anisotropic sample, it splits into two rays, the ordinary (o-ray) and extraordinary (e-ray), which emerge out of phase. The colors on a birefringence chart are determined by the different retardation experienced by each wavelength in the e-wave compared to the o-ray. As these rays recombine at the analyzer, constructive and destructive interference adjusts their intensities, resulting in the colors observed. The Michel-Levy Birefringence Chart helps identify sample thickness and type based on these interactions. Understanding these principles clarifies how color perception in birefringent materials is influenced by sample characteristics and light behavior.
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I'm not a physics whiz, so please be patient with me!

I understand that when polarized light passes through an anisotropic sample, it bifurcates into the o-ray and the e-eay. The two rays emerge out of phase to each other and when they hit the polarizer, they recombine. Due to the fact that the waves were out of phase, when recombined, it forms a new polarized wave (made up of various wavelengths). What I don't understand is what justifies the colors on the birefringence chart, specifically according to the second figure posted here according to the link posted below:

http://www.microscopy-uk.org.uk/mag...scopy-uk.org.uk/mag/artnov08/rd-crystals.html

Can someone explain figure 2 to me? From what I understand, the retardation of each specific color in the e-wave is different to its corresponding color in the o-ray (for instance, red light for the two waves has a higher retardation (No - Ne) than blue light when comparing where they are when the two rays emerge from the crystal). When two corresponding colors combine at the analyzer, some are amplified (constructively), while some are nullified (destructively) and most are somewhere in between. These colors (after having their intensity adjusted due to constructive/destructive interference) are put together and give us the color that we see (depending on where the viewing port is, which relates to sample thickness). Is this correct, or more likely completely incorrect?

Any help would be greatly appreciated. Something tells me that I'm missing something key here.
 

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Oh, I was referring to what happens after the e-wave and o-wave exit from an anisotropic sample at the stage area of a polarized light microscope. From what I understand (probably incorrectly), the individual wavelengths of light in each wave are retarded at different rates. When the two waves recombine, the amplitude for each wavelength is adjusted due to the constructive/destructive interference at the analyzer. All the wavelengths put together with their adjusted intensity gives us the color we see at the eyepiece. I'm just wondering if that is correct. The Michel-Levy Birefringence Chart can tell you how thick the sample is, or what type of sample it is, etc.
 
I think you are on the right track- let's see if we can step through the reasoning. First, we are illuminating a (thin) birefringent sample using 'white' linearly polarized light. Then, the light passes through a crossed polarizer. What will you see?
 
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