- #1
Particle-Wave
- 6
- 0
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.
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.