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Optical rotation and linear basis set 
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#1
Jul1213, 04:15 PM

P: 15

If I have a 45 degree linear polarized light which I then circularly polarize using a 1/4 wave plate and put this through an optical rotary crystal and then using the equivalent 1/4 wave plate but in the reverse oriention, will I get back a 45 degree linear polarized light?
Put another way, as circular polarized light can use a linear basis set 90 degrees out of phase, can the angle of the linear basis set with respect to the 1/4 wave plate orientation be rotated by the optical rotary crystal (while the phase remains circular) or is the basis set orientation always fixed? I would think the basis set orientation is fixed but just want to make sure. 


#2
Jul1313, 06:56 AM

Sci Advisor
P: 1,477

There is no reason why the basis set orientation ought to be fixed. By my understanding the optical rotary crystal can rotate the linear basis set. I'd go through the Jones algebra to be sure though.
Claude. 


#3
Jul1313, 06:03 PM

P: 15

However if one uses a circular basis set the light is totally right circularly (or left circularly depending on the 1/4 wave plate fast slow axis orientation) polarized when entering the optical rotary crystal. Lets assume right circularly polarized, so at best it could only interact with the right index of refraction having no left component. So the right circular polarized light's reorientation would only be the result of the right rotary index of refraction. That is to say we would get two different results depending on the basis set used. Where is the mistake? 


#4
Jul1413, 10:34 PM

P: 15

Optical rotation and linear basis set



#5
Jul1513, 07:54 AM

Sci Advisor
P: 5,523

For a linear rotator (say, sugar water), the Jones matrices are: [[cos(δ/2)+icos(2θ)sin(δ/2) isin(2θ)sin(δ/2)], [isin(2θ)sin(δ/2) cos(δ/2)icos(2θ)sin(δ/2)]] for cartesian basis states and [[cos(δ/2) i*exp(i2θ)sin(δ/2)], [i*exp(i2θ)sin(δ/2) cos(δ/2)]] for circular basis states, where θ is the azimuthal angle of the fast axis and δ the phase retardation.
For a circular rotator (say, the cholesteric liquid crystal phase), the cartesian Jones matrix is [[cos(δ/2) +/sin(δ/2)], [/+sin(δ/2) cos(δ/2)]] and for circular basis states [[exp(/+ iδ/2) 0],[0 exp(/+iδ/2)]]. 


#6
Jul1613, 04:02 PM

P: 15




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