Register to reply

Optical rotation and linear basis set

by galvin452
Tags: basis, linear, optical, rotation
Share this thread:
galvin452
#1
Jul12-13, 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.
Phys.Org News Partner Physics news on Phys.org
Physicists unlock nature of high-temperature superconductivity
Serial time-encoded amplified microscopy for ultrafast imaging based on multi-wavelength laser
Measuring the smallest magnets: Physicists measured magnetic interactions between single electrons
Claude Bile
#2
Jul13-13, 06:56 AM
Sci Advisor
P: 1,474
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.
galvin452
#3
Jul13-13, 06:03 PM
P: 15
Quote Quote by Claude Bile View Post
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.
For a linear basis set, one can take each individual linear orientation basis as two oppositely rotating basis set so that each orthogonal linear basis is rotated by the difference between the right and left index of refraction. This would rotate the basis set as you indicated. The amount of rotation dependent on the difference between the left and right rotary index of refraction.

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?

galvin452
#4
Jul14-13, 10:34 PM
P: 15
Optical rotation and linear basis set

Quote Quote by Claude Bile View Post
. I'd go through the Jones algebra to be sure though.
Claude.
I can't find the Jones matix is for an optical rotary crystal? Do you know it?
Andy Resnick
#5
Jul15-13, 07:54 AM
Sci Advisor
P: 5,510
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)]].
galvin452
#6
Jul16-13, 04:02 PM
P: 15
Quote Quote by Andy Resnick View Post
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)]].
Hi Andy, not sure whar you mean by a linear rotator vs a circular rotator. I know you can rotate the angle of linear polarized light which I assume is what you mean by a linear rotator. What is a circular rotator?


Register to reply

Related Discussions
Large optical image rotation for project General Engineering 3
Linear Algebra question regarding linear operators and matrix rep. relative to basis Calculus & Beyond Homework 6
Calculating the Optical Rotation Value Biology, Chemistry & Other Homework 1
A question about linear algebra...(change of basis of a linear transformation) Calculus & Beyond Homework 15
Linear algebra, basis, linear transformation and matrix representation Calculus & Beyond Homework 13