# Optics:Polarization in liquid

1. Apr 19, 2012

### Reut

1. The problem statement, all variables and given/known data

A cylinder with the length 20cm and radius of 5cm is filled with liquid. The refractive index of the cylinder, its surrounding and the liquid is 1.3. A polarization wave is given in the fluid:
E=Eo*exp( i *(ωt-kz)) $\hat{x}$ +c.c.
The wave is induced by light with a frequency different from ω as a result of a non linear process. The light sorce (freq' ω) is only as result of the polarization wave.

The cylinder is oriented in the Z direction, and the wave is E=0 from outside it

State the light intensity (I) at freq' ω on the surface of the cylinder.
Draw a graph of I(z,r=D/2) ,I(z=0,r) ,I(Z=L,r) and I(z,r=0)

2. Relevant equations

Wave equation, Vector poynting, Polarization

3. The attempt at a solution

I figured out from the wave equation that the polarization is linear.
As a result of the poynting vector i reached a conclusion that
I=ε$_{0}$ * c * |E|$^{2}$ *cos$^{2}$ (ωt-kz) $\hat{z}$ . (Im not entirely sure about my conclusion)

I dont understand 2 things:
First, how is I dependant of z,r when the wave is dependant of z,t ?
Second, what is the meaning of a wave induced by light with a frequency different from ω as a result of a non linear process?

*Note: English isn't my mother tongue so forgive me for any mistakes.

Thank you for your help

Last edited: Apr 19, 2012
2. Apr 19, 2012

### rude man

Well, if no one else answers:

Intensity is not a vector, so don't add the z unit vector to your expression for I. Otherwise your expression for I is correct.

I too don't understand the whole business about ω being non-linearly generated. It doesn't matter how it's generated. It's just a plane wave polarized in the x diection & traveling in the z direction.

The existence of the cylinder is also meaningless, as is the refractive index n. Since ω is "given in the fluid" there is no cause for taking n into account. (The orientation of the cylinder isn't given, and it doesn't need to be ...)

I don't know what the "c.c." stands for in the expression for E so I ignored it.

Maybe someone else will chime in here?

3. Apr 19, 2012

### Reut

Thanks for the comments

Im not sure about the the meaningless of the cylinder, just because i dont think they would give additional data without reason. The cylinder is oriented in the Z direction, and the wave is E=0 from outside it (sorry for not mentioning that earlier).

4. Apr 19, 2012

### rude man

I wouldn't bet on them not giving you extraneous information. Lots of wiseguy profs like to dio that.

Since the cylinder and its surroundings are all of the same n, then the cylinder is transparent to the wave. So it doesn't exist. Is there more info on the cylinder you haven't included?

5. Apr 19, 2012

### Reut

Thats it. As i said, the cylinder is oriented in the z direction, it has the length of 20cm and diameter of 10cm (radius 5). Do you have a clue how can the intensity be related to the radius?

6. Apr 19, 2012

### rude man

Nothing about attenuation thru the cylinder? That's the only parameter other than n that would make the cylinder a reality to the wave. Otherwise it's not there.

Anyway, my answer is: yes, the intensity as a function of r is I. Constant!

7. Apr 20, 2012

### Reut

The attenuation isn't given, im quite sure there's greater depth to the question that im missing. i just dont think they would ask us to plot graphs that are constants.
Just a thought: would your answer change if the wave was E=Eo*exp( i *(ωt)) $\hat{x}$ or E=Eo*exp( i *(ωt-1.1*kz)) $\hat{x}$ ?

8. Apr 20, 2012

### rude man

The first is not a traveling wave. The second just changes the wavelength (and the velocity) by 10%.

In any case, there is no cylinder. The entire universe, as far as the wave is concerned, is a medium with an index of refraction of n = 1.3. Unless you've left out other information. Are you quoting the problem or paraphrasing it?

And, again - do you know what the "+c.c." stands for? Maybe that's our problem ...

9. Apr 20, 2012

### Cthugha

c.c. is short for complex conjugated, so you have another term with the same real parts, but opposote sign of the imaginary parts. So all i change to -i and vice versa.

10. Apr 20, 2012

### rude man

Oh, OK.
BTW I was wrong when I said that changing from k to 1.1k would change the velocity. That was dumb. The velocity is c/n irrespective of frequency of the light.

k is actually fixed once n is known, which it is. Velocity v = c/n = λf where f = ω/2π.

11. Apr 20, 2012

### Reut

Ok thanks for the answers, if you come up with anything else im all ears.

BTW - the question is translated to English, i went over it again just to make sure i didn't forget anything. C.C. is complex conjugated thats correct.