 #1
nkinar
 76
 0
I am wondering if it is possible to construct a small electrooptic modulator which is able to create a large change in the refractive index of a material such as Lithium Niobate. The change in the refractive index will also cause a phase shift in the light beam.
An electooptic material changes its physical properties when an electric field is applied across the material. Lithium Niobate is often used as a material for electrooptic modulators since its electooptic coefficients are very large.
Suppose that we have the following setup. An ordinary polarized light beam is incident on a block of Lithium Niobate with a length of y (meters) and a height of d (meters). The light beam moves along the yaxis, which is taken in this system to be horizontal. An electric field is passed across the block of Lithium Niobate (along the z axis) by two electrodesone placed on the top of the block, and the other placed on the bottom. The direction of the electric field is vertical across the block, and perpendicular to the travel path of the light beam.
The change in the ordinary index of refraction of the Lithium Niobate is:
[tex]
\[
\Delta n_0 = \frac{{  n_0^3 r_{13} V}}{{2d}}
\]
[/tex]
Where
[tex]
\Delta n_0
[/tex]
= Change in the ordinary refractive index
[tex]
n_0
[/tex]
= ordinary refractive index of the Lithium Niobate
[tex]
n_0 = 2.1763
[/tex]
[tex]
r_{13}
[/tex]
= electrooptic coefficient of the Lithium Niobate
[tex]
r_{13} = 7.62 \times 10^{  12} {\rm{ m/V}}
[/tex]
[tex]
d
[/tex]
= distance over which the electric field acts (meters)
[tex]
V
[/tex]
= Voltage
From the classical properties of waves, the change in the phase of the light wave [tex]
\Delta \phi [/tex] is related to the angular wavenumber [tex] k [/tex], the change in the index of refraction [tex]\Delta n_0[/tex] and the distance traveled [tex]y [/tex] through the block by the expressions:
[tex]
\[
\Delta \phi = k\Delta n_0 y
\]
[/tex]
[tex]
\[
\Delta \phi = \frac{{2\pi \Delta n_0 y}}{\lambda }
\]
[/tex]
By substituting the change in the index of refraction into the above, it follows that:
[tex]
\[
\Delta \phi = \frac{{  \pi n_0^3 r_{13} Vy}}{{\lambda d}}
\]
[/tex]
Solving for the voltage ([tex]V[/tex]) :
[tex]
\[
V =  \frac{{\Delta \phi \lambda d}}{{\pi n_0^3 r_{13} y}}
\]
[/tex]
Now suppose that the wavelength of the light wave is [tex]\lambda = 450nm[/tex], and that the required phase shift is [tex]\Delta \phi = 2.1 \times 10^6 [/tex]. Note that the sign of the phase shift is positive. I am not completely certain as to whether this is physically possible, but I'll take a look at it mathematically. Suppose that we plug the following into the equation used to solve for voltage:
[tex]
\Delta \phi = 2.1 \times 10^{6}
[/tex]
[tex]
\lambda = 450nm
[/tex]
[tex]
d = 0.1 \times 10^{6} m
[/tex]
[tex]
n_{0} = 2.1763
[/tex]
[tex]
r_{13} = 7.62 \times 10^{12} m/V
[/tex]
[tex]
y = 1.0 \times 10^{1}m
[/tex]
Note that the distance ([tex]d[/tex]) over which the electric field acts is very small: only 0.1 micrometer! The horizontal distance traveled by the light wave through the Lithium Niobate block is 10 cm.
Evaluating the expression, I find that:
[tex]V = 3829.8V[/tex]
Now clearly it is not impossible to generate a voltage of 3kV. Note that as the separation distance ([tex]d[/tex]) increases, the applied voltage required to create the phase shift also increases as well. Increasing the separation distance shows that thousands or millions of volts are required to cause a phase shift this large in the incident light beam. This suggests that the separation distance between the electrodes must be very small for such a setup to be feasible.
This leads me to my questions:
(1) Is there an easier way of doing this: to create a large magnitude phase shift in a light beam?
(2) Can a negative voltage be applied to an electrooptic crystal? If so, is it physically reasonable to use a negative voltage to change the refractive index of the Lithium Niobate block? I think that the answer to this question is "yes," and the answer is supported by the mathematics above. I also think that it is possible for the light to have a positive phase shift.
(3) Is it possible to make such an apparatus? Is it possible to deposit a thin layer of Lithium Niobate (with a thickness of 0.1 micrometer) on a substrate, and attach electrodes to the top and the bottom of the layer to create an electric field?
(4) Does anyone know of a substance with a larger electrooptic coefficient? If [tex]r_{13}[/tex] increases (becomes larger), then the voltage required to create the phase shift drops appreciably.
(5) How might I manufacture such a device? Would it be possible to manufacture something like this? If so, what would be the cost to create this apparatus? Any guesses based on similar or related research? Is there a good monograph on creating tiny (micro) light guides?
(6) Suppose that I use a lens to send an image through the Lithium Niobate block. The lens is coupled to a piece of fiber optic. How would I interface the piece of fiber optic to the left side of the Lithium Niobate block?
(7) Could an image be focused by a lens and sent through the Lithium Niobate block? Note that d = 0.1 micrometer, which is a very small distance.
An electooptic material changes its physical properties when an electric field is applied across the material. Lithium Niobate is often used as a material for electrooptic modulators since its electooptic coefficients are very large.
Suppose that we have the following setup. An ordinary polarized light beam is incident on a block of Lithium Niobate with a length of y (meters) and a height of d (meters). The light beam moves along the yaxis, which is taken in this system to be horizontal. An electric field is passed across the block of Lithium Niobate (along the z axis) by two electrodesone placed on the top of the block, and the other placed on the bottom. The direction of the electric field is vertical across the block, and perpendicular to the travel path of the light beam.
Code:
electrode


=====================================
Light beam BLOCK OF LITHIUM NIOBATE
>
=====================================


electrode
<y>
The change in the ordinary index of refraction of the Lithium Niobate is:
[tex]
\[
\Delta n_0 = \frac{{  n_0^3 r_{13} V}}{{2d}}
\]
[/tex]
Where
[tex]
\Delta n_0
[/tex]
= Change in the ordinary refractive index
[tex]
n_0
[/tex]
= ordinary refractive index of the Lithium Niobate
[tex]
n_0 = 2.1763
[/tex]
[tex]
r_{13}
[/tex]
= electrooptic coefficient of the Lithium Niobate
[tex]
r_{13} = 7.62 \times 10^{  12} {\rm{ m/V}}
[/tex]
[tex]
d
[/tex]
= distance over which the electric field acts (meters)
[tex]
V
[/tex]
= Voltage
From the classical properties of waves, the change in the phase of the light wave [tex]
\Delta \phi [/tex] is related to the angular wavenumber [tex] k [/tex], the change in the index of refraction [tex]\Delta n_0[/tex] and the distance traveled [tex]y [/tex] through the block by the expressions:
[tex]
\[
\Delta \phi = k\Delta n_0 y
\]
[/tex]
[tex]
\[
\Delta \phi = \frac{{2\pi \Delta n_0 y}}{\lambda }
\]
[/tex]
By substituting the change in the index of refraction into the above, it follows that:
[tex]
\[
\Delta \phi = \frac{{  \pi n_0^3 r_{13} Vy}}{{\lambda d}}
\]
[/tex]
Solving for the voltage ([tex]V[/tex]) :
[tex]
\[
V =  \frac{{\Delta \phi \lambda d}}{{\pi n_0^3 r_{13} y}}
\]
[/tex]
Now suppose that the wavelength of the light wave is [tex]\lambda = 450nm[/tex], and that the required phase shift is [tex]\Delta \phi = 2.1 \times 10^6 [/tex]. Note that the sign of the phase shift is positive. I am not completely certain as to whether this is physically possible, but I'll take a look at it mathematically. Suppose that we plug the following into the equation used to solve for voltage:
[tex]
\Delta \phi = 2.1 \times 10^{6}
[/tex]
[tex]
\lambda = 450nm
[/tex]
[tex]
d = 0.1 \times 10^{6} m
[/tex]
[tex]
n_{0} = 2.1763
[/tex]
[tex]
r_{13} = 7.62 \times 10^{12} m/V
[/tex]
[tex]
y = 1.0 \times 10^{1}m
[/tex]
Note that the distance ([tex]d[/tex]) over which the electric field acts is very small: only 0.1 micrometer! The horizontal distance traveled by the light wave through the Lithium Niobate block is 10 cm.
Evaluating the expression, I find that:
[tex]V = 3829.8V[/tex]
Now clearly it is not impossible to generate a voltage of 3kV. Note that as the separation distance ([tex]d[/tex]) increases, the applied voltage required to create the phase shift also increases as well. Increasing the separation distance shows that thousands or millions of volts are required to cause a phase shift this large in the incident light beam. This suggests that the separation distance between the electrodes must be very small for such a setup to be feasible.
This leads me to my questions:
(1) Is there an easier way of doing this: to create a large magnitude phase shift in a light beam?
(2) Can a negative voltage be applied to an electrooptic crystal? If so, is it physically reasonable to use a negative voltage to change the refractive index of the Lithium Niobate block? I think that the answer to this question is "yes," and the answer is supported by the mathematics above. I also think that it is possible for the light to have a positive phase shift.
(3) Is it possible to make such an apparatus? Is it possible to deposit a thin layer of Lithium Niobate (with a thickness of 0.1 micrometer) on a substrate, and attach electrodes to the top and the bottom of the layer to create an electric field?
(4) Does anyone know of a substance with a larger electrooptic coefficient? If [tex]r_{13}[/tex] increases (becomes larger), then the voltage required to create the phase shift drops appreciably.
(5) How might I manufacture such a device? Would it be possible to manufacture something like this? If so, what would be the cost to create this apparatus? Any guesses based on similar or related research? Is there a good monograph on creating tiny (micro) light guides?
(6) Suppose that I use a lens to send an image through the Lithium Niobate block. The lens is coupled to a piece of fiber optic. How would I interface the piece of fiber optic to the left side of the Lithium Niobate block?
(7) Could an image be focused by a lens and sent through the Lithium Niobate block? Note that d = 0.1 micrometer, which is a very small distance.