Tag Archives: Optics

Fourth Generation Optics: Thin-Film Voltage-Controlled Polarization

Michael Benker
ECE591 Fundamentals of Optics & Photonics
April 20,2020


Dr. Nelson Tabiryan of BEAM Engineering for Advanced Measurements Co. delivered a lecture to explain some of the latest advances in the field of optics. The fourth generation of optics, in short includes the use of applied voltages to liquid crystal technology to alter the polarization effects of micrometer thin film lenses. Both the theory behind this type of technology as well as the fabrication process were discussed.


First Three Generations of Optics

A summary of the four generation of optics is of value to understanding the advancements of the current age. Optics is understood by many as one of the oldest branches of science. Categorized by applications of phenomena observable by the human eye, geometrical optics or refractive optics uses shape and refractive index to direct and control light.

The second generation of optics included the use of graded index optical components and metasurfaces. This solved the issue of needing to use exceedingly bulky components although it would be limited to narrowband applications. One application is the use of graded index optical fibers, which could allow for a selected frequency to reflect through the fiber, while other frequencies will pass through.

Anisotropic materials gave rise to the third generation of optics, which produced technologies that made use of birefringence modulation. Applications included liquid crystal displays, electro-optic modulators and other technologies that could control material properties to alter behavior of light.


Fourth Generation Optics

To advance technology related to optics, there are several key features needed for output performance. A modernized optics should be broadband, allowing many frequencies of light to pass. It should be highly efficient, micrometer thin and it should also be switchable. This technology is currently present.

Molecule alignment in liquid crystalline materials is essential to the theory of fourth generation optics. Polarization characteristics of the lens is determined by molecule alignment. As such, one can build a crystal or lens that has twice the refractive index for light which is polarized in one direction. This device is termed the half wave plate, which polarizes light waves parallel and perpendicular to the optical axis of the crystal. Essentially, for one direction of polarization, a full period sinusoid wave is transmitted through the half wave plate, but with a reversed sign exit angle, while the other direction of polarization is allowed only half a period is allowed through. As a result of the ability to differentiate a sign of the input angle to the polarization axis (full sinusoid polarized wave), the result is an ability to alter the output polarization and direction of the outgoing wave as a function of the circular direction of polarization of the incident wave.

The arrangement of molecules on these micrometer-thin lenses are not only able to alter the direction according to polarization, but also able to allow the lens to act as a converging lens or diverging lens. The output wave, a result of the arrangement of molecules in the liquid crystal lens has practically an endless number of uses and can align itself to behave as any graded index lens one might imagine. An applied voltage controls the molecular alignment.

How does the lens choose which molecular alignment to use when switching the lens? The answer is that, during the fabrication process, all molecular alignments are prepared that the user plans on employing or switching to at some point. These are termed diffraction wave plates.



Problem 1.


The second lens is equivalent to the first (left) lens, rotated 180 degrees. In the case of a polarization-controlled birefringence application, one would expect lens 2 to exhibit opposite output direction for the same input wave polarization as lens 1. For lens 1 (left), clockwise circularly polarized light will exit with an angle towards the right, while counterclockwise circularly polarized light exits and an angle to the left. This is reversed for lens 2.



Problem 2.


There are as many states as there are diffractive waveplates. If there are six waveplates, then there will be 6 states to choose from.


Focal Length of a Submerged Lens

Is the focal length of a spherical mirror affected by the medium in which it is immersed? …. of a thin lens? What’s the difference?



A spherical mirror may be either convex or concave. In either case, the focal length for a spherical mirror is one-half the radius of curvature.


The formula for focal length of a mirror is independent of the refractive index of the medium:




The thin lens equation, including the refractive index of the surrounding material (“air”):lens2

The effect of the refractive index of the surrounding material can be summarized as follows:

  • The focal length is inversely proportional to the refractive index of the lens minus the refractive index of the surrounding medium.
  • As the refractive index of the surrounding medium increases, the focal length also increases.
  • If the refractive index of the surrounding medium is larger than the refractive index of the thick lens, the incident ray will diverge upon exiting the lens.


Infinite Lateral Magnification of Lenses and Mirrors

Under what conditions would the lateral magnification (m=-i/o) for lenses and mirrors become infinite? Is there any practical significance to such condition?


Magnification of a lens or mirror is the ratio of projected image distance to object distance. Simply put, how much closer does the object appear as a result of the features of the lens or mirror? The object may seem larger or it may seem smaller as a result of it’s projection through a lens or mirror. Take for instance, positive magnification:


If the virtual image appears further than the real object, there will be negative magnification:


The formula for magnification is the following:


The question then is, how can there be an infinite ratio of image size to object size? Consider the equation for focal length:


For magnification to be infinite, the image distance should be infinite, in which case the object distance is equal to the focal length:


In this case, the magnification is infinite:


The meaning of this case is that the object appears as if it were coming from a distance of infinity, or very far away and is not visible. A negative magnification means that the image is upside-down.


Focal Length of a Lens as a function of light frequency

How does the focal length of a glass lens for blue light compare with that for red light? Consider the case of either a diverging lens or a converging lens.


This question really has three parts:

  • Focal Length of a lens
  • Effect of light frequency (color)
  • Diverging and Converging lens


Focal Length of the Converging and Diverging Lens

For the converging and diverging lens, the focal point has a different meaning. First, consider the converging lens. Parallel rays entering a converging lens will be brought to focus at the focal point F of the lens. The distance between the lens and the focal point F is called the focal length, f. The focal length is a function of the radius of curvature of both sides or planes of the lens as well as the refractive index of the lens. The formula for focal length is below,
(1/f) = (n-1)((1/r1)-(1/r2)).

This formula also works for a diverging lens, however the directions of the radius of curvature must be taken into account. If for instance the center of the circle for one side of the lens is to the left of the lens, one may chose that direction to be positive and the other direction to be negative; as long as one maintains the same standard for direction.


If the focal length of a lens is negative, meaning that the focal point is behind the lens, on the side at which the rays entered, this is a diverging lens.



Interaction of Color with Focal Length

The other part of this question dealt with how the focal length would change for one color such as blue versus another color such as red. The key to this relationship is the refractive index of the lens, as the refractive index can change with regards to the color (i.e. frequency).

The material from which the lens is made is not known, however as demonstrated by the following table, the refractive index is consistently higher for smaller wavelength colors.


Reviewing the focal length formula, it is understood from the inverse proportionality of the equation that as the refractive index increases, the focal length will decrease. Blue has a higher refractive index than red. Therefore, blue will have a smaller focal length than red.




Refractive Index as a Function of Wavelength

Previously, we discuss how the resultant wavelength and velocity in an optical system is said to be dependent on the refractive index. What we didn’t explain however is that the relationship between refractive index and wavelength more often involves a dependency of the refractive index according to the incident wavelength. After all, it is easier to change the wavelength of a light wave than it is to change the material that it is propagating through. So in fact, the refractive index will vary according to the wavelength of the incident wave. If the system is not monochromatic, the frequency may also change.


As we know from ray optics or geometrical optics is that the refractive index is used to determine how a ray will travel through an optical system. The relationship between wavelength and refractive index implies that an optical system with the same material will produce a different transmission angle (or perhaps a completely different result) for two rays of different wavelength.

Consider the range of refractive indexes for several different mediums with an altered wavelength and color (i.e frequency):


The differences in refractive indexes for these materials given different wavelengths and frequencies may seem small, however the difference is enough that rays of different wavelengths will interact slightly differently through optical systems.

Now, what if a ray managed to contain more than one wavelength? Or, if it were a blend of all colors? This case is called white light. If white light can contain a sum of a number of wavelengths and frequencies, each component of white light will behave according to it’s relative refractive index.

The classic example of this is of course the prism.



Refractive Index, Speed of Light, Wavelength and Frequency

The relationship between the speed of light in a medium and the refractive index is the following:


Therefore it can be understood that for a medium of higher refractive index, the speed of light in that medium will be slower. Light will not achieve a speed higher than c or 2.99 x 10^8 m/s. When light is traveling at this speed the refractive index of the medium is 1.00.

Now, what about the wavelength? Interestingly, one might begin to understand that the wavelength is the determining factor for color. In fact, this is not the case. Frequency is what defines the color of the light, which can vary from an invisible infrared range to the visible range to the invisible ultraviolet range. In a monochromatic system, the frequency of light (and therefore color) will stay the same. The velocity and wavelength will change with the refractive index.


As the above picture suggests, we might beleive that wavelength and frequency are forever tied together. The above example would in fact be incomplete at best, were we to consider that light can travel at more than one speed. However, let us review the relationship between wavelength and frequency. The following formula is normally presented for wavelength:


Now, here is the question: does c in this equation correspond to the speed of light in a vacuum, or does it correspond to the speed of the travelling light wave? Let’s consider, what does the speed of light in a vacuum have to say about the speed of light in water? It really doesn’t have much to say, does it? Which is why we can use instead, v to denote the speed of light.


Note that I’ve written the wavelength as a function of the speed of light in the medium. Taking this to it’s conclusions, we would understand that actually, the wavelength is not exclusively dependent on frequency and that multiple wavelengths may exist for one frequency. The determining factor in such a case is the refractive index, given that frequency is constant.


Given the wavelength, frequency and refractive index, the speed of the light wave may also be calculated.


Physically, one may picture that the frequency is the rate at which the peak of a wave passes by a point. A longer wavelength wave will need to move faster to keep at the same frequency.

The applications and implications of this physical relationship will be explored next.