Dispersion in Optical Fibers

Dispersion is defined as the spreading of a pulse as it propagates through a medium. It essentially causes different components of the light to propagate at different speeds, leading to distortion. The most commonly discussed dispersion in optical fibers is modal dispersion, which is the result of different modes propagating within a MMF (multimode fiber). The fiber optic cable supports many modes because the core is of a larger diameter than SMF (single mode fibers). Single mode fibers tend to be used more commonly now due to decreased attenuation and dispersion over long distances, although MMF fibers can be cheaper over short distances.

Let’s analyze modal dispersion. When the core is sufficiently large (generally the core of a SMF is around 8.5 microns or so), light enters are different angles creating different modes. Because these modes experience total internal reflection at different angles, their speeds differ and over long distances, this can have a huge effect. In many cases, the signal which was sent is completely unrecognizable. This type of dispersion limits the bandwidth of the signal. Often GRIN (graded index) fibers are employed to reduce this type of dispersion by gradually varying the refractive index of the fiber within the core so that it decreases as you move further out. As we have learned, the refractive index directly influences the propagation velocity of light. The refractive index is defined as the ratio of the speed of light to the speed of the medium. In other words, it is inversely proportional to the speed of the medium (in this case silica glass).


In order to mitigate the effects of intermodal distortion in multimode fibers, pulses are lengthened to overlap components of different modes, or even better to switch to Single mode fibers when it is available.

The next type of dispersion is chromatic dispersion. All lasers suffer from this effect because no laser is comprised of a single frequency. Therefore, different wavelengths will propagate at different speeds. Sometimes chirped Bragg gratings are employed to compensate for this effect. Doped fiber lasers and solid state lasers tend to have much thinner linewidths than semiconductor PIN lasers and therefore tend to have less chromatic dispersion, although the semiconductor lasers has several advantages such as lesser cost and smaller size.

Another dispersion type is PMD (Polarization mode dispersion) which is caused by different polarizations travelling at different speeds within a fiber. Generally, these travel at the same speed however spreading of pulses can be caused by imperfections in the material.

For SMF fibers, it is important to cover waveguide dispersion. It is important to note that since the cladding of the fiber is doped differently than the core, the core has a higher refractive index than the cladding (doping with fluorine lowers refractive index and doping with germanium increases it). As we know, a lower refractive index indicates faster speed of propagation. Although most of the light stays within the core, some is absorbed by the cladding. Over long distances this can lead to greater dispersion as the light travels faster in the core leading to different propagation velocities.

Image Resolution

Consider that we are interested in building an optical sensor. This sensor contains a number of pixels, which is dependent on the size of the sensor. The sensor has two dimensions, horizontal and vertical. Knowing the size of the pixels, we will be able to find the total number of pixels on this sensor.

The horizontal field of view, HFOV is the total angle of view normal from the sensor. The effective focal length, EFL of the sensor is then:

Effective Focal Length: EFL = V / (tan(HFOV/2)),

where V is the vertical sensor size in (in meters, not in number of pixels) and HFOV is the horizontal field of view. Horizontal field of view as an angled is halved to account that HFOV extends to both sizes of the normal of the sensor.

The system resolution using the Kell Factor: R = 1000 * KellFactor * (1 / (PixelSize)),

where the Pixel size is typically given and the Kell factor, less than 1 will approximate a best real case result and accounts for aberrations and other potential issues.

Angular resolution: AR = R * EFL / 1000,

where R is the resolution using the Kell factor and EFL is the effective focal length. It is possible to compute the angular resolution using either pixels per millimeter or cycles per millimeter, however one would need to be consistent with units.

Minimum field of view: Δl = 1.22 * f * λ / D,

which was used previously for the calculation of the spatial resolution of a microscope. The minimum field of view is exactly a different wording for the minimum spatial resolution, or minimum size resolvable.

Below is a MATLAB program that computed these parameters, while sweeping the diameter of the lens aperture. The wavelength admittedly may not be appropriate for a microscope, but let’s say that you are looking for something in the infrared spectrum. Maybe you are trying to view some tiny laser beams that will be used in the telecom industry at 1550 nanometer.

Pixel size: 3 um. HFOV: 4 degrees. Sensor size: 8.9mm x 11.84mm.


Spatial Resolution of a Microscope

Angular resolution describes the smallest angle between two objects that are able to be resolved.

θ = 1.22 * λ / D,

where λ is the wavelength of the light and D is the diameter of the lens aperture.

Spatial resolution on the other hand describes the smallest object that a lens can resolve. While angular resolution was employed for the telescope, the following formula for spatial resolution is applied to microscopes.

Spatial resolution: Δl = θf = 1.22 * f * λ / D,

where θ is the angular resolution, f is the focal length (assumed to be distance to object from lens as well), λ is the wavelength and D is the diameter of the lens aperture.



The Numerical Aperture (NA) is a measure of the the ability to of the lens to gather light and resolve fine detail. In the case of fiber optics, the numerical aperture applies to the maximum acceptance angle of light entering a fiber. The angle by the lens at its focus is θ = 2α. α is shown in the first diagram.

Numerical Aperture for a lens: NA = n * sin(α),

where n is the index of refraction of the medium between the lens and the object. Further,

sin(α) = D / (2d).

The resolving power of a microscope is related.

Resolving power: x = 1.22 * d * λ / D,

where d is the distance from the lens aperture to the region of focus.


Using the definition of NA,

Resolving power: x = 1.22 * d * λ / D = 1.22 * λ / (2sin(α)) = 0.61 * λ / NA.


Telescope Resolution & Distance Between Stars using the Rayleigh Limit

Previously, the Rayleigh Criterion and the concept of maximum resolution was explained. As mentioned, Rayleigh found this formula performing an experiment with telescopes and stars, exploring the concept of resolution. This formula may be used to determine the distance between two stars.

θ = 1.22 * λ / D.

Consider a telescope of lens diameter of 2.4 meters for a star of visible white light at approximately 550 nanometer wavelength. The distance between the two stars in lightyears may be calculated as follows. The stars are approximately 2.6 million lightyears away from the lens.

θ = 1.22 * (550*10^(-9)m)/(2.4m)

θ =2.80*10^(-7) rad

Distance between two objects (s) at a distance away (r), separated by angle (θ): s = rθ

s = rθ = (2.0*10^(6) ly)*(2.80*10^(-7)) = 0.56 ly.

This means that the maximum resolution for the lens size, star distance from the lens and wavelength would be that two stars would need to be separated at least 0.56 lightyears for the two stars to be distinguishable.


Diffraction, Resolution and the Rayleigh Criterion

The wave theory of light includes the understanding that light diffracts as it moves through space, bending around obstacles and interfering with itself constructively and destructively. Diffraction grating disperses light according to wavelength. The intensity pattern of monochromatic light going through a small, circular aperture will produce a pattern of a central maximum and other local minima and maxima.


The wave nature of light and the diffraction pattern of light plays an interesting role in another subject: resolution. The light which comes through the hole, as demonstrated by the concept of diffraction, will not appear as a small circle with sharply defined edges. There will appear some amount of fuzziness to the perimeter of the light circle.

Consider if there are two sources of light that are near to each other. In this case, the light circles will overlap each other. Move them even closer together and they may appear as one light source. This means that they cannot be resolved, that the resolution is not high enough for the two to be distinguished from another.


Considering diffraction through a circular aperture the angular resolution is as follows:

Angular resolution: θ = 1.22 * λ/D,

where λ is the wavelength of light, D is the diameter of the lens aperture and the factor 1.22 corresponds to the resolution limit formulated and empirically tested using experiments performed using telescopes and astronomical measurements by John William Strutt, a.k.a. Rayleigh for the “Rayleigh Criterion.” This factor describes what would be the minimum angle for two objects to be distinguishable.

Optical Polarizers in Series

The following problems deal with polarizers, which is a device used to alter the polarization of an optical wave.

  1. Unpolarized light of intensity I is incident on an ideal linear polarizer (no absorption). What is the transmitted intensity?

    Unpolarized light contains all possible angles to the linear polarizer. On a two dimensional plane, the linear polarizer will emit only that amount of light intensity that is found in the axis of polarization. Therefore, the Intensity of light emitted from a linear polarizer from incident unpolarized light will be half the intensity of the incident light.

  2. Four ideal linear polarizers are placed in a row with the polarizing axes vertical, 20 degrees to vertical, 55 degrees to vertical, and 90 degrees to vertical. Natural light of intensity I is incident on the first polarizer.

    a) Calculate the intensity of light emerging from the last polarizer.

    b) Is it possible to reduce the intensity of transmitted light (while maintaining some light transmission) by removing one of the polarizers?

    c) Is it possible to reduce the intensity of transmitted light to zero by removing a polarizer(s)?

    a) Using Malus’s Law, the intensity of light from a polarizer is equal to the incident intensity multiplied by the cosine squared of the angle between the incident light and the polarizer. This formula is used in subsequent calculations (below). The intensity of light from the last polarizer is 19.8% of the incident light intensity.

    b) My removing polarizer three, the total intensity is reduced to 0.0516 times the incident intensity.

    c) In order to achieve an intensity of zero on the output of the polarizer, there will need to exist an angle difference of 90 degrees between two of the polarizers. This is not achievable by removing only one of the polarizers, however it would be possible by removing both the second and third polarizer, leaving a difference of 90 degrees between two polarizers.





Jones Vector: Polarization Modes

The Jones Vector is a method of describing the direction of polarization of light. It uses a two element matrix for the complex amplitude of the polarized wave. The polarization of a light wave can be described in a two dimensional plane as the cross section of the light wave. The two elements in the Jones Vector are a function of the angle that the wave makes in the two dimensional cross section plane of the wave as well as the amplitude of the wave.


The amplitude may be separated from the ‘mode’ of the vector. The mode of the vector describes only the direction of polarization. Below is a first example with a linear polarization in the y direction.


Using the Jones Vector the mode can be calculated for any angle. See calculations below:


The phase differences of the Jones Vector are plotted for a visual representation of the mode. If both components of the differ in phase, the plot depict a circular or oval pattern that intersects both components of the mode on a two dimensional plot. The simplest of plots to understand is a polarization of 90 degree phase difference. In this case, both magnitudes of the components of the mode will be 1 and a full circle is drawn to connect these points of the mode. In the case of a zero phase difference, this is demonstrated at 45 degrees where both sin(45deg) and cos(45deg) equal 0.707. In this case, the phase difference is plotted as a straight line, indicating that polarization is of equal phase from each axis of the phase difference plot.




Optical Polarization, Malus’s Law, Brewster’s Angle

In the theory of wave optics, light may be considered as a transverse electromagnetic wave. Polarization describes the orientation of an electric field on a 3D axis. If the electric field exists completely on the x-axis plane for example, light is considered to be polarized in this state.

Non-polarized light, such as natural light may change angular position randomly or rapidly. The process of polarizing light uses the property of anisotropy and the physical mechanisms of dichroism or selective absorption, reflection or scattering. A polarizer is a device that utilizes these properties. Light exiting a polarizer that is linearly polarized will be parallel to the transmission axis of the polarizer.



Malus’s law states that the transmitted intensity after an ideal polarizer is

I(θ)=I(0)〖cos〗^2 (θ),

where the angle refers to the angle difference between the incident wave and the transmission axis of the polarizer.

Brewster’s Angle, an extension of the Fresnel Equation is a theory which states that the difference between a transmitted ray or wave into a material comes at a 90 degree angle to the reflected wave or ray along the surface. This situation is true only at the condition of the Brewster’s Angle. In the scenario where the Brewster’s Angle condition is met, the angle between the incident ray or wave and the normal, the reflected ray or wave and the surface normal and the transmitted ray or wave and the surface normal are all equal.


If the Brewster’s Angle is met, the reflected ray will be completely polarized. This is also termed the polarization angle. The polarization angle is a function of the two surfaces.







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.