Final examination
Theory

1. Explain the dispersion of light by a transparent dielectric material. The phenomen of dispersion is connected to the fact that the refractive index is dependent on the wavelength. Due to dispersion, the light waves from a complex radiation are bent by different angles as they enter a dispersive medium and they may be visualized separately. The index of refraction of a transparent dielectric medium is defined as the ratio of the speed of an electromagnetic wave in empty space to its speed in the medium. .
It is a measure of the slowing factor for light traveling in that medium
The refractive index can be expressed as a function of the electric and magnetic properties of the medium .
For most of the materials that are transparent to visible light and .
Hence the magnetic properties have a small effect on the light propagation. When an electromagnetic wave is incident on a medium, it electrically polarizes the molecules. This changes the value of which in its turn, determines the index of refraction. The process is wavelength dependent: different wavelengths will induce different polarizations of the molecules and, as a result, and will be different. So, the index of refraction changes with wavelength: The phenomenon described above is known as the dispersion of light. Equation is the dispersion relation.
For most of the materials, decreases with the wavelength (see figure1). This phenomenon is known as normal dispersion. The decreasing is more rapid at shorter wavelengths. This causes a Figure 1. Normal dispersion curved plot of versus. where, are empirical curve of flint glass constants whose values depend on the specific material.

2. The prism spectroscope( working principle, setup, application) A spectroscope is an instrument for visual observation. A prism’s spectroscope main component is an optical prism through which incident light is refracted and thus dispersed. A narrow entrance slit admits the light from the source then a collimator turns it into a parallel beam. This beam is incident on the prism and dispersed. Finally, a telescope takes the parallel beam which exits the prism and creates images of the entrance slit. Due to dispersion by refraction, they are located at different points for each wavelength present in the incident light. These images are called spectral lines. Together they make up the spectrum of the incident light. . A discontinuous (discrete) spectrum is composed of individual colored lines separated by dark regions while a continuous spectrum is a band in which a continuum of colors from red to violet is displayed. Occasionally dispersion is an unwanted effect. For instance, single lens components can produce some coloring which deteriorates the performance of an optical instrument. This problem is known as chromatic aberration. Solving it means keeping the red and violet rays (the extremes of the optical range) close together as they travel through the optical instrument. This is done by using a combination of two lenses, called an achromatic doublet, instead of single lenses.

3. Discuss the total internal refraction of a white light beam on a glass/air interface by taking into consideration the normal dispersion of light in glass.
The total internal reflection is a phenomenon which occurs when the refractive index of the medium in which light is transmitted is lower than the refractive index of the medium in which light is incident, .
Consider the situation described in figure 2. As the angle of incidence increases, the transmitted ray bends away more and more from the normal (rays 1, 2 and 3). At a particular value of the angle of incidence which is called critical angle (), the angle of refraction becomes as large as and the ray can no longer be transmitted through the interface into the second medium (ray 4). Rays that are incident on the boundary surface under angles higher than the critical one will be back reflected from the interface (ray 5).

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Figure 2. Total internal reflection of light

4. Find the formula for critical angle for total internal reflection of light and discuss it’s significance The critical angle is obtained by applying Snell’s law of refraction: . Then .
If the second medium is air (), then the critical angle is given by .
Notice that due to dispersion, different wavelengths have different critical angles. For a white light ray (wavelengths in the visible range) incident on a glass/air interface, the critical angle ranges in an interval . So, the smallest critical angle corresponds to the highest refractive index, which is in normal dispersion. On the other hand, the largest critical angle corresponds to the smallest refractive index, which is . Then total internal reflection of a white light beam occurs if the angle of incidence is larger than the largest critical value, that is , otherwise some wavelengths will be transmitted into the second medium.

5. What are evanescent waves? Find an expression for evanescent wave function. Examining the possibility that some fraction of the incident light intensity is still transmitted in the second medium, in spite of the fact that the angle of incidence is such that total internal reflection does occur. For this we will consider that the plane of incidence of the light is , with lying along the interface (see figure 3).

Figure 3. Evanescent waves at the interface between an optically dense medium and a medium less dense

For the time being we will pretend that there exists a transmitted ray which obeys Snell’s law of refraction which is in a direction that makes an angle with the normal at the incidence point (keeping however in mind that ). For reasons you will understand in the following this transmitted light beam is not represented in figure 3.
Then the components of the transmitted wave vector and those of the position vector are The oscillations of the electric field intensity in the transmitted wave are described by an equation (in the complex representation) (*) in which we used the scalar product . As expected, due to the fact that , Snell’s law of refraction gives and accordingly, an imaginary value of the cosine function of the refraction angle: .
We will however take this imaginary expression in equation (*) which describes the electric field oscillations in the transmitted wave. Thus

This is the equation of a wave which propagates in the direction (meaning along the interface) whose wavelength is (see figure 3). The amplitude of this wave is exponentially decreasing in the direction normal to the interface. Experiments show that this type of waves, which look like the result of pure mathematical manipulation, does exist in reality. They are called “evanescent” (meaning “which tend to be vanishing”) because of the fast decreasing amplitude.

6. The optical fiber (structure, light incidence requirements, applications)
Optical fibers use total internal reflection to transmit light signals and images. A typical optical fiber is made of a high refractive index core and a low refractive index coating (cladding), as shown in figure 6. Some fibers are provided with EMA (extra mural absorption), to suppress evanescent waves. Light needs to be incident approximately parallel to the fiber axis, so that it can strike the walls at angles higher than the critical angle of the core/cladding interface and be thus totally reflected. As a result, light is trapped inside; once it is input, it will continue to reflect almost without loss off the walls of the fiber (see figure 4). Optical fibers (single or bundles) are the main components of fiber optic cables which are commonly used in telecommunications.

Figure 4. Optical path of light in an optical fiber

As we have already mentioned, for the light to be confined inside the fiber, the angle of incidence to the core/cladding interface has to be greater than the limiting angle for total internal reflection. The critical angle corresponds to a refraction angle of 90°.

Figure 5. The structure of an optical fiber

In figure 5, the refractive index of the core is denoted by and the refractive index of the cladding is . According to Snell’s law of refraction, at the point where light strikes the fiber wall, . To undergo total internal reflection light should be input at an angle less than a maximum value corresponding to .
The main applications of fiber optics are in the field of medicine and communications. In medicine, optical fibers are used to visualize various internal organs. The fibers are tiny, flexible, have little loss and allow for straightforward construction of the images. All these qualities make them extremely suitable for medical investigations. In communications, optical fibers are used to transmit information at a much higher rate than systems using radio frequencies, since the rate of transmission is related to the signal frequency. Also, the volume of information is huge: a single fiber the thickness of a hair can transmit audio information equivalent to 32000 voices speaking simultaneously. If we add to that the fact that optical fibers are not subject to electrical interference and are long-life devices, you will understand why they are preferred to the classic telecommunication systems. Original patents on fiber optic transmission were granted to John Logie Baird in the 1930's.

7. Determine the numerical aperture of an optical fiber and discuss the physical significance of the formula.
To undergo total internal reflection light should be input at an angle less than a maximum value corresponding to . The value of is called numeric aperture NA.
The numeric aperture is easily computed as a function of the refractive indices of the optical fiber. From Snell’s law and [ ]: If , where is a small quantity, the last member above can be rewritten as The value of NA gives the input angle range for a ray of light to be transmitted in the fiber.

8. Discuss the physical significance of the optical path concept.
Let us consider that light propagates in space between two points P and Q. The geometrical path is the actual distance traveled by light, i.e. the distance that separates the two points, denoted by . The optical path length is defined as the geometrical path length times the refractive index of the medium in which light propagates and is denoted by , .
The optical path length in vacuum is equal to the geometrical path, . In any other medium the optical path length is longer.
Light travels at its highest speed in a vacuum. In a medium, it is slowed down and it travels at a speed which is times smaller than . Say that light takes seconds to travel a distance in a medium of refractive index . In the same time interval light would travel a distance times longer if it propagated in vacuum. Indeed, we have .
So, the optical path length is the distance that light would travel in vacuum, in the time necessary to travel a geometrical path in a medium whose optical properties are described by the refractive index . Obviously, the higher the refractive index, the smaller the speed is and the optical path length is thus longer.

9. Demonstrate that the optical significance of the optical path length of light travelling in a material determines the time of propagation, the phase difference and the number wavelengths that span the geometrical path of light

Light travels at its highest speed in a vacuum. In a medium, it is slowed down and it travels at a speed which is times smaller than . Say that light takes seconds to travel a distance in a medium of refractive index . In the same time interval light would travel a distance times longer if it propagated in vacuum. Indeed, we have . (**)
So, the optical path length is the distance that light would travel in vacuum, in the time necessary to travel a geometrical path in a medium whose optical properties are described by the refractive index . Obviously, the higher the refractive index, the smaller the speed is and the optical path length is thus longer.
Note that this physical quantity takes into account both the actual distance traveled by light and its speed. It is a crucial concept for understanding light propagation in a dispersive medium. Since any medium is more or less dispersive, we understand that optical path length is extremely useful for any piece of optics.
The optical path length determines the propagation time - see equation (**), the difference in phase introduced by light traveling a geometrical path : and the number of wavelengths that span a geometrical path : .
Note that the propagation of light in a medium can be described in terms of propagation in vacuum, simply by taking the precaution of replacing the geometrical paths by the corresponding optical paths.

10. Formulate Fermat’s Priciple, discuss it’s physical significance and give its analytical expression.
Fermat’s principle originally stated that of all the geometrically possible optical paths that light could take between two points P and Q in a medium, the actual path is that of least time. In other words, the propagation time has a minimum value on the real path. The modern formulation of this principle is that of all the geometrically possible optical paths that light could take between two arbitrary points P and Q in a medium, the actual optical path is that of a stationary value. By “stationary” we mean a minimum or maximum value. According to the propagation time is proportional to the optical path length, so there is no contradiction between the two statements. It is just that the modern formulation is more general. If the refractive index is a function of position, , then the optical path length is defined as , where C is the curve that connects arbitrary P and Q. Fermat’s principle states that the real curve corresponds to a stationary value of the optical path length integral, that is . Note that, once found, the actual optical path length stays the same for any pair of points. Hence if light propagates from P to Q on , it will take the same path from Q to P. This is what the principle of reversibility of light rays states. Now we can see that it results as a direct consequence of Fermat’s principle.

11. Use Fermat’s Priciple to demonstrate the law of reflection To derive the law of reflection we can apply Fermat’s principle. Consider that light gets from P to Q after perfect reflection by a mirror (surface ). There are an infinity geometrically possible paths which meet this condition. However path POQ is the actual one (see figure 6). Light travels only in air, whose refractive index is almost unity. The total optical path is the sum of the optical path from P to the plane of the mirror and the optical path from there to Q.

O
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Q

Figure 6.

With the denotations in the figure, we have where is the overall optical path length and was used. For the correct position of point O, should exhibit a maximum value, i.e. .
It results .
Now we go back to the figure and notice that in and the cosine functions of the complementary angles of and can be expressed as

Consequently, if we recall equation [ ], we see that which is the law of reflection.

12. Use Fermat’s Priciple to demonstrate Snell’s law of refraction. To derive the law of refraction we can apply Fermat’s principle. Consider that P and Q lie in two different dielectric transparent media separated by the surface (see figure 8), which are described by the indices of refraction and .

O
P
Q

Figure 7.

The total optical path is thus the sum of the optical path of the ray in the medium where light is incident and the optical path in the medium where light is refracted. .
We apply Fermat’s principle: which results in the following equation .
This is the law of refraction. As expected, by applying Fermat’s principle we obtained that the actual path of light that crosses the boundary surface between two dielectric media is subject to Snell’s law of refraction.

13. State the Huygens Principle discuss its physical significance and mention its flaws.
Huygens’ Principle states that ”every point on a primary wave front serves as a source of secondary spherical wavelets, which advance with a speed equal to that of the primary wave front at each point in space; the primary wave front at some later time is the envelope of these wavelets”.
Basically this is a geometrical technique for finding the path of a wave, regardless its wavelength. All the points on the primary wave front are thus considered sources of secondary wavelets spreading out in all directions, with a speed equal to the wave speed in the medium. One can actually draw arcs of a circle with a radius, then draw the surface tangent, or envelope, to find the new wave front at time (figure 8).

According to Huygens' Principle, the light rays associated with a plane light wave in empty space propagate in straight lines. It is fairly straightforward to account for the laws of reflection and refraction using Huygens' Principle. However, the principle fails to explain why, for example, an expanding spherical wave continues to expand outward from its source, rather than converging back toward the source. In other words, Huygens overlooks the fact that the wave front formed by the back half of the wavelets implies a light disturbance traveling in the opposite direction. Also, it does not account for diffraction of the light into the region of geometric shadow, when light is incident on obstacles.

Figure 8.

14. State the Huygens-Fresnel Principle, discuss its physical significance and give its analytical expression.

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The Huygens-Fresnel Principle states: Every unobstructed point of a wave front is a source of spherical secondary wavelets with the same frequency as that of the primary. The amplitude of the resultant wave at any forward point is the superposition of these wavelets, considering their amplitudes and relative phase. In other words, the secondary waves mutually interfere and this accounts for light diffraction, as we are going to show in the following chapters.

Figure 9.

According to Huygens-Fresnel Principle, the equation which expresses the electric field intensity in the spherical wavelet emitted by an arbitrary elementary area of an arbitrary wave front at a point of observation P, located at the distance from , at any time , can be written as , where and are the amplitude and initial phase of the electric field oscillation, and are the angular frequency and the wave number of the wavelets (figure 9). The quantity is a constant of proportionality which depends on the angle between the direction of the normal to the elementary surface and the direction of the vector position of the observation point. It describes how waves propagate at maximum intensity in the forward direction and cannot propagate back to the source. Thus, and The factor is due to the stated spherical character of the wavelets. Their amplitude thus diminishes even when the wave propagates in a nonabsorbing medium. The resultant wave at P is obtained by superimposing all the secondary wavelets generated by the wave front S: .
This equation is an analytic expression of the Huygens-Fresnel Principle. It is used to compute the intensity of the diffracted light at any point in space and any time. The calculation of the integral is however far from being easy. It presumes knowing the angle for any infinitesimal surface belonging to the primary front wave S, at any point P in space. For an arbitrary surface, the calculations involve sophisticated approximating techniques. Fortunately, where highly symmetric front surfaces are concerned, the calculation reduces to a mere algebraic summation, as we are going to see.

15. Describe the linear state of polarization. What are the ratio of amplitudes and the phase difference between two perpendicularly oscillating components of a linearly polarize wave?

The existence of polarization phenomena is a direct consequence of light being a transverse wave. In a plane electromagnetic wave, the electric field vector oscillating in an arbitrary direction can always be resolved into two components parallel to the coordinate axes in the plane of its oscillation which is perpendicular to the direction of propagation. Let label the direction of travel of the light and and be directions in the plane of oscillation. Then the components of the electric field vector of the light wave will vibrate as , , where a difference in phase between the two perpendicular disturbances was introduced. The electric field vector is thus .
According to the ratio of the amplitudes and the difference in phase between the component vibrations, an electromagnetic wave can be linearly, circularly or elliptically polarized. With linear polarization, the direction of the electric field vector resides in a fixed plane and it stays constant at any point in space, at any time. The plane of polarization of the electromagnetic wave is defined by the direction of the electric field vector and by the direction of propagation. Since it has a fixed position, this type of polarization is also called plane polarization.

The electric field makes a constant angle to the axis, as in the figure 11.
Figure 10.The electric field vector in a plane linearly polarized wave

Figure11. The electric field vector in a plane linearly polarized wave

y z z y The electric field vector oscillates in space and in time at the frequency of the light wave. If the direction of is constant in space, then the changes in should be in-step along and directions. Thus, for we have and, the electric field vector oscillates as in the direction described by the angle (see figure 10). Also, the amplitude of the oscillation is given by .

Similarly, for the component disturbances are out of phase by radians and the electric field vector of the linearly polarized wave oscillates as in a direction which makes an angle with the Oy axis (see figure 11). In the figures below the time evolution of the electric field intensity vector is represented for (figure 12) and (figure 13), respectively. It is convenient to use the origin of the axis of propagation as an observation point. At that location, the oscillations in the component disturbances are described by the equations: for

and, respectively for .

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Figure 12. Time evolution of the electric field components in linearly polarized light ()

Notice that the two projections and acquire simultaneously the maximum, zero and minimum values, such that the angle stays constant at any instant of time. The time evolution will be similar at any arbitrary point , since the component

disturbances will exhibit there a supplementary phase (the same for each of them), which does not change the value of the difference in phase .

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Figure 13 Time evolution of the electric field components in linearly polarized light
()

In conclusion, any arbitrary linearly polarized plane electromagnetic wave will always be a superposition of two orthogonal linearly polarized waves, which have a difference in phase given by [ ] or [ ]. A linear state of polarization is sometimes designated as a P -state, meaning a plane state of polarization.

16. Describe the circular state of polarization. What are the ratios of amplitudes and the phase difference between the two perpendicularly oscillating components of a right an left circularly polarize wave?
With circular polarization, the direction of the electric field vector is no longer fixed. In fact, it rotates in time, so that the tip of sweeps out a circular helix whose axis coincides to the direction of propagation of the wave. The frequency of rotation is the wave frequency. In circularly polarized light two different states of polarization are possible: right and left. If an observer looking toward the source (the wave is approaching him) can detect a clockwise rotation of , the wave is right circularly polarized — R -state. If the rotation is counterclockwise, the wave is left circularly polarized — L-state. An arbitrary right circular state of polarization can be expressed in terms of two orthogonal P-states of equal amplitude , whose difference in phase is given by .
Thus the component disturbances are and .
Let us consider the evolution in time of the component orthogonal oscillations of the wave at and .
Figure [ 2 ]

Notice that when reaches its maximum value, is null and vice versa, so that the tip of the light vector in the R-state of polarization rotates clockwise on a circle of radius while traveling in the direction as well (see figure 14)
The L-state of polarization is a superposition of two orthogonal P-states of equal amplitudes and a difference in phase .
Thus (1) and (2).
The resultant L-state wave field is a superposition of (1) and (2), i.e. a vector of constant amplitude which rotates counterclockwise at the wave frequency (see figure 15).

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Figure 14. Time evolution of the electric field components in right circularly polarized light,
; the Oy axis is horizontal, the Oz axis is vertical, the positive Ox direction points toward the reader and is showing the direction of propagation

Figure 15. Time evolution of the electric field components in left circularly polarized light,
; the Oy axis is horizontal, the Oz axis is vertical, the positive Ox direction points toward the reader and is showing the direction of propagation

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17. Show that a linearly polarized wave can be through as a superposition of a right and a left circularly polarized wave. By overlying an R-state wave field and an L-state wave field, a P-state wave field of the same frequency is obtained. Indeed, let us define a superposition of a right and left state of polarization according to [ ] [ ]
Then

It results that any P-state wave field can be described as a superposition of an R- and L-state, provided that their amplitudes are equal.

18. Draw the graph which represents the dependence of the reflectance coefficients Rll and Ron the angle of the incidence and discuss its physical significance

The intensity coefficients of reflection in the parallel and perpendicular directions are defined as irradiance ratios. They are also named reflectance coefficients or simply reflectance. The corresponding Fresnel equations are and At normal incidence , we obtain and ,

Angle of incidence
Reflectance
Figure 16 which means that the reflectance of light whose polarization is parallel to the plane of incidence is equal to the reflectance of light with normal polarization. For air-glass interface , the reflectance coefficients are given by [ ]
Only 4% of the incident light is thus reflected and the rest is transmitted. In figure 16, the dependence of the reflectance on the incidence angle is presented. The intensity reflectance of the perpendicular state of polarization increases monotonously from 4%- the value corresponding to the normal incidence - to 100% at grazing incidence, that is .

19. Describe the applications of light polarization by reflection: polarizing lenses, Brewster’s angle polarizer, Brewster’s winow.

Equations and yield that the zero reflectance of the parallel polarization occurs at a particular angle of incidence , where

Figure 17.
( is called the Brewster angle of incidence after the name of the scientist who discovered this phenomenon). As a result, light reflected at this angle will be totally polarized in a direction which is perpendicular to the plane of incidence since. As far as the transmitted light is concerned, it is partially polarized, predominantly in the parallel direction, as figure 17 indicates. Let us calculate the angle as a function of the refractive indices of the two media.
According to Snell’s law , where equation was used to eliminate the angle of refraction . Then For air-glass interface Brewster’s angle is given by The polarization of light by reflection is used in some optical devices intended to obtaining polarized light or to the transmission of polarized light. Brestwer’s angle polarizer is a device which is used to obtain totally polarized light. It is composed of a stack of glass plates disposed so that the natural light is incident at Brewster’s angle (see figure 18). The light reflected by the first plate is totally polarized in a direction perpendicular to the plane of incidence. The transmitted light is partially polarized in a direction parallel to the plane of incidence. It is subsequently incident on the second plate at the Brewster incidence, such that the light transmitted by the second plate will exhibit a higher degree of polarization in a direction parallel to the plane of incidence. The process is repeated at the incidence on the next plate. As a result, the light transmitted by the stack of plates is practically totally polarized in a direction parallel to the plane of incidence. The intensity of that light is however much lower, due to the fact that some energy is lost at each reflection.

Figure 18.

Brester’s window is a device which transmits 100% polarized light, without losses by reflection. It is composed of a glass plate placed at Brewster’s angle with respect to the vertical direction. It is used in laser systems as an exit window for polarized laser light. As you can see in figure 19, laser light which is vertically polarized in the plane of incidence impinges on the exit window of a laser system at Brewster’s angle. Since there is no light polarized in the horizontal direction, the entire incident light is transmitted into the plate and subsequently exits the window, while maintaining the incident vertical state of polarization.

Figure 19.

20. What is the polarization of light by absorbtion and how is it obtained? Write Malus law an discuss its physical significance.

According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer.
i.e I ∞ cos2θ

Suppose the angle between the transmission axes of the analyzer and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyzer. If E0 is the amplitude of the electric vector transmitted by the polarizer, then intensity I0 of the light incident on the analyzer is I ∞ E02 The electric field vector E0 can be resolved into two rectangular components i.e E0 cosθ and E0 sinθ.

The analyzer will transmit only the component ( i.e E0 cosθ ) which is parallel to its transmission axis. However, the component E0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,
I ∞ ( E0 x cosθ )2 I / I0 = ( E0 x cosθ )2 / E02 = cos2θ I = I0 x cos2θ
Therefore, I ∞ cos2θ.
This proves law of malus.
When θ = 0° ( or 180° ), I = I0 cos20° = I0

That is the intensity of light transmitted by the analyzer is maximum when the transmission axes of the analyzer and the polarizer are parallel. When θ = 90°, I = I0 cos290° = 0 That is the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other.

21. Linearly polarizing sheets (ideal and real sheets, construction, working principle, applications); discuss Malus law when applied to real polarizing sheets.

By definition if natural light is incident on an ideal linear polarizer only light is a P-state will be transmited that P-state will have an orientation parallel to a specific direction called the transision axis of polarizer. Only the component of the optical field parallel to the transmission axis will pas through the device essentially unaffected. If the polarizer is rotate about the z axis, the reading of the detector will be unchanged because of the complete symmetry of unpolarize light. We are dealing with waves, so because of the very high frequency of light our etector will measure only the incient irraiance. Since the irradiance is proportional to the square of the aplitue of the electric field we need only concern ourselves with that aplitude.

Now suppose that we introduce a second identical ideal polarizer, or analyser, whose transmition axis is vertical. If the amplitude of the electric field transmited by the first polarizer is E01, only its component, E01cos, parallel to the transmission axis of the analyser will be passed on to the detector(assuming no absorbtion). The irradiance reaching the dedector, when the angle between the polarizer and analyzer is 0, is given by : I=I0cos2
This is known as Malus law.

20. Describe the working principle of a liquid crystal display by using the concept of optical activity.
A liquid crystal display (LCD) is a thin, flat electronic visual display that uses the light modulating properties of liquid crystals (LCs). LCs do not emit light directly. Liquid crystals have long cigar-shaped molecules that can move about and consequently. Their molecules strongly interact to sustain a large-scale orientational order, there is a large friction between molecules. Each pixel of an LCD typically consists of a layer of molecules aligned between two transparent electrodes, and two polarizing filters, the axes of transmission of which are (in most of the cases) perpendicular to each other. With no actual liquid crystal between the polarizing filters, light passing through the first filter would be blocked by the second (crossed) polarizer. In most of the cases the liquid crystal has double refraction. The liquid crystal between two transparent electrodes. If the long molecules are aligned parallel to a set of microgrooves on the inside faces of the two electrodes, when the voltage is applied the molecules rotate into alignment with the field. If the LC molecules are aligned horizontally on the left window and vertically on the right window, and they gradually twist(plane upon plane) from one to other, when the voltage is applied across the cell, the molecules align with the electric field. Before applying an electric field, the orientation of the liquid crystal molecules is determined by the alignment at the surfaces of electrodes. If the applied voltage is large enough, the liquid crystal molecules in the center of the layer are almost completely untwisted and the polarization of the incident light is not rotated as it passes through the liquid crystal layer. This light will then be mainly polarized perpendicular to the second filter, and thus be blocked (no light passing). By controlling the voltage applied across the liquid crystal layer, light can be allowed to pass through in varying amounts thus constituting different levels of gray. This electric field also controls (reduces) the double refraction properties of the liquid crystal. Using color filters we get colors.

21. Give a qualitative description of the temporal an spatial coherence of light Coherence is a very important concept in optics and it is related to the ability of light to yield interference and diffraction effects. The coherence term is used to describe a light field which exhibits a fixed phase relationship between the electric field values at different locations in the field and/or at different times. In other words it describes the correlation between the phases of the electric field values. When the light waves behave in a completely predictable way, as in a plane harmonic light wave, then the coherence is complete. Since waves are oscillatory motions in time and space, two types of coherence can be defined: temporal coherence and spatial coherence. Temporal coherence describes the correlation between the phases at different times, when observed at the same location. As we are going to see in the following, temporal coherence is related to the spectral purity of light. Ideal monochromatic light waves exhibit complete temporal coherence. On the other hand, spatial coherence is a measure of the correlation between the phases of the light waves at different locations, when observed at the same time. Spatial coherence of light is related to the dimension of the light source. Ideal point sources exhibit complete spatial coherence.

22. Can you observe interference patterns when sunlight is incient on a window pane? Justify your answer

I consider the window pane as a thin film with a refractive index , on which a white light beam is incident. The thickness of the film is assumed to be constant and small. The incident light is partially reflected and partially refracted at the air/glass interface. The refracted beam reaches the lower surface of the film where it subsequently undergoes reflection and refraction. Light reaching the observer is a superposition of the coherent light beams reflected from the upper (air/glass) interface and lower (glass/air) interface - see the figure below.

The superposition results in an interference pattern. According to the difference in optical path length between the light beams reflected from the top and bottom interfaces, some wavelengths reinforce each other and thus interfere constructively. Other wavelengths undergo destructive interference and cannot be seen. For interference to occur, the difference in optical path length should be within the coherence length of incident light. This explains why the optical film needs to be thin.
Notice that the difference in optical path length between ray 1 and ray 2 is almost equal to , since the plate is so thin. However, when we compute the difference in phase, we need to recall adding π radians. This is the shift in phase which the beams reflected by the upper (air/glass) interface undergo upon reflection. Note that the rays which are reflected by the lower (glass/air) are not shifted in phase.
The observer’s eye collects light from different points of the film. As a result, he sees a wide range of colors. Consider that the reflected beams originating in region A interfere destructively for red wave lengths, for instance. Then A looks blue-greenish. On the other hand, the difference in optical path length of the reflected rays originating in region B is shorter so the blue–green wavelengths interfere destructively and B looks red.

23. Describe the pump as a main component of a laser system.
The pump is an external source which provides the necessary energy to obtain the population inversion. Its nature depends on the type amplifying medium a certain laser uses; it can be optical, electrical or chemical (there is a type of laser whose pumping energy is the result of a nuclear fusion reaction, but that is quite exotic). Optical pumping is done by using flashtubes, arc lamps or even laser diodes. It is a type of pumping suitable for solid state and liquid state lasers. By absorbing the incident photons, the atoms or molecules in the gain medium make the transition to high energy levels, called pumping levels, then get to populate the upper metastable laser level by fast nonradiative transitions. Electrical pumping is adequate for gas, semiconductor and some excimer lasers. It is done by using continuous electric discharges (gas lasers), electric currents (semiconductor lasers) or electron beams (excimer lasers). Again the laser atoms or ions use the external electric energy to populate the upper laser level. In chemical lasers the population inversion is obtained by using the energy resulted from a chemical reaction. The process is known as chemical pumping.

24. Describe the amplifying mediun as a main component of a laser system.
The amplifying media are extremely diverse (it is significant that more than half of the known chemical elements were proved to have laser levels). The wavelength of the laser output is essentially dependent on the gain medium it uses. Because of its importance, most of the commonly used lasers are named after the amplifying medium. Solid, liquid and gaseous gain media are available. As a general rule, the amplifying medium is complex. It is composed of a laser host medium and laser atoms / ions. The role of the host medium is to take over the “excess” energy of the pumped laser atoms. This makes the required population inversion possible. The laser atom/ion needs to have specific metastable laser levels which represent upper laser levels (if they were short lived, the gain medium could not sustain the population inversion long enough for amplification).

25. Describe the optical resonator as a main component of a laser system.
The optical cavity is essentially an optical resonator made of two mirrors, one perfectly reflecting (as close as possible to 100% reflectivity) and another weakly transmitting (99% reflectivity). The space between the two mirrors is fully or partially filled with the amplifying medium. The role of the resonator is to direct the useful photons in a motion back and forth along the axis of the laser system, allowing for the build up of the laser output (which consist in photons that escape through the partially transmitting mirror). The geometry and alignment of the optical cavity have a crucial effect on the exact distribution of the intensity pattern across the output laser beam.

26. Compare a three-level laser system with a four-level one. The population inversion in a three level laser requires very intense pumping that is high input external energy because the lower laser level is the ground state which naturally tends to be highly populated. A different type of laser where the lower laser level is actually an excited state also exists. Such a laser works on a four-level scheme. An example is the helium- neon laser.

27. Discuss the working of ruby laser.
In a ruby laser the population inversion is achieved by optical pumping. A helical flash lamp (with Xenon, for example) surrounds the amplifying medium consisting in a ruby rod. The short () bursts of intense light supply the necessary energy of excitation. Ruby is a red, transparent crystal of aluminum tri-oxide () which contains about 5% of chromium ions (). The simplified energy level patterns of (laser ions) are presented in the figure below. As you can see, the ions have two absorption bands in the green and blue regions of the visible range (this explains why the ruby is red) which can be used as pump levels. Below those there exist two metastable states, collectively labeled in the figure. Because the pumping levels are short-lived, the laser ions which are pumped there will soon decay to , by giving off some of their energy to the host medium () in fast, radiationless relaxation processes. The population of level thus builds up and a population inversion between and (the ground state) occurs. The above mentioned levels are in fact the upper and the lower levels of the laser transition. Since it basically involves three levels (the ground state, the pump level and the metastable laser level) the ruby laser belongs to the category of three-level lasers.
Some of the atoms lying in the energy state will spontaneously emit photons of frequency and thus decay to the lower level. As we have already mentioned, the direction of the photons which are obtained by spontaneous emission is arbitrary, so we expect that a number of them will be lost through the side walls of the laser system. However, there will always be some photons which are emitted in the right direction, along the axis of crystal. We call them “useful photons” because they initiate stimulated emission processes when passing by ions lying in the upper laser level. The optical cavity is created by silvering the flat ends of the ruby bar. As long as the pump maintains the population inversion, the photons traveling back and forth along the axis of the optical cavity will be multiplied by further stimulated emission processes. A small fraction of them escape the cavity through the weakly transmitting mirror. They build up the output laser beam which consists of short pulses (a few milliseconds) with energies up to 100 J in modern ruby lasers.

28. Discuss the working of a He-Ne laser.

The helium- neon laser is a laser hich works on a four-level scheme.. In the figure below a simplified diagram of the energy levels involved in its working is presented.

Helium is the host medium while the neon atoms constitute the laser medium. The pump energy is produced by an electrical discharge in the low-pressure gas mixture. The helium atoms lying in the ground state are excited to the high pump energy level denoted by . Because neon happens to have an energy level close to it, excited atoms of helium give off their energy in resonant collisions to unexcited neon atoms and then return to their ground state . The process of pumping is continued as long as the discharge exists.
As for the atoms of neon which are excited to (which is a metastable upper laser level), they build up the population inversion between and . The latter acts as a lower laser energy level and it is also an excited state. Consequently, it is fast depleted in spontaneous decay processes. The laser transition occurs between and . It gives rise to photons with , which represents the red radiation of what is perhaps the most common type of laser. The He-Ne laser is classified as a four level laser because it essentially involves the following four levels: the ground energy level (close together for both types of atoms), the pump level of the helium atoms, the upper and lower laser levels belonging to the neon atoms.
The properties which make lasers a unique source of light consist in monochromaticity, high coherence, directionality and high levels of intensity. The remarkable number of laser applications is due to these unmatched properties of laser light. The interaction of laser light with matter is used in materials processing (cutting, drilling, heat treating, welding etc), medicine (cutting, vaporizing, cauterizing, coagulating, destroying unwanted cells, healing, treating ocular disorders), applied science (laser spectroscopy, imagery, laser-driven nonpolluting sources of energy, etc). Another wide range of applications use lasers to transmit, detect, store and process information. Telecommunications, information processing, optical sensing, optical storing, metrology, holography are only some of these fields.