Optics

 

 

 

1. The Path of Light

Light travels in a straight path, until it encounters a different medium. In fact, it travels straight at a constant speed, c, called (appropriately enough) the speed of light, where c = 300,000 km/sec

Light through a hole

All visible objects either emit light or reflect light. Some do both, most just reflect. The way that we "see" things is by the light that is reflected off of them into our eyes. When an object either emits light or reflects it, the light travels in all directions from the object. You can confirm this by just walking around your computer and looking at it: you can see it from every angle. Only a tiny fraction of the light from the object gets into our eyes at one time, however. Even then, we are usually focusing on one part of the object and not the whole thing at once (unless it's far away).

Imagine a tree off in the distance. We may focus our eyes on the top branches of the tree but light from the trunk and the branches facing us is still traveling into our eyes. The same is happening with the pinhole and the light source. But why is the image upside down on the screen? Here is where we have to think about the path of light.

We aim our pinhole viewer at the tree and notice that the tree's image on the tape screen is upside down. Imagine a ray of light from the top branch of the tree traveling toward the pinhole. Since the pinhole is so small, only this ray and a few others from the top branch can pass through the hole. The light rays haven't encountered any different media, so they travel in a straight path from the tree branch to the screen. Notice in the image above that the rays from the top branch and from the trunk have to cross at the pinhole in order to pass through. This is what causes the image of the tree (or light bulb, or any other object) to be upside down. Our pinhole is acting as a converging lens!

Light and Electrons

Electrons in matter can be described as occupying energy levels. Electrons can be raised to higher energies by absorbing energy from a photon (destroying light) or move to lower energies by giving off a photon (creating light) whose energy is equal to the difference in energy between the two levels. Figure 1 shows two electron levels separated in energy by 2 electron volts (E = E₂ - E₁ = 2 eV).

Figure 1. Two electron energy levels in a solid: a) electron occupying the lower level and the upper level empty, b) photon is absorbed by the electron which moves to the upper level, c) photon is created by electron moving from upper to lower level with energy = E₂-E₁

An incident photon (orange light of 620 nanometers or 2 eV) gives up its energy (and vanishes) to the electron which moves to the upper level (Fig 1b). In Fig 1c, an electron in the upper level moves to the lower level by giving up an energy of 2 eV and emitting orange light. Energy is conserved.

Electrons in atoms and in matter - from gases to solids - are held in place and require a boost in energy to remove them. In atoms the negatively charged electrons are held (bound) by the positively charged protons. From the quantum theory, we know that the electrons are bound to the atom with fixed energies, the binding energies. In the semiconductor element silicon the binding energies for the innermost electrons are nearly 2000 electron volts (eV) - more exactly, 1.839 kiloelectron volts (keV). This means that an energy of at least 1.839 keV must be given to an innermost electron to remove it from the silicon atom. These 1 keV energies lie in the x-ray portion of the spectrum.

The outermost electrons in silicon have a binding energy of 3 electron volts. This is at the upper end of the visible light energy spectrum. Violet light has an energy close to 3.1 eV with a wavelength of 400 nanometers (Energy in electron volts equals 1240 divided by the wavelength in nanometers). In solid silicon the outermost electrons are shared between atoms and these electrons are bound to the solid by energies of 1.1 electron volts. This means that visible light is absorbed by silicon and infrared light of energy less than 1.1 eV and wavelengths greater than 1100 nanometers can pass through silicon without absorption.

1.1. theory of rays.

 

Fig.1.1.

 

Light shows to be concentrated in cylinders, so-called light bundles. In a isotropic medium, light shows to be propagating along curvatures, which can be seen as light bundles with cross section approaching zero, and otherwise as lines along which energy propagates (light rays). A light bundle then is a collection of infinite rays, bordereing each other.

For the wave function to satisfy Helmholtz’equation we write:

 

                                             (1.1)

 

in which A(r) is a slowly changing function and S(r)  is a yet unknown function called eikonaal. The factor k₀= ω(ε₀μ₀)^1/2 is the wave factor in vacuum and depends via k=nk₀ on the wave number in the medium. If F and G are 2 functions of r, then  = . With this the equation of Helmholtz with (1.1) :

 

                   (1.2)

which with k=nk₀ can be ordered to poers of k₀, so that  , because the exponential function does not disappear identically, this leads to

                                  (1.3)

Because  k₀ is very large (for visible light k₀ =10⁷ m^-1) each term in (1.3) is of different order, if A and S don’t vary too quick with r, then (1.3) decomposes into

 

                                (Laplace)                                                          (1.4)

 

             (Transport)                                                       (1.5)

 

                          (Eikonal)                                                          (1.6)                

 

Eq.(1.5) describes the propagating of energy along the rays and is called the transport equation. Eq. (1.6) describes the path of rays through space and is called eikonal.

For light rays in a isotropic medium , each point of a lighting object is perpendicular on the wave fronts of emited waves of that point, and form a ray-congruence. (law of Malus).

The optical wavelength between 2 points P₁ and P₂ is to be defined as:

                                                                                (1.7)

and is extreme along the path followed. (principle of Fermat).

Light follws a path along which the travel time is extreme (minimal). Example: along a road on which the sun shines, the air above the ground is increasingly cooler, so that the breaking index n, increases. For an observerW, the dashed path has a smaller optical path, than the straight line, and it seems that B comes from under the road for the observer.

 

Fig.1.2.

 

 

2. The Reflection of Light

 

What is it about objects that let us see them? Why do we see the road, or a pen, or a best friend? If an object does not emit its own light (which accounts for most objects in the world), it must reflect light in order to be seen. The walls in the room that you are in do not emit their own light; they reflect the light from the ceiling "lights" overhead. Polished metal surfaces reflect light much like the silver layer on the back side of glass mirrors. A beam of light incident on the metal surface is reflected.

Reflection involves two rays - an incoming or incident ray and an outgoing or reflected ray. In Figure 1 we use a single line to illustrate a light ray reflected from the surface. The law of reflection requires that two rays are at identical angles but on opposite sides of the normal which is an imaginary line (dashed in Fig. 1) at right angles to the mirror located at the point where the rays meet. We show in Fig. 1 that the angles of incidence i and reflection i' are equal by joining the two angles with an equal sign.

Fig.2. 1 Light reflected from a metal surface with angle of incidence i equal to the angle of reflection i'. The dashed line (normal) is perpendicular to the surface.

 

 

All reflected light obeys the relationship, called Snell's Law, that the angle of incidence equals the angle of reflection. Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. We call the reflection from a smooth, mirror-like surface specular (as shown in Figure 2a). When the surface of water is wind-blown and irregular, the rays of light are reflected in many directions. The law of reflection is still obeyed, but the incident rays (Fig. 2b) strike different regions which are inclined at different angles to each other. Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

Fig. 2.2.  Light reflection from a) smooth surface (specular reflection ) and b) rough surface (diffuse reflection). In both cases the angle of incidence equals the angle of reflection at the point that the light ray strikes the surface.

Light is also reflected when it is incident on a surface or interface between two different materials such as the surface between air and water, or glass and water. Each time a ray of light strikes a boundary between two materials - air/glass or glass/water - some of the light is reflected. The laws of reflection are obeyed at all interfaces. The amount of reflected light at the interface depends on the differences in refraction between the two adjoining materials.

Reflection and Scattering

We first follow a beam of light which strikes a partially transparent material (Fig.2.3).

Figure 2.3. The beam of light passing through a block of partly transparent substance.

Every time light encounters a boundary between two transparent materials with different indices of refraction some of the light is reflected (about 4% reflection at the air/glass boundary) and some is transmitted with a change in direction due to refraction (see Readings on Reflection and Refraction). If the first surface is smooth then a mirror-like or specular reflection will occur. If the surface is rough, diffuse reflection will occur. As the light penetrates into materila containing small particles, the light can be scattered by the particles or be absorbed. In the sky colors we are interested in reflection and scattering.

Blue Sky White Clouds

The light in the sky is simply scattered sunlight. The scattering by small particles (in a clear sky, it is the air molecules, N₂, O₂, and others) is greater at the violet end of the spectrum. The color of the sky is composed of violet -- our eyes are not very sensitive to violet -- lots of blue, and decreasing amounts of green, yellow, and red. The sum appears to our eyes as "sky blue". If you observe the sky with an optical spectrometer, you note the blue sky has a broad band of colors, extending from the blue and fading toward the red.  The scattering is due to particles whose size is smaller than the wavelength of the light.

On a normally sunny day with a blue sky, 80% of the light comes directly from the sun and only 20% from the blue sky. It is the sunlight that reflects from the clouds. The water droplets in the clouds may be many times larger than the wavelength of visible light. There are many droplets and many surfaces to reflect the light. The clouds scatter almost all the light and look white even though the clouds are nearly transparent.

Beer Foam and Salt Piles

If there are many air-particle interfaces, little penetration of the light occurs and almost no color is seen. The foam on a liquid such as beer is white even though the liquid itself is colored; here, the walls of the foam bubbles are so thin that little of the light is absorbed by the liquid while reflection and scattering occur at each of the many liquid-air interfaces. A similar argument can be made for piles of small transparent solids, such as salt, sugar, and glass. Though the walls of the solids are not thin, there are so many particles in close proximity that the light cannot penetrate the pile without being scattered and reflected many times. The sum of all the scattered light results in a perception of a white pile, though each particle is itself transparent.

 

The Refraction of Light

 

 

Ever notice how your leg looks bent as you dangle it in the water from the edge of a pool? Why do fish seem to radically change position as we look at them from different viewpoints in an aquarium? What makes diamonds sparkle so much?

These are all questions that can be addressed with the important concept of refraction, the bending of light as it encounters a medium different than the medium through which it has been traveling. This meeting place of two different media is called the interface between the media. All refraction of light (and reflection) occurs at the interface.

What happens at the interface to make light refract or reflect? When light is incident at a transparent surface, the transmitted component of the light (that which goes through the interface) changes direction at the interface. Another component of the light is reflected at the surface. As shown in Figure 2.4, the refracted beam changes direction at the interface and deviates from a straight continuation of the incident light ray.

Figure 2.4. Light in air incident on glass surface where it is partly reflected at the interface and partly transmitted into the glass. The direction of the transmitted ray is changed at the air/glass surface. The angle of refraction r is less than the angle of incidence i.

The change of direction of light as it passes from one medium to another is associated with a change in velocity and wavelength. The energy of the light is unchanged as it passes from one media to another. When visible light in air enters a medium such as glass, the velocity of light decreases to 75% of its velocity in air and in other materials the decrease can be even more substantial. For example, in linseed oil, the velocity decreases to 66% of its velocity in air. Figure 2.5 displays in bar chart format the velocity of light in different media. The 100% value is the velocity of light in vacuum. For air, the velocity is 99.7% of the speed in vacuum. For some pigments such as titanium (Ti) white, the velocity decreases to 40%.

Figure 2.5. Bar chart of the velocity of visible light in different media. The value of 100% refers to the velocity of light in vacuum.

waves

Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. Light is refracted when it crosses the interface from air into glass in which it moves more slowly. Since the light speed changes at the interface, the wavelength of the light must change, too. The wavelength decreases as the light enters the medium and the light wave changes direction. We illustrate this concept in Figure 2.6 by representing incident light as parallel waves with a uniform wavelength . As the light enters the glass the wavelength changes to a smaller value '. Wave "a" passes the air/glass interface and slows down before b, c, or d arrive at the interface. The break in the wave-front intersecting the interface occurs when waves "a" and "b" have entered the glass, slowed down and changed direction. At the next wave-front in the glass, all four waves are now traveling with the same velocity and wavelength .

Figure 2.6.  Light waves of wavelength incident on glass change direction and wavelength when transmitted into the glass.

The waves are continuous and remain connected as they pass from one medium to another. We can think of it like a long line of people running into the ocean. As the first few people run into the water, they're slowed down because it's harder to run in water. Thus, they bunch up and stay bunched up as they run through the water. When everyone in the line has entered the water, we would see a line of people all running in the same direction, but the line would be shorter and the people would be bunched close together. If they run back to the beach, the first few people would clear the water and run faster. Eventually everyone will have cleared the water and would be running at the original pace with the original spacing between persons.

In this analogy we can think of the whole line of people as the "light wave" and the people themselves as the "crests" of the wave. The distance from one person to her neighbor would be the wavelength of the wave and the water would be the medium into which the light wave is traveling. Why, then does the light wave change direction when it enters the new medium?

We can extend our analogy and imagine two lines of people running into the ocean from the beach. The lines are close together and each person in a line is matched up with another person in the other line. This is analogous to waves a, b, c, and d in Fig. 3 above. When line "a" hits the water first, the line slows down. In order to maintain the one-to-one relationship with the other line, both lines must turn when they hit the water. Which way do they turn? Towards the normal - the imaginary line that runs perpendicular to the interface between the two media (the water and the beach); a pier is a good example of something normal to the water/beach interface.

So the two lines must turn towards the normal when they hit the water. The greater the change in velocity and wavelength, the greater the change in direction. Figure 2.7. shows the change in direction for light in air incident at 45° on water with refracted angle of 32° and on titanium white (a paint pigment) with a refracted angle of 16°. These angles correspond to the differences in velocity shown in Fig.   .

Figure 2.7. Light incident at 45° on water and Ti white. The angles of refraction (32° for water, 16° for Ti white) depends on the optical properties. The reflected components are not shown.

We can characterize the change in velocity by a number called the refractive index of the material.

 

The Refractive Index

The ratio of the velocity of light in vacuum to the velocity of light in a medium is referred to as the medium's refractive index, denoted by the letter n. The velocity of light in a vacuum is 3.0 x 10⁸ m/s or about 186,000 miles/s. If we go back to our beach/ocean analogy we can think of the refractive index of the water as something like its density. Air is not very dense at all (its refractive index is 1.003), so the people run through quite easily; but when they run into water, which is denser than air (and has a refractive index of 1.333), the people slow down and their line bends at the ocean/beach interface. For light, the index of refraction n equals the ratio of the velocities of light in vacuum (c) to that in the medium (v), that is n = c/v.

Refractive indices are most easily determined from the measured values of the incident angle and the angle of refraction and their geometric relationship. Values of the refractive indices for the media shown in Fig. 2 are given below.

Values of Refractive Index

The path of light in air incident on and transmitted through a glass plate is shown in Figure 2.8. The angle of the incident ray to the normal is 45° and equals that of the reflected ray. The transmitted ray is refracted at an angle of 28° to the normal and exits the glass at an angle of 45° to the normal, an angle equal to that of the incident ray. This explains why, for example, the image we see through a flat-glass window pane is unchanged from that seen through an open window.

Figure 2.8. Light incident on a glass plate. The reflected part of the ray is shown along with the light path for the refracted component.

Light incident normal to a glass plate does not change direction as the transmitted light continues normal to the surface (air/glass interface). The light is not refracted (that is, no change in angle) but the wavelength and velocity do change. Light does reflect as it encounters the air/glass interface (about 4% in this case).

The paths of light traversing different media are reversible. The same relations are obeyed in Fig. 2.8, for example, if the light were incident on the bottom of the glass plate. Similarly, in Fig. 2.7, if the light started in the water, it would be refracted at the water/air interface and would retrace the same reversible path as for light incident from air.

Total Internal Reflection

If light is inside a material such as glass with a larger refractive index n₂ than that n₁ of the material outside such as air, there is an angle, the critical angle of incidence, beyond which the light is reflected back into the material and does not escape. This is total internal reflection. The critical angle i_c is given by

sin i_c = n₁ / n₂

For light exiting glass, n₂ = 1.5, the relation becomes

sin i_c = 1.0 / 1.5

and the value of the critical angle of incidence is 41°. Light in glass at any angle of incidence to the normal 41° or greater will be reflected from the glass-air interface back into the glass. The bluer area in Fig. 6 is the angular region where light is reflected back into the glass. The greater difference in the two indices of refraction, the smaller the amount of light can escape.

Figure 2.9. The internal reflectance at an air/glass interface for light rays from a point source in glass. Light rays incident at angles to normal at greater than the critical angle (here, 41° for glass to air) do not leave the material and are reflected at the glass/air interface.

Light fibers and diamonds are both materials with a high refractive index and are used for their properties of "retaining" light.

 

3. Wave propagation in homogenous media.

 

 

3.1.Spherical and plane waves.

 

Spherical scalar waves in a loss free medium.

We suppose the whole of space to be filled with a isotrope homogeneous medium.Somewhere in space there Is a pointform well which transmits light in all directions in the same way. It is to be expected that the fields and so the scalar wave function will be spherical symmetrical around the source. If we introduce spherical coordinates R,q,f around origin,then on ground of symmetry : U=U(R,t)

 

Fig.3.1.

 

If we introduce the complex function U(R,w) with

 

U(R,t) = Re [U(R,w) exp(—iwt)]                                                                                 (3. 1)

 

leads to the equation

           

Ѳ U+ k²U=0                                                                                                               (3.2)

 

where the wave vector is determined by

           

k² = w²em = w²/c         (Re(k)³0),                                                                                                         (3.3)

 

with c=(em)^-1/2 .We must write V² out in spherical coordinates; This becomes easily because the differentiations to q and f can be left out. With this (3.2) leads to

 

R^-2 (d/dR) (R² dU/dR] + k²U=0                                                                                    (3.4)

 

or for R=0 (outside the well)

 

d²U/dR² + (2/R) dU/dR + k²U= 0                                                                                 (3.5)

 

Because

 

d²U/dR² + (2/R)dU/dR ={(d/dR)(RdU/dR +U)}/R=

 

                ={(d/dR)[d(RU)/dR ]}/R ={d²(RU)/dR² }/R                                                     (3.6)

 

,eq (3.5) can also be written as

 

{d² (RU)/dR² }/R + k² U = 0                                                                                         (3.7)

 

or

           

d² (RU)/dR +k² (RU) = 0                                                                                             (3.8)

 

This equation has the general solution

 

RU= A exp(ikR) + B exp (-ikR)                                                                                    (3.9)

so that

U(R,w)= A exp(ikR)/R + B exp(-ikR)/R                                                                        (3.10)

 

Substituting in (3.1) gives

 

U(R,t)=R^-1 [Re(A)cos(wt—kR)+Im(A)sin(wt—kR) + Re(B)cos(wt+kR)+Im(B)sin(wt +kR)] 

(3.11)

The first two terms between the brackets represent waves in positive R-direction which propagate with velocity w/k=c=(em)^-1/2. These are phenomena which move away from the source (divergent waves). The last two terms represent waves which move towards the source (convergent waves) with the same velocity c. If there no other sources (or mirrors) the second term is no physical solution (emitting condition of Sommerfeld).The wave fronts in our problem move away from the source with velocity of light.

 

3.2. Spherical and electromagnetic waves.

 

We consider a point source at the origin within a infinite large, isotrope and homogeneous medium. For E we find a expression of the form

 

E(R,w)=A exp(ikR/R) + B exp (-ikR/R)                                                                          (3.12)

 

where A and B are arbitrary and only depend on q and f via the laws of Maxwell and the divergence relations. If we now suppose that the medium has losses (s = 0) then

 

ѲE +k²E=0                                                                                                                (3.13) 

 

where k² =w²em + iwms. The vector H(R,w) is ,with (1.6)—(1.10):

 

H=(-i/wm)Ñ x E                                                                                                           (3.14)

 

From (3.12) and with k=(w²em + iwms) = k₁  + ik₂ (k₁ and k₂ real)

 

E(R,t)=R^-1 [exp(—k₂R){Re(A) cos(wt—k₁R)+Im(A) sin (wt—k₁R)}+

 

           + exp(k₂R){Re(B) cos(wt +k₁R)+Im(B) sin (wt+k₂R)}]                                    (3.15)

 

The first form between brackets gives the character of sphere-like waves from the source and the exponential function before damps the wave in the direction of propagation.The second form represents sphere like waves towards the source and will be zero if only one source present: B=0. So

 

E(R,w)= A exp (ikR)/R = (A_Ri_R + A_qi_q +A_fi_f) exp (ikR)/R                                            (3.16)

 

is the complex representation of sphere waves. When the charge density is zero from  : Ñ.E=0, which with (3.16) leads to

 

Ñ.E(R,w)=R^-2 ¶(R²E_R)/¶R +{1/R sin(q)}¶{E_q sin(q)}/¶q  + {1/R sin(q)}¶E_f¤¶f =

 

        = [exp(ikR)/R²][A_R(1 + ikR)+A_qcot(q)+¶A_q/¶q + (1/sin(q)¶A_f/¶f]=0                    (3.17)

 

This must apply for all R,q and f outside the source; the second factor is zero and so A_R=0. This also follows from Ñ.H=0. For H(R,w) from (3.14)

 

H(R,w)=(-i/wm)[ i_q{—R^-1 ¶(RE_f)/¶R} + i_q {R^-1 ¶(RE_q)/¶R}]=

                                                                        

            =  (k/wm)(— A_fi_f +A_qi_q) exp(ikR)/R                                                               (3.18)

We can see that

 

E(R, w). H(R, w)=0                                                                                                   (3.19)

 

so the complex fields are orthogonal and perpendicular to i_r ;that is why these waves are called transversal.

Fig. 3.1. Plane waves                  

 

3.3. Plane waves.

 

The complex wave functions contain the factor

 

F(R,w) = A exp (ikR)/R                                                                       (3.20)

 

where R is the distance of the source at the origin 0 to a point of observation P.We now introduce a second cartesian reference system at some other place and suppose that 0 in that system has the coordinates and P has the coordinates x,y,z. So

 

R={(x₀ —x)² +(y₀—y)² +(z₀-z )² }^1/2                                                (3.21)

 

When in (3.20)

 

x₀² +y₀² +z₀² >> x² +y² +z²                                                                   (3.22)

then with

 

R₀= (x₀² +y₀² +z₀² )^1/2                                                                           (3.23)

 

by approximating the root by the first two terms of the binomial series becomes:

 

R=R₀ [i -2(xx₀ +yy₀ + zz₀ )/R₀² ]= R -(xx₀ +yy₀ + zz₀)/R₀.                   (3.24)

 

Fig.3.2.

 

The factors x₀/R_0, y₀/R₀ and z₀/R₀ give the cosine of direction of the line through the new origin where 0 disappears to infinite. They are to be seen as the components of the unit vector g along that line. If we (from origin) introduce the radius vector r =x_i +y_i +z_i to P, then from (3.24)

 

R =R₀ -g.r                                                                                                       (3.25)

 

We now substitute (3.24) in (3.19),.For large values of R₀, with -g.r =D.

 

F=A{exp(ikR₀)/R₀}{exp(ikD)/(1 +D/R₀)}=

 

A{exp(ikR₀)/R₀}{l+ ikD- ½ k²D².…..}{1 -D/R₀+ D²/R₀² -….. }=

 

A{(exp(ikR₀)/R₀}{1+ D(ik- l/R₀) + D²( ½ k² - ik/R₀ +1/R₀² +…..}  (3.26)

 

For light waves k=10⁷ while l/R₀ becomes small for large R .So that