Introduction in Optics I:

I-1) Electro-magnetic Waves

The discussion of electric and magnetic fields can be classified in two general categories. The first includes fields that do not vary with time. The electrostatic field of a distribution of charges at rest and the magnetic field of a steady current in a conductor are examples of fields, which, while they may vary from point to point in space, do not vary with time at any individual point. For such situations it is possible to treat the electric field and magnetic fields independently, without worrying about interactions between the fields.

The second category includes situation in which the fields do vary with time, and in all such cases it is not possible to treat the fields independently. Faraday’s law tells us that a time-varying magnetic field acts as source of electric field. This field is manifested in the induced electromotive forces (emf’s) in inductances and transformers. Similarly, in developing the general formulation of Ampere’s law, which is valid for charging capacitors and similar situations as well as for ordinary conductors, we found it necessary to regard a changing electric field as a source of magnetic field. Therefore, when either field is changing with time, a field of the other kind is induced in adjacent regions of space. We are led to consider the possibility of an electromagnetic disturbance, consisting of time-varying electric and magnetic fields, which can propagate through space from one region to another, even when there is no matter in the intervening region. Such a disturbance, if it exists will have the properties of a wave, and the appropriate descriptive term is electromagnetic wave. Such waves do exist; radio and television transmission, x-ray and in the current context most important: light. Figure I-1-1 visualizes such electromagnetic wave propagation.

x
 

z
 

y
 

 

Figure I-1-1: Electromagnetic monochromatic wave. E and B correspond to the electric and magnetic field. Note the transverse character of the wave.

 

The magnitudes of the field vectors E and B are in phase and are related by E=cB, with

 

,                                                (I-1-1)

 

where, c (=2.9979246´108 m s-1) is the speed of light in the vacuum, where m0  (=1.2566´10-6 Vs V-1 m-1) and e0 (=8.8542´10-12 As V-1 m-1) are the permeability and permittivity (i.e., the dielectric constant) of the vacuum. The space (x) and time (t) dependence of the electric and magnetic field is described by

Ey=Ey0sin(kx-wt)                                      (I-1-2)

 

and

 

Bx=Bx0sin(kx-wt),                                      (I-1-3)

 

where Ey0 and Bz0 is the amplitude of the E and B field, respectively; k =2p/l is the wave number and w =2pn is the angular frequency which depend on the wavelength l and frequency n, respectively. As shown in figure I-1-1, in vacuum (and air) the E and B fields at any point are in phase. In a dissipative medium, however, a phase shift between the fields takes place. In good conductors, the magnetic field is much larger than the electric field and exhibits a phase delay of approximately 45o. In non-dissipative media, as e.g. glass for visible light, the E and B fields behave similar as in vacuum and are in phase.     

The energy per photon of a monochromatic (=one color) wave is given by

,                                             (I-1-4)

 

where h is Planck’s constant (=6.626´10-34 J s), n is the frequency and l is the wavelength. The electromagnetic spectrum is shown in figure I-1-2. Specifically in semiconductor optics the energy is expressed in eV rather than in J. Hence, it is convenient to apply the following relation to convert nm (10-9 m) into eV,

 

.                                              (I-1-5)

Example: One of the possible emissions of an Argon laser is at 514.5 nm. What is the energy of the emission in nm?

Solution: E=1240/514.5=2.41 eV. 

 

          The electromagnetic waves cover an extremely broad spectrum of wavelengths, as shown in figure I-1-2. We can detect only a very small segment of this spectrum directly through our sense of sight from approximately 750 to 430 nm.   

         

 

Figure I-1-2: The electromagnetic spectrum. 

 

I-2 Refraction and Reflection

We shall begin our introduction in optical phenomena with reflection and refraction at a boundary surface that has been formed by the meeting of two different media. The velocity of light in a medium is below the velocity of light in the vacuum and is given by v=c/n, where n is the refractive index of the medium.  We will see that the refractive index does not only determine the light velocity in a medium but it is also an essential parameter for the reflection. Let’s consider that we investigate the directions of the incident, reflected and refracted rays of monochromatic light. We fill find the following results illustrated by Fig. I-2-1:

1.     The incident, reflected and refracted beams and the normal to the surface, all lie in the same plane.

2.     The angle of reflection fr is equal to the angle of incidence fa (fr=fa).

3.     For a given pair of substances, a and b, on opposite sides of the surface of separation, the ratio of the sine of the angle fa (between the beam in substance a and the normal) and the sine of angle fb (between the beam in substance b and the normal) is a constant (sin fa/sinfb=constant).

 

Figure I-2-1: Incident, reflected and refracted light rays at the interface of media a and b.

 
 

  

If the a beam of monochromatic light travels in vacuum (or air), making an angle of incident f0 with the normal to the surface of a substance a, we write

 

,                                          (I-2-1)

 

where na is the refractive index of substance a. The refractive index is always greater than unity and depends not only on the substance but on the wavelength of the light.

 

I-3 Snell’s law of refraction

Applying equation (I-2-1) to the to substances a and b in figure I-3-1, we have sinf0/sinfa=na and sinf0/sinfb=nb. Dividing the second equation by the first, we obtain sinfa/sinfb=nb/na and from this the best known form of Snell’s law of refraction,

 

nasinfa=nbsinfb.                                         (I-3-1)

 

The angles in figure I-3-1 are independent of the thickness and space between the two plates and are the same when the space shrinks to nothing, as in figure I-3-2.

 

Parallel
 

 

 

Figure I-3-1: The transmission of light through parallel plates of different substances. The incident and emerging rays are parallel.

 

 

 

Figure I-3-2: The figure shows the light rays at the interface of substances a and b without space between the plates. The angles are the same as these in figure I-3-1.

Example: In figure I-3-3 material a is water and b is glass with index of refraction of 1.52. If the incident ray makes an angle of 60o with the normal, find the directions of the reflected and refracted rays.

 

Figure I-3-3: Reflection and refraction of light passing from water to glass.

 

 

Solution: Using equation (I-3-1), we find (1.33)(sin600)=(1.52)(sinqb) and qb=arcsin[(1.33)(sin60o)/1.52]=49.3o.

 

I-4 Total Internal Reflection

Figure I-4-1 shows a number of rays diverging from a point source P in a medium a of index na and striking the surface of a second medium b of index nb, where na>nb.

 

Figure I-4-1: Total internal reflection. The angle of incidence fa, for which the angle of refraction is 90o, is called the critical angle.

The angle of incidence for which the refracted ray emerges tangent to the surface is called critical angle fcrit. At this angle fb=90o and Snell’s law becomes nasinfa=nb, since sin90o=1. We then have with fa=fcrit

 

.                                              (I-4-1)

 

For a glass/air interface with n=1.52 for the glass, sinfcrit=1/1.52 and it follows fcrit=41.1o.  The fact that fcrit is less than 45o makes it possible to use a triangular prism with angles 45o, 45o, and 90o as a totally reflecting surface Such a prism is called Porro prism and is shown in figure I-4-2 (a). An application of total internal reflection is shown in figure I-4-2 (b).

 

Figure I-4-2: (a) A 45o-45o-90o Porro prism. (b) A combination of two Porro prisms is often used in binoculars.

 
 

 

 

Example: A persiscope uses two totally reflecting 45o-45o-90o prisms. It springs a leak, and the bottom prisms is covered with water. Explain why the periscope no longer works.

 

Solution: The critical angle for water (nb=1.33) on glass (na=1.52) is fcrit=arcsin(1.33/1.52)=61.0o. The 45o angle of incidence is less than the 61o critical angle for a totally reflecting prism, so total internal reflection does not occur at the glass/water interface. Most of the light is transmitted into the water, and very little is reflected back into the prism.

         

 
A very important application of total internal reflection is the fiber-optic cable shown in figure I-4-3 (a). Figure I-4-3 (b) shows the working principle of the cable. When a beam of light enters at one end of the transparent fiber, the light is totally reflected internally and is trapped within the rod.

 

(b)
 

(a)
 
 

 

 


Figure I-4-3: (a) Fiber-optic cable, used to transmit a modulated laser beam for communication purposes. (b) The so-called light pipe. The light is trapped by internal reflection, provided that the angles shown exceed the critical angle.

 

I-5 Dispersion

Ordinarily, white light is a superposition of waves with wavelengths extending through-out the visible spectrum. The speed of light in vacuum is the same for all wavelengths, but the speed in a material substance is different for different wavelengths. Therefore the index of refraction of a material depends on the wavelength. The dependence of the index of refraction on the wavelength is called dispersion. Figure I-5-1 shows the variation of the refractive index with the wavelength for different optical materials. The value of n usually decreases with increasing wavelength and thus increases with increasing frequency. Light of longer wavelength usually has greater speed in a material than light of shorter wavelength.

          The brilliance of diamond is due in part to its large dispersion and in part to its unusually large refractive index (2.417). When you experience the beauty of a rainbow, you are seeing the combined effects of dispersion and total internal reflection.

 

Figure I-5-1: Variation of the refractive index with wavelength.
 

 

Figure I-5-2 shows the ray of white light incident on a prism. The deviation (change of direction) produced by the prism increases with increasing the refractive index and frequency (i.e., the energy, see equation I-1-4). 

Figure I-5-2: Schematic representation of dispersion by a prism.

 
 

 

 

Refractive Index
by
Alphabetical Listing of Material

 

Refractive Index
by
Increasing RI Value

Material

RI

 

Material

RI

 

 

 

 

 

Air (STP)

1.00029

 

Vacuum

1.00000

Amethyst (Quartz)

1.54 (+1.55)

 

Air (STP)

1.00029

Beryl (Emerald)

1.57 (+1.60)

 

Water

1.333

Citrine

1.55

 

Glass

1.517

Corundum (Ruby, Sapphire)

1.76 (+1.77)

 

Quartz

1.54 (+1.55)

Emerald (Beryl)

1.57 (+1.60)

 

Amethyst (Quartz)

1.54 (+1.55)

Diamond

2.417

 

Rock Crystal (Quartz)

1.54 (+1.55)

Garnet (Pyropes)

1.73-1.75

 

Citrine

1.55

Garnet (Almandine)

1.76-1.83

 

Beryl (Emerald)

1.57 (+1.60)

Garnet (Rhodolite)

1.76

 

Emerald (Beryl)

1.57 (+1.60)

Glass

1.517

 

Topaz

1.61 (+1.62)

Peridot (Olivine)

1.65 (+1.69)

 

Tourmaline

1.62 (+1.64)

Quartz

1.54 (+1.55)

 

Peridot (Olivine)

1.65 (+1.69)

Rock Crystal (Quartz)

1.54 (+1.55)

 

Garnet (Pyropes)

1.73-1.75

Ruby (Corundum)

1.76 (+1.77)

 

Garnet (Rhodolite)

1.76

Sapphire (Corundum)

1.76 (+1.77)

 

Garnet (Almandine)

1.76-1.83

Topaz

1.61 (+1.62)

 

Ruby (Corundum)

1.76 (+1.77)

Tourmaline

1.62 (+1.64)

 

Sapphire (Corundum)

1.76 (+1.77)

Vacuum

1.00000

 

Corundum (Ruby, Sapphire)

1.76 (+1.77)

Water

1.333

 

'High' Zircon

1.96 (+2.01)

High' Zircon

1.96 (+2.01)

 

Diamond

2.417

 

 

 

 

 

 

 Table I-5-1: Refractive index of various materials (from http://www.3dlapidary.com/HTML/Materials3.htm). Note: refractive index listings which have two numbers [ex. 1.54 (+1.55)] denote materials with double refraction properties.

 

 

I-6 Polarization

Polarization occurs to all transverse waves. Figure I-6-1 illustrates the idea of polarization by showing a transverse wave as it travels long a rope toward a slit. The wave is said to be linearly polarized, which means that its vibration always occur along one direction.

 

 

Figure I-6-1: The principle of polarization: A transverse wave is linearly polarized when its vibrations always occur along one direction. (a) The rope passes a slit parallel to the vibrations, but (b) does not pass trough a slit that is perpendicular to the vibrations.   

 

Linearly polarized light can be produced from unpolarized light with the aid of certain materials. One commercially available material goes under the name of Polaroid. As shown in figure I-6-2, such materials allow only the component of the electric field along one direction to pass through, while absorbing the field component perpendicular to this direction.

Figure I-6-2: Linearly polarized light produced by a polarizing filter. 

 
  

 

Light from ordinary sources is not polarized. The “antennas” that radiate light waves are the molecules that makes up the sources. The waves emitted by any one molecule may be linearly polarized. However, any actual light source contains a tremendous number of molecules with random orientations, so the light emitted is a random mixture of waves that are linearly polarized in all-possible directions.

In figure I-6-3 unpolarized light is incident on a polarizer. The blue line represents the polarizing axis. The E vectors of the incident wave exhibit random directions. The polarizer transmits only the components of E parallel to the polarizing axis. The intensity of the transmitted light is exactly

 

Figure I-6-3: Unpolarized light is incident on the polarizer. The intensity of the transmitted linearly polarized light, measured by the photocell, is the same for all orientations of the polarizer.

 

half that of the incident unpolarized light, no matter how the polarizing axis is oriented. Here’s why: We can resolve the E field of the incident wave into a component parallel to the polarizing axis and a component perpendicular to it. Because the incident light is a random mixture of all states of polarization, these two components are, on average, equal. The (ideal) polarizer transmits only the component that is parallel to the polarizing axis, so half of the incident intensity is transmitted.

          What happens when the linearly polarized light emerging from a polarizer passes through a second polarizer, as shown in figure I-6-4? To find the transmitted intensity at intermediate values of the angle f, we bear in mind that the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave. The ratio of the transmitted to incident amplitude is cosf, so the ratio of transmitted to incident intensity is cos2f. Thus the intensity of the light transmitted through the analyzer is

 

I=Imaxcos2f,                                               (I-6-1)

 

where Imax is the maximum intensity of the light at f=0. Equation (I-6-1) is called Malus’s law.

Figure I-6-4: The analyzer transmits only the component that is parallel to its polarization axis.

 

Example: In figure I-6-4 the incident unpolarized light has the intensity I0. Find the intensity transmitted by the first polarizer and the second if the angle between the axes of the two filters is 30o.

Solution: As explained above, the intensity after the first filter is I0/2. According to equation (I-6-1) with 30o, the second polarizer reduces the intensity by a factor cos230o=3/4. Thus the intensity transmitted by the second polarizer is I0/2´(3/4)=(3/8)´I0.

 

Example: What value of f should be used in figure I-6-4, so that the average intensity of the polarized light reaching the photocell is one-tenth the average intensity of the unpolarized light?

Solution: Using equation (I-6-1), we find I0/10=(I0/2)cos2f. Solving this relation for f yields f=arccos(1/5)(1/2) =63.4o. 

 

A further possibility to create either partially or totally polarized light is by reflection. In figure I-6-5, unpolarized light is incident on a reflecting surface between two transparent optical materials. The plane containing the incident and reflected rays and the normal to the surface is called the plane of incidence.

 

Figure I-6-5: When light is incident at the polarizing angle, the reflected light is linearly polarized. 

 

At one particular angle of incidence, called the polarizing angle qp, only the light for which the E vector is perpendicular to the plane of incidence is reflected. The reflected light is therefore linearly polarized perpendicular to the plane of incidence (i.e., parallel to the reflecting surface).

In 1812, Sir David Brewster noticed that when the angle of incidence is equal to the polarizing angle qp, the reflected and refracted ray are perpendicular to each other. The situation is shown in figure I-6-6. In this case qb=90o-qp. Using equation (I-3-1), we find sinqp/sin(90o-qp)=sinqp/cosqp=nb/na and finally

 

                                               (I-6-2)

 

This relation is known as Brewster’s law.

 

Figure I-6-6: When light is incident at the polarizing angle, the reflected and refracted rays are perpendicular to each other. The circles represent an E-component perpendicular to the plane of the figure.

 
 

 

Light and other electromagnetic radiation can also have circular or elliptical polarization, i.e., the E describes a circular or elliptical rotation. In this context polarization by birefringence is important. Birefringence occurs in calcite and other noncubic materials (hence also in various semiconductors) and some stressed plastics and cellophane. Most materials are isotropic, that is, the speed of light passing through the material is the same in all directions. Because of their atomic structure, birefringent materials are anisotropic. The speed of light depends on its direction of propagation through the material. When a light ray is incident on such materials it may be separated into two rays called the ordinary and extraordinary ray. There is one particular direction in a birefringent material in which both rays propagate with the same speed. This direction is called the optic axis of the material. However, when light is incident at an angle to the optic axis, as shown in figure I-6-7, the rays travel in different directions and emerge separated in space.

Figure I-6-7: A light ray incident on a birefringent material is split into two beams, called the ordinary (o ray) and extraordinary ray (e ray), that have mutually perpendicular polarizations. 

 
 

If light is incident on a birefringent plate perpendicular to its crystal face and perpendicular to the optic axis, the two rays travel in the same direction but different speeds. The rays emerge with a phase difference that depends on the thickness of the plate and on the wavelength of the incident light. In a quarter-wave plate, the thickness is such that there is a 90o phase difference between the waves of a particular wavelength when they emerge. In a half-wave plate, the rays emerge with a phase difference of 180o.

Suppose that the incident light is linearly polarized such that E is 45o to the optic axis, as illustrated in figure I-6-8. The ordinary and extraordinary rays start out in phase and have equal amplitudes. With a quarter-wave plate, the emerge with a phase difference of 90o, so the resultant components of E are Ex=E0coswt and Ey=E0cos(wt+90o)=E0sinwt, where w=2pn is the angular frequency and t represents the time. The electric field vector thus rotates in a circle and the wave is circularly polarized.

 

 

Figure I-6-8: Polarized light is incident on a birefringent crystal such that E makes 45o angle with the optic axis, which is perpendicular to the light beam. If the crystal is a quarter-wave plate the light behind the crystal is circularly polarized.

 

Figure I-6-9 shows the propagation of circular polarized light. If the advancing wave revolves clockwise (looking toward the source), then it’s said to be right-circularly polarized; if counterclockwise, it’s left-circularly polarized. The magnitude of E remains constant while revolving once around with every advance of one wavelength. 

Figure I-6-9: Right-circular light. As the wave advances, the electric field vector E rotates clockwise once around per wavelength. The magnitude of E is constant.

 
 
 

 

 


  

With a half-wave plate, the wave emerge with a phase difference of 180o, so the resultant electric field is linearly polarized with components Ex=E0sinwt and Ey=E0sin(wt+180o)=-E0sinwt. Hence, the direction of the wave polarization is rotated by 90o relative to that of the incident wave, as shown in figure I-6-10.

Figure I-6-10: If the birefringent crystal is a half-wave plate, the direction of the polarization of the emerging light is rotated by 90o.

 
 

If the phase difference between the two components of E is something other than a quarter wavelength, or if the two component wave have different amplitudes, the resulting wave is elliptically polarized.

 

I-7 Huygens’ Principle

The principles governing reflection and refraction of light rays, discussed in I-2 and I-3, were discovered experimentally long before the wave nature of the light was firmly established. These principles however can be derived from wave considerations and thus shown to be consistent with the wave nature of light. To establish this connection we use a principle called Huygens’ principle (Christiaan Huygens in 1678). According to this principle, each point on a given wavefront can be considered to be a point source of secondary wavelets.

Figure I-7-1 shows the plane wavefront AA’ striking a mirror at point A. As can be seen from the figure, the angle f1 between the wavefront and the mirror is the same as the angle of incidence q1. From figure I-7-1 it is

 

Figure I-7-1: Plane wave reflected at a plane mirror.

 

readily shown that the angle of reflection equals the angle of incidence. Figure I-7-2 shows an enlargement of a portion of figure I-7-1 showing AP, which is part of the original wavefront. The reflected BB’’ makes an
 

Figure I-7-2: Geometry of Huygen’s construction of the law of reflection.
 
 

anglef1’ with the mirror that is equal to the angle of reflection q1’ between
reflected ray and the normal to the mirror. The triangles ABP and BAB’’ are

both right triangles with a common side AB and an equal side AB’’=BP=ct. Hence, these triangles are congruent and the angles f1 and f1’ are equal, implying that q1’=q1.

Figure I-7-3 shows a plane wave incident on an air/glass interface. We apply Huygen’s construction to find the wavefront of the transmitted wave. The new wavefront BB’ is not parallel to the original wavefront AP because the speeds v1 and v2 are different. From the triangle APB, sinf1=v1t/AB or AB=v1t/sinf1=v1t/sinq1 using the fact that f1=q1. Similarly, from triangle AB’B, sinf2=v2t/AB or AB=v2t/sinf2=v2t/sinq2, where q2=f2 is the angle of refraction.  Equating the two values for AB, we obtain

 

.                                           (I-7-1)

 

Substituting v­1=c/n1 and v2=c/n2 in this equation delivers Snell’s law, n1sinq1=n2sinq2.

Figure I-7-3: Application of Huygens’ principle to the refraction of plane waves.
 

 

I-8 Thin films

You have probably noticed the colored bands in a soap bubble or in the film on the surface of oily water. The bands are due to the interference of light reflected from the top to the bottom surfaces of the film. The different colors arise because of the variation in the thickness of the film, causing interference for different wavelength at different points. Such an interference effect is shown in figure I-8-1.

Figure I-8-1: Interference in a vertical soap film of variable thickness.
 

We consider now a thin film of uniform thickness d and index of refraction n shown in figure I-8-2. To determine whether the reflected light rays interfere constructively or destructively, we must note the following fact: A wave traveling in a medium of low refractive index (air) undergoes a 180o phase change upon reflection from a medium of higher refractive index. There is no phase change in the reflected wave if it reflects from a medium of lower refractive index.

 

d
 

Figure I-8-2: Interference in light reflected from a thin film is due to a combination of rays reflected from the upper and lower surfaces.
 
   

          Ray 1 is reflected from the upper surface A undergoes a phase change of 180o with respect to the incident wave. On the other hand, ray 2, which is reflected from the lower surface B undergoes no phase change with respect to the incident wave. Therefore, ray 1 is 180o out of phase with ray 2 corresponding to path difference of ln/2. However, we must consider that ray 2 travels an extra distance equal to 2d before the waves recombine. Hence, if 2d=ln/2=l/(2n) the phase difference between both rays is 360o and the waves recombine in phase and constructive interference takes palace. In general, the condition for constructive interference is expressed as

 

2nd=(m+1/2)l    (m=0,1,2,…)                           (I-8-1)

 

and for destructive interference we have

 

2nd=ml.      (m=0,1,2,…)                            (I-8-2)

 

Thin films are of considerable importance for the formation semiconductor devices. Almost all optoelectronic devices are composed of the combination of various thin films. Concerning research and development, by means of optical spectroscopy not only the optical or optoelectronic features of semiconductors are investigated but also other features as the film thickness. Hence, in many cases optical characterization methods accompany the manufacturing steps of electronic and optoelectronic devices.      

          For optical thickness measurements, equations (I-8-1) and (I-8-2) can be used to determine the film thickness. According to equation (I-8-2), the fringe of order m lies at l1 and that of order (m+1) at l2. Hence, we have ml1=(m+1)l2 so that m=l2/(l1-l2). With (I-8-2) we find, 2nd=l1l2/(l1-l2) and the thickness of the film is

 

,                                          (I-8-3)

 

where l1 and l2 is the wavelength of two adjacent maxima or minima in the spectrum.

 

Example: Figure I-8-3 shows the transmission spectrum of a thin CdS film on glass. The transmittance starts at the band-gap of the material (»500 nm) and pronounced fringes at 582 and 631 nm are observed. More details concerning semiconductor will follow in Semiconductor Optics I & II.

 Figure I-8-3: Transmittance of thin film CdS on glass at room temperature.

 

Solution: The thin film CdS exclusively causes the fringes in figure I-8-3. The glass substrate does not influence the interference effect. Hence, we insert the wavelengths of the two indicated maxima and nCdS=2.5 in equation (I-8-3) and get the thickness of the film,

 

.

 

The calculation of the transmitted and reflected intensities of thin films requires the consideration of the internal reflections. Figure I-8-4 shows the concept. I0 is the intensity of the incident beam, R is the reflection coefficient of the surface and backface, a is the absorption coefficient and x the thickness of the film. The transmitted and reflected intensities are summed up with a geometrical series delivering the following results for the transmittance and reflectance,

 

                                        (I-8-4)

and

.                                     (I-8-5)

 

 

 

Figure I-8-4: The transmitted and reflected intensities through a thin film.

 

For many applications, the formulas

 

                           (I-8-6)

 

and

                                      (I-8-7)

 

are accurate enough.

 

Example: Calculate the transmittance of the film with a thickness of 1 mm, an absorption coefficient of 100 cm-1 and an reflection coefficient of 0.2.

 

Solution:  Tr=(1-0.2)2exp(-100 cm-1´10-4 cm-1)»(1-0.2)2=0.64.We see, in case of effective absorption the reflection of the surface determines the transmission features of the film.