Interference of light in thin films. Stripes of equal slope and equal thickness. Newton's rings. Practical application of interference. Application of light interference Performing light interference using thin film
Interference fringes of equal inclination. When a thin film is illuminated, a superposition of waves from the same source occurs, reflected from the front and rear surfaces of the film. This may cause light interference. If the light is white, then the interference fringes are colored. Interference in films can be observed on the walls of soap bubbles, on thin films of oil or petroleum floating on the surface of water, on films that appear on the surface of metals or mirrors.
Let us first consider a plane-parallel plate of thickness with a refractive index (Fig. 2.11). Let a plane light wave fall on the plate, which can be considered as a parallel beam of rays. The plate throws upward two parallel beams of light, one of which was formed due to reflection from the upper surface of the plate, the second - due to reflection from the lower surface. Each of these beams is shown in Fig. 2.11 with only one beam.
When entering and exiting the plate, beam 2 undergoes refraction. In addition to two beams and , the plate throws upward beams resulting from three-, five-, etc. multiple reflection from plate surfaces. However, due to their low intensity, they can be ignored.
Let us consider the interference of rays reflected from the plate. Since a plane wave falls on the plate, the front of this wave is a plane perpendicular to rays 1 and 2. In Fig. 2.11 straight line BC represents a section of the wave front by the plane of the figure. The optical path difference acquired by rays 1 and 2 before they converge at point C will be
| , | (2.13) |
where is the length of the segment BC, and is the total length of the segments AO and OS. The refractive index of the medium surrounding the plate is assumed to be equal to unity. From Fig. 2.11 it is clear that 
, . Substituting these expressions into (2.13) gives . Let's use the law of light refraction: 
; and take into account that , then for the path difference we obtain the following expression: 
.
When calculating the phase difference between oscillations in the rays and it is necessary, in addition to the optical path difference D, to take into account the possibility of a phase change upon reflection at point C. At point C, the wave is reflected from the interface between an optically less dense medium and an optically more dense medium. Therefore, the phase of the wave undergoes a change by p. At a point, reflection occurs from the interface between an optically denser medium and an optically less dense medium, and a phase jump does not occur in this case. Qualitatively, this can be imagined as follows. If the thickness of the plate tends to zero, then the formula we obtained for the optical path difference gives . Therefore, when the rays are superimposed, the oscillations should increase. But this is impossible, since an infinitely thin plate cannot influence the propagation of light at all. Therefore, the waves reflected from the front and back surfaces of the plate must cancel each other out during interference. Their phases must be opposite, that is, the optical path difference D at d→0 should tend to . Therefore, you need to add or subtract to the previous expression for D, where λ 0 is the wavelength in vacuum. The result is:
 .
| (2.14) |
So, when a plane wave falls on a plate, two reflected waves are formed, the path difference of which is determined by formula (2.14). These waves can interfere if the optical path difference does not exceed the coherence length. The last requirement for solar radiation leads to the fact that interference when illuminating the plate is observed only if the thickness of the plate does not exceed a few hundredths of a millimeter.
In practice, interference from a plane-parallel plate is observed by placing a lens in the path of the reflected beams, which collects the beams at one of the points of the screen located in the focal plane of the lens. The illumination at this point depends on the optical path difference. At , we get maxima, and at , we get minima of intensity. Therefore, the condition for intensity maxima has the form:
 ,
| (2.15) |
and the minimums:
 .
| (2.16) |
These relations are obtained for reflected light.
Let a thin plane-parallel plate be illuminated with scattered monochromatic light. Let us place a lens parallel to the plate, in the focal plane of which we place the screen (Fig. 2.12). Scattered light contains rays from a wide variety of directions. Rays parallel to the plane of the pattern and incident on the plate at an angle , after reflection from both surfaces of the plate, will be collected by the lens at a point and create illumination at this point, determined by the value of the optical path difference. Rays coming in other planes, but incident on the plastic at the same angle, will be collected by the lens at other points located at the same distance from the center of the screen as the point . The illumination at all these points will be the same. Thus, rays incident on the plate at the same angle will create on the screen a set of equally illuminated points located in a circle with the center at point O. Similarly, rays incident at a different angle will create on the screen a collection of equally illuminated points located in a circle of a different radius . But the illumination of these points will be different, since they correspond to a different optical path difference.
As a result, a set of alternating dark and light circular stripes with a common center at point O will appear on the screen. Each stripe is formed by rays incident on the plate at the same angle. Therefore, the resulting interference fringes in this case are called fringes of equal inclination.
According to (2.15), the position of the intensity maxima depends on the wavelength, therefore, in white light, a set of stripes shifted relative to each other, formed by rays of different colors, is obtained, and the interference pattern will acquire a rainbow color.
To observe fringes of equal inclination, the screen must be located in the focal plane of the lens, just as it is positioned to obtain objects at infinity. Therefore, they say that bands of equal inclination are localized at infinity. The role of the lens can be played by the lens of the eye, and the role of the screen can be played by the retina.
Interference fringes of equal thickness. Let us now take a wedge-shaped plate. Let a parallel beam of rays fall on it (Fig. 2.13). But now the rays, reflected from different surfaces of the plate, will not be parallel. 
Two almost merging beams before falling on the plate after reflection from the upper and lower surfaces of the wedge intersect at point . After reflection, two practically merging rays intersect at point . It can be shown that the points and lie in the same plane passing through the vertex of the wedge ABOUT.
If you position the screen E so that it passes through the points and, an interference pattern will appear on the screen. At a small angle of the wedge, the difference in the path of the rays reflected from its upper and lower surfaces can be calculated with a sufficient degree of accuracy using the formula 
obtained for a plane-parallel plate, taking as the thickness of the wedge at the point where the rays fall on it. Since the difference in the path of the rays reflected from different parts of the wedge is now unequal, the illumination will be uneven - light and dark stripes will appear on the screen. Each of these stripes arises as a result of reflection from sections of the wedge with the same thickness, as a result of which they are called stripes of equal thickness.
Thus, the interference pattern resulting from the reflection of a plane wave from a wedge turns out to be localized in a certain region near the surface of the wedge. As you move away from the top of the wedge, the optical path difference increases, and the interference pattern becomes less and less distinct.
 Rice. 2.14 |
When observed in white light, the stripes will be colored, so that the surface of the plate will have a rainbow color. In real conditions, when observing, for example, rainbow colors on a soap film, both the angle of incidence of the rays and the thickness of the film change. In this case, bands of a mixed type are observed.
Stripes of equal thickness can easily be observed on a flat wire frame that has been dipped in a soap solution. The soap film that covers it is covered with horizontal interference fringes, resulting from the interference of waves reflected from different surfaces of the film (Fig. 2.14). Over time, the soap solution drains and the interference fringes move down.
If you follow the behavior of a spherical soap bubble, you will easily find that its surface is covered with colored rings, slowly sliding towards its base. The displacement of the rings indicates a gradual thinning of the walls of the bubble.
Newton's rings
A classic example of strips of equal thickness are Newton's rings. They are observed when light is reflected from a plane-parallel glass plate and a plane-convex lens with a large radius of curvature in contact with each other (Fig. 2.15). The role of a thin film, from the surface of which waves are reflected, is played by the air gap between the plate and the lens (due to the large thickness of the plate and lens, interference fringes do not arise due to reflections from other surfaces). With normal incidence of light, stripes of equal thickness look like circles; with inclined light, they look like ellipses.
Let's find the radii of Newton's rings obtained when light is incident normally to the plate. In this case and . From Fig. 2.15 it is clear that , where is the radius of curvature of the lens, is the radius of the circle, all points of which correspond to the same gap. The value can be neglected, then . To take into account the change in phase by p that occurs during reflection from the plate, you need to add to the path difference: , that is, at the point of contact between the plate and the lens, a minimum intensity is observed due to the change in phase by p when the light wave is reflected from the plate.
 Rice. 2.16 |
In Fig. Figure 2.16 shows a view of Newton's interference rings in red and green light. Since the wavelength of red light is longer than that of green light, the radii of the rings in red light are larger than the radii of rings with the same number in green light.
If, in Newton’s installation, the lens is moved upward parallel to itself, then due to the increase in the thickness of the air gap, each circle corresponding to a constant path difference will be contracted towards the center of the picture. Having reached the center, the interference ring turns into a circle, disappearing as the lens moves further. Thus, the center of the picture will alternately become light and dark. At the same time, new interference rings will appear at the periphery of the field of view and move towards the center until each of them disappears in the center of the picture. As the lens moves continuously upward, the rings of the lowest orders of interference disappear and rings of higher orders appear.
Example
Optics coating
Coating of optics is done to reduce the reflectance of the surfaces of optical parts by applying one or more non-absorbing films to them. Without antireflective films, losses due to light reflection can be very large. In systems with a large number of surfaces, such as complex lenses, light loss can reach 70% or more, which degrades the quality of images generated by such optical systems. This can be eliminated by clearing the optics, which is one of the most important applications of interference in thin films.
When light is reflected from the front and back surfaces of a film deposited on an optical part, the reflected light will produce a minimum intensity as a result of interference, and therefore, the transmitted light will have a maximum intensity for that wavelength. At normal incidence of light, the effect will be maximum if the thickness of the thin film is equal to an odd number of quarters of the wavelength of light in the film material. Indeed, in this case, the loss of half the wavelength upon reflection does not occur, since on both the upper and lower surfaces of the film the wave is reflected from the interface between an optically less dense and an optically more dense medium. Therefore, the condition for maximum intensity will take the form 
. From here we get 
.
By changing the thickness of the antireflection film, you can shift the minimum reflection to different parts of the spectrum.
We often observe iridescent coloring of thin films, for example, oil films on water, oxide films on metals, which appear as a result of the interference of light that is reflected by two surfaces of the film.
Interference in thin films
Consider a plane-parallel thin plate whose refractive index is n and whose thickness is b. Let a plane monochromatic wave fall on such a film at an angle (let’s assume that this is one beam) (Fig. 1). On the surface of such a film, at some point A the beam is divided. It is partially reflected from the upper surface of the film, and partially refracted. The refracted ray reaches point B, is partially refracted into the air (the refractive index of air is equal to one), is partially reflected and goes to point C. Now it will again be partially reflected and refracted, and exits into the air at an angle. The rays (1 and 2) that come out of the film are coherent if their optical path difference is small in comparison with the coherence length of the incident wave. If a converging lens is placed on the path of rays (1 and 2), they will converge at some point D (in the focal plane of the lens). In this case, an interference pattern will appear, which is determined by the optical difference in the path of the interfering beams.
The optical difference in the path of rays 1 and 2, which appears in the rays when they travel the distance from point A to plane CE, is equal to:
where we assume that the film is in a vacuum, so the refractive index is . The appearance of the value is explained by the loss of half the wavelength when light is reflected from the media interface. With title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: -3px;"> половина волны будет потеряна в точке А, и при величине будет стоять знак минус. Если , то половина волны будет потеряна в точке В и при будет стоять знак плюс. В соответствии с рис.1:!}
where is the angle of incidence inside the film. From the same figure it follows that:
Let us take into account that for the case under consideration the law of refraction is:
Considering the loss of half the wavelength:
For the case in which title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: -3px;">, получим:!}
According to the condition for interference maxima, at point D we will observe a maximum if:
The minimum intensity will be observed at the point in question if:
The interference phenomenon can only be observed if twice the film thickness is less than the coherence length of the incident wave.
Expressions (8) and (9) show that the interference pattern in films is determined by the thickness of the film (for us b), the wavelength of the incident light, the refractive index of the film substance and the angle of incidence (). For the listed parameters, each inclination of the rays () corresponds to its own interference fringe. The stripes resulting from the interference of rays incident on the film at the same angles are called stripes of equal inclination.
Examples of problem solving
EXAMPLE 1
| Exercise | What must be the minimum thickness of a soap film (refractive index) that is in the air in order for the light of wavelength m reflected from it to be maximally amplified as a result of interference? Assume that the light hits the film normally. |
| Solution | As a basis for solving the problem, we use the formula that we obtained within the theoretical part of this section. Maximum interference will be observed if: where m=1, for minimum film thickness. Let us take into account that according to the conditions of the problem, light falls on the surface of the film along the normal, that is, in addition, we note that in expression (1.1), by placing a plus sign in front of , we took into account that the refractive index of the soap film is greater than the refractive index of air. Thus, from formula (1.1) we obtain:
Expressing b, we have: Let's do the calculations:
|
| Answer | m |
Interference of light amplitude division method in many respects it is easier to observe than in experiments with wavefront division. One of the ways using this method is Paul's experience .
In the Field experiment, light from a source S is reflected by two surfaces of a thin transparent plane-parallel plate (Fig. 8.7).
Anywhere P, located on the same side of the plate as the source, two rays arrive. These rays form an interference pattern.
To determine the type of stripes, you can imagine that the rays come out of virtual images S 1 and S 2 sources S, created by the surfaces of the plate. On a remote screen located parallel to the plate, the interference fringes have the form of concentric rings with centers perpendicular to the plate passing through the source S. This experience places less stringent requirements on source size S than the experiments discussed above. Therefore it is possible as S use a mercury lamp without an auxiliary screen with a small hole, which provides a significant luminous flux. Using a mica sheet (0.03 - 0.05 mm thick) you can get a bright interference pattern directly on the ceiling and walls of the auditorium. The thinner the plate, the larger the scale of the interference pattern, i.e. greater distance between stripes.
Equal slope strips
Particularly important is the special case of interference of light reflected by two surfaces of a plane-parallel plate, when the observation point P is at infinity, i.e. observation is carried out either with an eye accommodated to infinity, or on a screen located in the focal plane of a collecting lens (Fig. 8.8).
In this case, both rays coming from S To P, generated by one incident ray and after reflection from the front and rear surfaces of the plate are parallel to each other. Optical path difference between them at a point P same as on line DC:
Here n– refractive index of the plate material. It is assumed that there is air above the plate, i.e. . Because 
, 
(h– thickness of the plate, and – angles of incidence and refraction on the upper face; ), then for the path difference we obtain
It should also be taken into account that when a wave is reflected from the upper surface of the plate, in accordance with the Fresnel formulas, its phase changes by π. Therefore, the phase difference δ of the folded waves at the point P is equal to:
,
where is the wavelength in vacuum.
According to the last formula, light stripes are located in places for which 
, Where m – interference order. The band corresponding to this interference order is caused by light incident on the plate at a very specific angle α. Therefore, such stripes are called interference stripes of equal slope
.
If the lens axis is located perpendicular to the plate, the fringes have the form of concentric rings with the center at the focus, and in the center of the picture the interference order is maximum.
Stripes of equal inclination can be obtained not only in reflected light, but also in light transmitted through the plate. In this case, one of the rays passes straight, and the other after two reflections on the inside of the plate. However, the visibility of the stripes is low.
To observe strips of equal inclination, instead of a plane-parallel plate, it is convenient to use Michelson interferometer (Fig. 8.9). Let's consider the circuit of the Michelson interferometer: z1 and z2 are mirrors. The translucent mirror is silvered and divides the beam into two parts - beams 1 and 2. Beam 1, reflecting from z1 and passing, gives, and beam 2, reflecting from z2 and further from, gives. The plates are identical in size. is installed to compensate for the difference in the path of the second beam. The rays are both coherent and interfere.
Interference from the wedge. Stripes of equal thickness
We examined interference experiments in which the division of the amplitude of a light wave from a source occurred as a result of partial reflection on the surfaces of a plane-parallel plate. Localized bands with an extended source can also be observed under other conditions. It turns out that for a sufficiently thin plate or film (the surfaces of which do not have to be parallel and generally flat), one can observe an interference pattern localized near the reflecting surface. The bands that appear under these conditions are called stripes of equal thickness . In white light, the interference fringes are colored. Therefore, this phenomenon is called colors of thin films. It is easy to observe on soap bubbles, on thin films of oil or gasoline floating on the surface of water, on films of oxides that appear on the surface of metals during hardening, etc.
Let us consider the interference pattern obtained from plates of variable thickness (from a wedge).
The directions of propagation of the light wave reflected from the upper and lower boundaries of the wedge do not coincide. Reflected and refracted rays meet, so the interference pattern when reflected from a wedge can be observed without using a lens if the screen is placed in the plane of the intersection points of the rays (the lens of the eye is placed in the desired plane).
Interference will be observed only in the 2nd region of the wedge, since in the 1st region the optical path difference will be greater than the coherence length.
The result of interference at points and screen is determined by the well-known formula 
, substituting into it the thickness of the film at the point of incidence of the beam ( or ). The light must be parallel (): if two parameters change simultaneously b and α, then there will be no stable interference pattern.
Since the difference in the path of the rays reflected from different parts of the wedge will be unequal, the illumination of the screen will be uneven, and there will be dark and light stripes on the screen (or colored when illuminated with white light, as shown in Fig. 8.11). Each of these stripes arises as a result of reflection from sections of the wedge with the same thickness, which is why they are called stripes of equal thickness .
In Fig. Figure 8.12 shows a frame in which two glass plates are clamped. One of them is slightly convex, so that the plates touch each other at some point. And at this point something strange is observed: rings appear around it. In the center they are almost not colored, a little further they shimmer with all the colors of the rainbow, and towards the edge they lose color saturation, fade and disappear.
This is what the experiment that laid the foundation for modern optics looks like in the 17th century. Newton studied this phenomenon in detail, discovered patterns in the arrangement and color of the rings, and also explained them on the basis of the corpuscular theory of light.
Ring strips of equal thickness, observed in the air gap between the convex spherical surface of a lens of small curvature and the flat surface of the glass in contact(Fig. 8.13), called Newton's rings.
The common center of the rings is located at the point of contact. In reflected light, the center is dark, since when the thickness of the air gap is much smaller than the wavelength , the phase difference of the interfering waves is due to the difference in the conditions of reflection on the two surfaces and is close to π. Thickness h air gap is related to the distance r to the point of contact (Fig. 8.13):
.
Here the condition is used. When observed along the normal, the dark stripes, as already noted, correspond to the thickness, so for the radius m-th dark ring we get
(m = 0, 1, 2, …).
If the lens is gradually moved away from the glass surface, the interference rings will be drawn towards the center. As the distance increases, the picture takes on the same form, since the place of each ring will be occupied by a ring of the next order. Using Newton's rings, as in Young's experiment, it is possible to approximately determine the wavelength of light using relatively simple means.
Stripes of equal thickness can also be observed using a Michelson interferometer if one of the mirrors z1 or z2 (Fig. 8.9) is deflected by a small angle.
So, stripes of equal slope obtained by illuminating a plate of constant thickness () diffused light, which contains rays of different directions. Stripes of equal thickness observed when illuminating a plate of variable thickness(wedge) () parallel beam of light. Stripes of equal thickness are localized near the plate.
Optics coating. The phenomenon of interference is used to improve the quality of optical devices and obtain highly reflective coatings. The passage of light through each refractive surface of the lens is accompanied by the reflection of 4% of the incident flux (with a refractive index of glass of 1.5). Since modern lenses consist of a large number of lenses, the number of reflections in them is large, and therefore the loss of light flux is large. To eliminate this and other shortcomings, the so-called optics are cleared. To do this, thin films with a refractive index lower than that of the lens material are applied to the free surfaces of the lenses. When light is reflected from the air-film and film-glass interfaces, interference of the reflected rays occurs. The film thickness d and the refractive indices of the glass and film n are selected so that the reflected waves cancel each other. To do this, their amplitudes must be equal, and the optical path difference must be equal. The calculation shows that the amplitudes of the reflected rays are equal if. Since, the loss of a half-wave occurs on both surfaces; hence the minimum condition (the light falls normally)
Usually accepted then
Since it is impossible to achieve simultaneous suppression for all wavelengths (the refractive index depends on the wavelength), this is done for color c (the eye is most sensitive to it). Therefore, lenses with coated optics have a bluish-red tint.
Interference filters. Multipath interference can be achieved in a multilayer system of alternating films with different refractive indices (but the same optical thickness). When light passes through, a large number of reflected interfering rays arise, which, given the optical thickness of the films, will mutually intensify, i.e. the reflection coefficient increases. Such reflectors are used in laser technology and are also used to create interference filters.
Interferometers. The phenomenon of interference is used in very precise measuring instruments - interferometers. In Fig. shows a diagram of a Michelson interferometer. A beam of light from a source S falls on a plate coated with a thin layer of silver (due to which the reflection coefficient is close to 0.5). The further path of the interfering rays is clear from the figure. On the path of beam 1, a plate exactly like, but not silvered, is placed. It equalizes the paths of rays 1 and 2 in the glass. The interference pattern is observed using a telescope.
The interference pattern corresponds to the interference in the air layer formed by the mirror and the virtual image of the mirror in the translucent plate. The nature of the interference pattern depends on the position of the mirrors and on the divergence of the light beam incident on the device. If the beam is parallel, and the planes form a wedge, then interference fringes of equal thickness are observed, located parallel to the edge of the air wedge. With a diverging beam of light and a parallel arrangement of planes, stripes of equal inclination are obtained, having the form of concentric rings.
The Fabry-Perot interferometer consists of two parallel glass or quartz plates separated by an air gap (Fig.). The intensities of the rays emerging from the device are related as
Accordingly, the amplitude ratios will be as follows
The phase of oscillation with increasing beam number changes by the same amount, determined by the optical difference in the paths of adjacent beams.
When a diverging beam of light is passed through the device, stripes of equal inclination appear in the focal plane of the lens, having the form of concentric rings.
The applications of interferometers are very diverse. They are used for precise (about 10 7 m) length measurements, angle measurements, determining the quality of optical parts, studying fast processes, etc.
Translation by Alexander Zhdanov
Thin film interference occurs when incident light waves reflected from the top and bottom of the thin film interfere with each other to form a new wave. By examining this reflected wave, information can be revealed about the surface from which the components of this wave were reflected, including the thickness of the film or the value of the effective refractive index of the film material. Thin films have many commercial applications, including anti-reflective coatings, mirrors, and optical filters.
Consider light incident on a thin film and reflected from the top and bottom boundaries. It is necessary to calculate the optical path difference of the reflected light to determine the interference condition. This condition may change after considering the possible phase shifts that occur during reflection. If the incident light is monochromatic, then the interference patterns appear in the form of light and dark stripes. Light stripes correspond to regions where constructive interference occurs between reflected waves, and dark stripes correspond to regions of destructive interference. Just like film thickness varies from one location to another, interference can vary from constructive to destructive. A good example of this phenomenon is "Newton's rings", which demonstrate the interference pattern that occurs when light is reflected from a spherical surface adjacent to a flat surface. If the incident light is broadband, or white, like light from the sun, interference patterns appear as colorful stripes. Different wavelengths of light create constructive interference for different film thicknesses. Different sections of the film appear in different colors depending on the local film thickness.
A thin film is a layer of material ranging in thickness from subnanometer to micron. When light hits the surface of the film, it either passes through or reflects off the top surface. Light that passes through the upper boundary reaches the lower surface and can be refracted or reflected again. Fresnel's equations provide a quantitative description of how much light will pass or be reflected at the boundary. Light reflected from the top and bottom surfaces will exhibit interference properties. The degree of constructive or destructive interference between two light waves depends on the difference in their phase. This difference, in turn, depends on the thickness of the film layer, the refractive index of the film, and the angle of incidence of the original wave on the film. In addition, a phase shift of 180° or Pi in radians can occur upon reflection at the boundary, depending on the ratio of the refractive indices of the materials on either side of the boundary. This phase shift occurs if the refractive index of the medium is less than the refractive index of the material through which the light passes (propagates). In other words, if n 1 
