Phase difference in the interference of thin films. Application of light interference

If n> n 0, .

At the point R will be the maximum if or (9.2)

Minimum if or (9.3)

When the film is illuminated with white light, for some wavelengths, the condition of the maximum reflection is satisfied, for some others - the minimum. Therefore, in reflected light, the film appears colored.

Interference is observed not only in reflected light, but also in light passing through the film. the optical path difference for transmitted light differs from that for reflected light by, then the interference maxima in the reflected light correspond to minima in the transmitted light, and vice versa. Interference is observed only if the doubled thickness of the plate is less than the length coherence incident wave.

1. Equal slope stripes(interference from a plane-parallel plate).

Def. 9.1. Interference fringes resulting from the superposition of rays incident on a plane-parallel plate at the same angles are called stripes of equal inclination.

The rays / / and / //, reflected from the upper and lower edges of the plate, are parallel to each other, since the plate is plane-parallel. That. rays 1 "and I""Intersect" only at infinity, therefore they say that stripes of equal slope localized at infinity. To observe them, a collecting lens and a screen (E) located in the focal plane are used.

The beams / "and /" / will gather in focus F lenses (in the figure, its optical axis is parallel to the beams G and / "), other rays (ray 2), parallel to the ray /, will come to the same point, - the total intensity increases. 3, tilted at a different angle, will gather in a different so-called. R focal plane of the lens. If the optical axis of the lens is perpendicular to the surface of the plate, then the stripes of equal inclination will look like concentric rings centered at the focus of the lens.

Objective 1. A beam of monochromatic light is normally incident on a thick glass plate covered with a very thin film. The reflected light is attenuated as much as possible due to interference. Determine the film thickness.

Given: Solution:

Because the refractive index of air is less than the refractive index of the film, which in turn is less than the refractive index of glass, then in both cases the reflection occurs from an optically denser medium than the medium in which the incident beam travels. Therefore, the phase of the oscillations is changed to twice and the result will be the same as if there were no phase change.

Minimum condition:, where is not taken into account,, and. Assuming,,, etc.

2.

Stripes of equal thickness (interference from a plate of variable thickness).

Let a plane wave fall on the wedge (the angle a between the lateral faces is small), the direction of propagation of which coincides with parallel rays / and 2.P Consider the rays / / and / // reflected from the upper and lower surfaces of the wedge. With a certain relative position of the wedge and the lens, the rays / / and 1" intersect in some so. A, which is an image of a point V.

Since the rays / / and / // are coherent, they will interfere. If the source is located far from the surface of the wedge and the angle a is small enough, then the optical path difference between the beams / / and / // can be calculated by the formula (10.1), where as d the thickness of the wedge is taken at the place where the ray falls on it. Beams 2" and 2", formed by beam splitting 2, wedge falling to another point, are collected by the lens, incl. A". The optical path difference is determined by the thickness d ". A system of interference fringes appears on the screen. Each of the bands arises due to reflection from places of the plate that have the same thickness.

Def. 9.2. Interference fringes resulting from interference from places of the same thickness are called. strips of equal thickness.

Since the upper and lower edges of the wedge are not parallel to each other, the rays / / and / // {2" and 2"} intersect near the plate. Thus, stripes of equal thickness are localized near the wedge surface. If light falls on the plate normally, then stripes of equal thickness are localized on the upper surface of the wedge. If we want to get an image of the interference pattern on the screen, then the collecting lens and the screen must be positioned in relation to the wedge so that the image of the upper surface of the wedge can be seen on the screen.

To determine the width of the interference fringes in the case of monochromatic light, we write down the condition for two adjacent interference maxima ( m th and m + 1- th orders) according to the formula 9.2: and , where . If the distances from the edge of the wedge to the considered interference fringes are equal and, then, and, where the small angle between the edges of the wedge (refractive angle of the wedge), i.e. ... Due to its smallness, the refractive angle of the wedge should also be very small, since otherwise, stripes of equal thickness will be so closely spaced that they cannot be distinguished.

Objective 2. A beam of rays of monochromatic light falls on the glass wedge normally to its edge. The number of fringes per 1 cm is 10. Determine the refractive angle of the wedge.

Given: Solution:

A parallel beam of rays, falling normally to the edge of the wedge, is reflected from both the upper and lower edges. These beams are coherent, so a stable interference pattern is observed. Because If fringes are observed at small wedge angles, then the reflected rays will be practically parallel.

Dark stripes will be observed in those areas of the wedge for which the difference between the walking rays is equal to an odd number of half-waves: or, Since , then . Let an arbitrary dark strip of the number correspond to a certain thickness of the wedge at this place, and the dark strip of the number corresponds to the thickness of the wedge at this point,. According to the condition, 10 stripes fit into, then, since , then .

Newton's rings.

Newton's rings are an example of stripes of equal thickness. They are observed when light is reflected from an air gap formed by a plane-parallel plate and a plano-convex lens with a large radius of curvature in contact with it. A parallel beam of light falls on the flat surface of the lens and is partially reflected from the upper and lower surfaces of the air gap between the lens and the plate, i.e. reflected from optically denser media. In this case, both waves change the phase of the oscillations by and no additional path difference arises. When the reflected rays are superimposed, stripes of equal thickness appear, which at normal incidence of light have the form of concentric circles.

In reflected light, the optical path difference ati = 0: R) determine and, conversely, by the known find R ..

For strips of equal inclination and strips of equal thickness the position of the maxima depends on... The system of light and dark stripes is obtained only when illuminated with monochromatic light. When viewed in white light, a set of stripes shifted relative to each other is obtained, formed by beams of different wavelengths, and the interference pattern becomes rainbow-colored. All reasoning was carried out for reflected light. Interference can be observed and in transmitted light, and in this case, no loss of a half-wave is observed - the optical path difference for the transmitted and reflected light differs by / 2, i.e. with. the maxima of interference in reflected light correspond to minima in transmitted light, and vice versa.

We often see iridescent coloration of thin films, such as oil films on water, oxide films on metals, which appear as a result of the interference of light that reflects the two surfaces of the film.

Interference in thin films

Consider a plane-parallel thin plate, the refractive index of which is equal to n, and the thickness is equal to b. Let a plane monochromatic wave fall on such a film at an angle (let us assume that this is one ray) (Fig. 1). On the surface of such a film, at some point A, the beam is divided. It is partially reflected from the top surface of the film, partially refracted. The refracted ray reaches point B, partially refracts into air (the refractive index of air is equal to one), partially reflects and goes to point C. Now it will be partially reflected and refracted again, and will be released into the air at an angle. Beams (1 and 2) that emerged from the film are coherent if their optical path difference is small in comparison with the long coherence of the incident wave. In the event that a converging lens is placed in 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 path difference of the interfering beams.

The optical difference between the paths of rays 1 and 2, which appears in the rays when they pass the distance from point A to the plane CE, is equal to:

where we assume that the film is in vacuum, therefore the refractive index. The appearance of the quantity is explained by the loss of half the wavelength when light is reflected from the boundary between the media. With title = "(! LANG: 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's take into account that for the considered case the law of refraction:

Given the loss of half a wavelength:

For the case where title = "(! LANG: Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: -3px;">, получим:!}

According to the condition for the interference maxima, at point D we will observe a maximum if:

The minimum intensity will be observed at the point under consideration if:

The phenomenon of interference can only be observed if the doubled 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 film thickness (here b), the wavelength of the incident light, the refractive index of the film material, and the angle of incidence (). For the listed parameters, each ray tilt () has its own interference fringe. The fringes resulting from the interference of rays falling on the film at the same angles are called equal slope fringes.

Examples of problem solving

EXAMPLE 1

Stripes of equal slope. Interference fringes are called stripes of equal slope, if they arise when light is incident on a plane-parallel plate (film) at a fixed angle as a result of the interference of rays reflected from both surfaces of the plate (film) and coming out parallel to each other.

Stripes of equal inclination are localized at infinity, therefore, to observe the interference pattern, the screen is placed in the focal plane of the collecting lens (as for obtaining an image of infinitely distant objects) (Fig. 22.3).

Rice. 22.3.

The radial symmetry of the lens causes the interference pattern on the screen to appear as concentric rings centered at the focus of the lens.

Let from air (n, ~ 1) onto a plane-parallel transparent plate with a refractive index n 2 and thickness d a plane monochromatic light wave with a wavelength X(fig.22.3).

At the point A light beam SA partly reflected and partly refracted.

Reflected beam 1 and reflected at the point V Ray 2 coherent and parallel. If you bring them to a point with a collecting lens R, then they will interfere in reflected light.

We will take into account reflection feature electromagnetic waves and, in particular, light waves when they fall from a medium with a lower dielectric constant (and a lower refractive index) on the interface between two media: when a wave is reflected from an optically denser medium ( n 2> n,) its phase changes by n, which is equivalent to the so-called "half-wave loss" (± A / 2) during reflection, i.e. the optical path difference A changes by X / 2.

Therefore, the optical path difference of the interfering beams is determined as

Using the law of refraction (sin 0 = "2 sind"), as well as the fact that i, = 1, AB- BC = d/ cos O "and AD - AC sin fs-2d tgO "sin O, you can get

Consequently, the optical path difference A of the waves is determined by the angle O, which is uniquely related to the position of the point R in the focal plane of the lens.

According to formulas (22.6) and (22.7), the position of the light and dark stripes is determined by the following conditions:

So for data X, d and n 2 each tilt of 0 rays relative to the plate corresponds to its own interference fringe.

Strips of equal thickness. Let a plane monochromatic light wave fall on a transparent thin plate (film) of variable thickness - a wedge with a small angle and between the side faces in the direction of parallel rays 1 and 2 (fig.22.4). The intensity of the interference pattern formed by coherent rays reflected from the upper

from the thickness of the wedge at a given point (d and d " for rays 1 and 2 respectively).

Rice. 22.4. Observation of stripes of equal and lower surfaces of the wedge, depends

Coherent ray pairs (G and G, 2 and 2") intersect near the surface of the wedge (respectively, points O and O ") and are collected by the lens on the screen (respectively at points R and R").

Thus, a system of interference fringes appears on the screen - strips of equal thickness, each of which arises from reflection from sections of the wedge with the same thickness. Stripes of equal thickness are localized near the wedge surface (in the plane 00", marked with a dotted line).

When light beams from an extended light source fall on a transparent wedge almost normally, then the optical path difference

and depends only on the thickness of the wedge d at the point of incidence of the rays. This explains the fact that the fringes on the surface of the wedge have the same illumination at all points on the surface where the thickness of the wedge is the same.

If T is the number of light (or dark) interference fringes per wedge segment of length /, then the angle at the top of the wedge (sina ~ a), expressed in radians, is calculated as

where d] and d 2- the thickness of the wedge, on which they are located, respectively To-Me and (k + t) th interference fringes; Oh- the distance between these stripes.

Newton's rings. Newton's rings are a classic example ring strips of equal thickness, which are observed when monochromatic light with a wavelength X is reflected from the air gap formed by a plane-parallel plate and a plano-convex lens with a large radius of curvature in contact with it.

Rice. 22.5.

A parallel beam of light falls normally on the flat surface of the lens (Fig. 22.5). Stripes of equal thickness have the form of concentric circles with the center of contact of the lens with the plate.

Let us obtain the condition for the formation of dark rings. They arise where the optical path difference D of the waves reflected from both surfaces of the gap is equal to an odd number of half-waves:

where X / 2 is associated with the "loss" of the half-wave upon reflection from the plate.

We use both last equations. Therefore, in reflected light, the radii of the dark rings

Value T= 0 corresponds to the minimum of the dark spot in the center of the picture.

Similarly, we find that the radii of the light rings are determined as

These formulas for the radii of the rings are valid only in the case of ideal (point) contact of the spherical surface of the lens with the plate.

Interference can also be observed in transmitted light, and in transmitted light, the interference maxima correspond to the interference minima in reflected light, and vice versa.

Optics enlightenment. The lenses of optical instruments contain a large number of lenses. Even a slight reflection of the light of each

Rice. 22.6.

from the surfaces of the lenses (about 4% of the incident light) leads to the fact that the intensity of the transmitted light beam is significantly reduced. In addition, lens flare and background stray light occur in lenses, which reduces the efficiency of optical systems. In prismatic binoculars, for example, the total loss of luminous flux reaches -50%, but conditions can be created at the boundaries of the media when the intensity of light transmitted through the optical system will be maximum. For example, thin transparent films are applied to the lens surface. dielectric thickness d with refractive index n b (Fig.22.6). At d - NX / 4 (N- odd number) ray interference G and 2, reflected off the top and bottom surfaces of the film will give a minimum of reflected light intensity.

Typically, optical antireflection is performed for the middle (yellow-green) region of the visible spectrum. As a result, lenses appear purple in reflected light due to the mixing of red and purple. Modern technologies for the synthesis of oxide films (for example, by the sol-gel method) make it possible to create new antireflection protective coatings in optoelectronics based on the elements of the metal - oxide - semiconductor structure.

When a light wave is incident on a thin transparent plate (or film), reflection occurs from both surfaces of the plate. As a result, two light waves are generated, which, under certain conditions, can interfere.

Let a plane light wave fall on a transparent plane-parallel plate, which can be considered as a parallel beam of rays (Fig. 122.1). The plate casts up two parallel beams of light, of which one was formed due to reflection from the upper surface of the plate, the second due to reflection from the lower surface (in Fig. 122.1, each of these beams is represented by only one ray). When entering and leaving the plate, the second beam undergoes refraction. In addition to these two beams, the plate will throw up the beams arising as a result of three-, five-, etc., multiple reflections from the surfaces of the plate. However, due to their low intensity, we will not take these beams into account. We will also not be interested in the beams that have passed through the plate.

The path difference acquired by beams 1 and 2 before they converge at point C is

where is the length of the BC segment, is the total length of the AO and OC segments, is the refractive index of the plate.

The refractive index of the medium surrounding the plate is assumed to be equal to unity. From fig. 122.1 shows that the thickness of the plate). Substitution of these values ​​into expression (122.1) gives that

Making a replacement and taking into account that

easy to bring the formula to form

When calculating the phase difference between the oscillations in beams 1 and 2, it is necessary, in addition to the optical path difference, to take into account the possibility of changing the phase of the wave upon reflection (see § 112). At point C (see Fig. 122.1), reflection occurs from the interface between a medium that is optically less dense, with a medium that is optically denser. Therefore, the phase of the wave undergoes a change to. At point O, reflection occurs from the interface between the optically denser medium and the optically less dense medium, so that there is no phase jump. As a result, an additional phase difference arises between beams 1 and 2. It can be taken into account by adding to (or subtracting from it) half the wavelength in vacuum. As a result, we get

So, when a plane wave falls on the plate, two reflected waves are formed, the path difference of which is determined by formula (122.3). Let us find out the conditions under which these waves turn out to be coherent and can interfere. Let's consider two cases.

1. Plane-parallel plate. Both plane reflected waves propagate in the same direction, forming an angle with the normal to the plate equal to the angle of incidence.

These waves can interfere if the conditions for both temporal and spatial coherence are met.

For temporal coherence to take place, the path difference (122.3) must not exceed the coherence length; equal (see formula (120.9)). Therefore, the condition must be met

In the resulting ratio, half can be neglected in comparison with The expression has a value of the order of unity. Therefore, one can write

(the doubled plate thickness should be less than the coherence length).

Thus, the reflected waves will be coherent only if the plate thickness does not exceed the value determined by relation (122.4). Putting, we get the limiting value of the thickness equal to

Let us now consider the conditions for the observance of spatial coherence. We put in the path of the reflected beams, the screen E (Fig. 122.2). The rays arriving at point P and are spaced apart in the incident beam at a distance. If this distance does not exceed the radius of coherence pco of the incident wave, beams 1 and 2 will be coherent and will create illumination at point P, determined by the value of the path difference corresponding to the angle of incidence.Other pairs of rays traveling at the same angle will create the same illumination at the other points of the screen. Thus, the screen will be uniformly illuminated (in the particular case when the screen is dark). When the inclination of the beam changes (i.e., when the angle changes), the illumination of the screen will change.

From fig. 122.1 it is seen that the distance between incident rays 1 and 2 is

If we accept then for it turns out a for

For a normal fall for any.

The radius of coherence of sunlight has a value of the order of 0.05 mm (see (120.15)). At an angle of incidence of 45 °, we can set Consequently, for the occurrence of interference under these conditions, the relation

(122.7)

(compare with (122.5)). For an angle of incidence of the order of 10 °, spatial coherence will be maintained at a plate thickness not exceeding 0.5 mm. Thus, we come to the conclusion that, due to the limitations imposed by temporal and spatial coherence, interference when the plate is illuminated by sunlight is observed only if the plate thickness does not exceed a few hundredths of a millimeter. When illuminated with light with a higher degree of coherence, interference is also observed when reflected from thicker plates or films.

In practice, interference from a plane-parallel plate is observed by placing a lens in the path of the reflected beams, which collects the rays at one of the points of the screen located in the focal plane of the lens (Fig. 122.3). Illumination at this point depends on the value of the quantity (122.3). At, maxima are obtained, at - intensity minima (is an integer). The condition for the maximum intensity has the form

Let a thin plane-parallel plate be illuminated by scattered monochromatic light (see Fig. 122.3). Place the lens parallel to the plate, in the focal plane of which we place the screen. The diffused light contains rays of 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 point P and create illumination at this point, determined by the value of the optical path difference. Rays going in other planes, but falling on the plate at the same angle, will be collected by the lens at other points located at the same distance from the center of the screen O as the point P. The illumination at all these points will be the same. Thus, rays falling on the plate at the same angle will create on the screen a set of equally illuminated points located in a circle centered at O. As a result, a system of alternating light and dark circular stripes with a common center at point O will appear on the screen. Each stripe is formed by rays falling on the plate at the same angle. If the lens is positioned differently relative to the plate (the screen in all cases must coincide with the focal plane of the lens), the shape of the stripes of equal inclination will be different.

Each point of the interference pattern is due to rays that form a parallel beam before passing through the lens. Therefore, when observing stripes of equal inclination, the screen should be located in the focal plane of the lens, i.e., as it is positioned to obtain images of infinitely distant objects on it. In accordance with this, it is said that stripes of different inclination are localized at infinity. The lens can play the role of the lens, and the retina of the eye can play the role of the screen. In this case, in order to observe stripes of equal inclination, the eyes must be accommodated in the same way as when examining very distant objects.

According to formula (122.8), the position of the 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 becomes rainbow-colored. The ability to observe an interference pattern in white light is determined by the eye's ability to distinguish shades of light of close wavelengths. Rays differing in wavelength by less than 20 A are perceived by the middle eye as having the same color. Therefore, to estimate the conditions under which interference from plates in white light can be observed, one should set equal to 20 A. It is this value that we took when evaluating the plate thickness (see (122.5)).

2. Plate of variable thickness. Take a plate in the form of a wedge with an apex angle (fig. 122.4).

Let a parallel beam of rays fall on it. Now the rays reflected from different surfaces of the plate will not be parallel. Two practically merging beams before falling on the plate (in Fig. 122.4 they are shown as one straight line indicated by a number) intersect after reflection at point Q. Two practically merging beams 1 "intersect at a point It can be shown that points Q, Q" and other similar points lie in the same plane passing through the vertex of the clia O. Ray V reflected from the lower surface of the wedge and ray 2 reflected from the upper surface will intersect at point R, which is closer to the wedge than Q. Similar rays G and 3 will intersect at point P, which is farther from the wedge surface than

The directions of propagation of waves reflected from the upper and lower surfaces of the clia do not coincide. Temporal coherence will be observed only for the parts of waves reflected from the places of the wedge, for which the thickness satisfies the condition (122.4). Let us assume that this condition is fulfilled for the entire wedge. Also, assume that the coherence radius is much larger than the wedge length. Then the reflected waves will be coherent in the entire space above the wedge, and at any distance of the screen from the wedge, an interference pattern in the form of stripes parallel to the top of the wedge O will be observed (see the last three paragraphs of § 119). This, in particular, is the case when the wedge is illuminated with the light emitted by the laser.

With limited spatial coherence, the area of ​​localization of the interference pattern (that is, the area of ​​space in which the screen can be observed on it the interference pattern) is also limited. If the screen is positioned so that it passes through the points (see screen E in Fig. 122.4), an interference pattern will appear on the screen even if the spatial coherence of the incident wave is extremely small (at the points of the screen, rays intersect, which before falling on the wedge coincided).

At a small wedge angle, the difference in the path of the rays can be calculated with a sufficient degree of accuracy by formula (122.3), taking as b the thickness of the plate at the place where the rays fall on it. Since the path difference for the rays reflected from different parts of the wedge is now not the same, the illumination of the screen will be uneven - light and dark stripes will appear on the screen (see in Fig. 122.4 the dotted curve showing the illumination of the screen E). Each of these stripes results from reflection from sections of the wedge with the same thickness, as a result of which they are called stripes of equal thickness.

When the screen is displaced from position E in the direction from the wedge or towards the wedge, the degree of spatial coherence of the incident wave begins to affect. If in the screen position indicated in fig. 122.4 through E, the distance between the incident beams 1 and 2 will become of the order of the coherence radius, the interference pattern on the E screen will not be observed. Similarly, the picture disappears at the screen position indicated by

Thus, the interference pattern resulting from the reflection of a plane wave from the wedge turns out to be localized in a certain region near the surface of the wedge, and this region is narrower, the lower the degree of spatial coherence of the incident wave. From fig. 122.4 it can be seen that as one approaches the top of the wedge, the conditions for both temporal and spatial coherence become more favorable. Therefore, the distinctness of the interference pattern decreases when moving from the top of the wedge to its base. It may happen that the pattern is observed only for the thinner part of the wedge. For the rest of the screen, uniform illumination appears.

Practically stripes of equal thickness are observed by placing a lens near the wedge and a screen behind it (Fig. 122.5). The lens can play the role of the lens, and the retina can play the role of the screen. If the screen behind the lens is located in a plane conjugate to the plane indicated in Fig. 122.4 through E (accordingly, the eye is accommodated on this plane), the picture will be the clearest. When the screen, on which the image is projected, is moved (or when the lens is moved), the picture will deteriorate and disappear altogether when the plane conjugated with the screen goes beyond the localization region of the interference pattern observed without a lens.

When viewed in white light, the stripes will be colored so that the surface of the plate or film appears to be rainbow colored. For example, thin films of oil or oil spreading on the surface of the water, as well as soap films, have such a color. The tarnishing colors that appear on the surface of steel products during their hardening are also caused by the interference from the transparent oxide film.

Let us compare the two cases of interference we have considered when reflecting from thin films. Stripes of equal inclination are obtained when a plate of constant thickness is illuminated) with diffused light, which contains rays of different directions, varies in more or less wide limits). Stripes of equal inclination at infinity are localized. Stripes of equal thickness are observed when a plate of variable thickness is illuminated (changes) by a parallel beam of light). Stripes of equal thickness are localized near the plate. In real conditions, for example, when observing rainbow colors on a soap or oil 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.

Note that interference from thin films can be observed not only in reflected light, but also in transmitted light.

Newton's rings. Newton's rings are a classic example of stripes of equal thickness. They are observed when light is reflected from a plane-parallel thick glass plate and a plano-convex lens with a large radius of curvature in contact with each other (Fig. 122.6). The role of a thin film, from the surfaces of which coherent waves are reflected, is played by the air gap between the plate and the lens (due to the large thickness of the plate and the lens due to reflections from other surfaces, interference fringes do not appear). With normal incidence of light, stripes of equal thickness have the form of concentric circles, with oblique incidence - ellipses. Let us find the radii of Newton's rings, resulting from the incidence of light along the normal to the plate. In this case, the optical path difference is equal to twice the thickness of the gap (see formula (122.2); it is assumed that in the gap). From fig. 122.6 it follows that are the radii of the dark rings. The value corresponds, i.e., the point at the point of contact between the plate and the lens. At this point, a minimum of intensity is observed, due to a change in phase on when the light wave is reflected from the plate.

Optics enlightenment. Interference due to reflection from thin films underlies the antireflection coating of optics. The passage of light through each refractive lens surface is accompanied by a reflection of approximately 4% of the incident light. In complex lenses, such reflections occur repeatedly and the total loss of luminous flux reaches a noticeable value. In addition, reflections from lens surfaces cause glare. In coated optics, to eliminate light reflection, a thin film of a substance with a refractive index different from that of the lens is applied to each free surface of the lens. The thickness of the film is selected so that the waves reflected from both of its surfaces cancel each other out. A particularly good result is achieved if the refractive index of the film is equal to the square root of the refractive index of the lens. Under this condition, the intensity of both waves reflected from the film surfaces is the same.

When illuminating a thin transparent plate or film, you can observe the interference of light waves reflected from the top and bottom surfaces of the plate (Fig. 26.4). Consider a plane-parallel plate of thickness / with a refractive index NS ) on which a plane monochromatic wave with a wavelength is incident at an angle a X. Let us assume for definiteness that a ray falls on a plate from air with a refractive index

and the plate lies on a substrate with a refractive index

Rice. 26.4

This situation occurs, for example, when there is interference in a thin plate or film surrounded by air.

Let us find the optical path difference of the interfering beams 2 and 3 between the point A and plane CD. It is this difference that determines the interference pattern, since the collecting lens (or eye) located further downstream only converges two interfering beams into one. It should be taken into account that, in accordance with experiment, the reflection from an optically denser medium at the point A leads to a phase change to X / 2(to the opposite), and reflection from an optically less dense medium at the point V does not lead to a change in the phase of the wave. Thus, the optical path difference of the interfering beams 2 and 3 is gained, equal to

From aAVO it follows that

From aACD taking into account the law of refraction - = NS we have

J sin p

AD = AC sina = 2 / 10sina = 2 / tgPsina = 2w / tgpsinp = 2rc / sin 2 p / cosp.

Then the optical path difference is

It is more convenient to analyze this formula if the angle of refraction is expressed through the angle of incidence from the law of refraction:

From the maximum condition (26.19) we have

In turn, the minimum condition (26.20) gives

(in the last formula, the numbering of integers is shifted by one to simplify the form of the formula).

According to the formulas, depending on the angle of incidence of monochromatic light, the plate may appear light or dark in reflected light. If the plate is illuminated with white light, then the conditions of maximum and minimum can be met for individual wavelengths and the plate looks colored. This effect can be observed on the walls of soap bubbles, on films of oil and oil, on the wings of insects and birds, on the surface of metals when they are quenched (discoloration).

If monochromatic light is incident on a plate of variable thickness, then the conditions for maximum and minimum are determined by the thickness /. Therefore, the plate looks covered with light and dark stripes. In this case, in the wedge, these are parallel lines, and in the air gap between the lens and the plate - rings (Newton's rings).

Directly related to interference in thin films is optical enlightenment. Calculations show that the reflection of light leads to a decrease in the intensity of the transmitted light by several percent, even at almost normal incidence of light on the lens. Considering that modern optical devices contain a sufficiently large number of lenses, mirrors, beam-splitting elements, the loss of the intensity of the light wave without the use of special measures can become significant. To reduce reflection losses, optical parts are coated with a film with a specially selected thickness / and refractive index n and. The idea of ​​reducing the intensity of reflected light from the surface of optical parts consists in interference damping of the wave reflected from the outer surface of the film by the wave reflected from the inner surface of the film (Fig. 26.5). To accomplish this, it is desirable that the amplitudes of both waves are equal and the phases differ by 180 °. The light reflectance at the boundary of the media is determined by the relative refractive index of the media. So if Rice. 26.5

light travels from air to a lens with a refractive index n y then the condition of equality of the relative refractive indices at the entrance to the film and exit from it is reduced to the relation

The thickness of the film is selected based on the condition that the additional incursion of the light phase is equal to an odd number of half-waves. In this way, it is possible to weaken the reflection of light dozens of times.