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

If n>n0, .

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

Minimum if or (9.3)

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

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

1. Stripes of equal slope(interference from a plane-parallel plate).

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

Beams / / and / // , reflected from the upper and lower faces of the plate, are parallel to each other, since the plate is plane-parallel. That. rays 1" and I""intersect" only at infinity, so they say that bands of equal slope are localized at infinity. For their observation, a converging lens and a screen (E) located in the focal plane are used.

Beams /" and /" / will gather in focus F lenses (in the figure, its optical axis is parallel to the rays G and /"), other rays will also come to the same point (beam 2), parallel to the beam /, - the overall intensity increases. Rays 3, inclined at a different angle, will gather in a different t. R focal plane of the lens. If the optical axis of the lens is perpendicular to the surface of the plate, then the bands of equal slope will look like concentric rings centered at the focus of the lens.

Task 1. A beam of monochromatic light falls normally on a thick glass plate covered with a very thin film. Reflected light is maximally attenuated due to interference. Determine 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 a medium that is optically denser than the medium in which the incident beam passes. Therefore, the phase of the oscillations changes twice by and the result will be the same as if there were no phase change.

Minimum condition: , where 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 a wedge (the angle a between the side faces is small), the direction of propagation of which coincides with the parallel rays / and 2. P Let us 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 t. A, which is the image of a point V.

Since the beams / / and / // are coherent, they will interfere. If the source is located far from the wedge surface and the angle a 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 point where the beam falls on it. Rays 2" and 2", formed by beam division 2, falling to another point of the wedge, are collected by a 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 on the plate that have the same thickness.

Def. 9.2. Interference fringes resulting from interference from places of the same thickness, called. stripes of equal thickness.

Since the upper and lower faces of the wedge are not parallel to each other, the rays / / and / // {2" and 2"} intersect near the plate. In this way, bands of equal thickness are localized near the surface of the wedge. If the 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 converging lens and the screen must be positioned in such a way with respect to the wedge that the image of the upper surface of the wedge is visible on the screen.

To determine the width of the interference fringes in the case of monochromatic light, we write the condition for two adjacent interference maxima ( m th and m+1-th order) according to the formula 9.2: and , where . If the distances from the edge of the wedge to the interference fringes under consideration are equal and , then , and , where is a small angle between the faces of the wedge (the refractive angle of the wedge), i.e. . In view of the smallness, the refractive angle of the wedge must also be very small, since otherwise, bands of equal thickness will be so closely spaced that they cannot be distinguished.

Task 2. A beam of monochromatic light is incident on a glass wedge normal to its face. The number of interference fringes per 1 cm is 10. Determine the refractive angle of the wedge.

Given: Solution:

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

Dark stripes will be observed in those sections of the wedge for which the difference in the path of the rays is equal to an odd number of half-waves: or, Because. , then . Let an arbitrary dark strip of the number correspond to a certain thickness of the wedge at this place , and let the dark strip of the number correspond to the thickness of the wedge at this place ,. According to the condition, 10 bands fit into , then, because , then .

Newton's rings.

Newton's rings are an example of bands 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 in contact with it with a large radius of curvature. 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, with normal incidence of light, have the form of concentric circles.

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

Both for strips of equal slope and for 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 observed in white light, a set of bands shifted relative to each other, formed by rays of different wavelengths, is obtained, and the interference pattern acquires an iridescent color. All reasoning was carried out for reflected light. Interference can be observed and in transmitted light moreover, in this case there is no loss of a half-wave - the optical path difference for transmitted and reflected light will differ by /2, t. interference maxima 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 with refractive index n and thickness 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 partly reflected from the upper surface of the film, partly refracted. The refracted beam 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 is partially reflected and refracted again, exiting into the air at an angle. Beams (1 and 2) that have emerged from the film are coherent if their optical path difference is small compared to the long coherence of the incident wave. In the event that a converging lens is placed on the paths of rays (1 and 2), then they will converge at some point D (in the focal plane of the lens). In this case, an interference pattern will arise, which is determined by the optical path difference of the interfering rays.

The optical path difference of beams 1 and 2, which appears for the beams when they pass the distance from point A to the plane CE, is equal to:

where we assume that the film is in vacuum, so the refractive index is . The occurrence of the quantity is explained by the loss of half the wavelength when light is reflected from the interface 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 us take into account that for the considered case the law of refraction:

Given the loss of half a wavelength:

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

The intensity minimum will be observed at the considered point if:

The interference phenomenon 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 (we have b), the wavelength of the incident light, the refractive index of the film substance, and the angle of incidence (). For the listed parameters, each beam tilt () corresponds to its own interference fringe. The bands resulting from the interference of rays incident on the film at the same angles are called bands of equal slope.

Examples of problem solving

EXAMPLE 1

Stripes of equal slope. The interference fringes are called stripes of equal slope, if they arise when light falls 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 emerging parallel to each other.

Fringes of equal slope are localized at infinity, therefore, to observe the interference pattern, the screen is placed in the focal plane of the converging 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 look like concentric rings centered at the focal point of the lens.

Let from air (n, ~ 1) onto a plane-parallel transparent plate with a refractive index i 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 a point V Ray 2 coherent and parallel. If a converging lens brings them to a point R, then they will interfere in the 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 permittivity (and a lower refractive index) onto the interface between two media: when a wave is reflected from an optically denser medium ( p 2> n), its phase changes by n, which is equivalent to the so-called "loss of a half-wave" (± A / 2) during reflection, i.e. optical path difference A changes to X/2 .

Therefore, the optical path difference of interfering rays is defined as

Using the law of refraction (sin 0 = "2 sind"), and also that i, = 1, AB-BC = d/ cosO" and AD - AC sin fs-2d tgO "sin Oh, 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 light and dark bands is determined by the following conditions:

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

Stripes 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 a 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 beams reflected from the top

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

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

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

Thus, a system of interference fringes appears on the screen - stripes of equal thickness, each of which occurs upon reflection from wedge sections with the same thickness. Stripes of equal thickness are localized near the surface of the wedge (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. This explains the fact that the interference 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- the number of light (or dark) interference fringes per wedge segment of length /, then the angle at the wedge apex (sina ~ a), expressed in radians, is calculated as

where d] and d2- the thickness of the wedge, which are located respectively To-Me and (k + t)-I interference fringes; Oh is the distance between these bands.

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 an 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 a flat lens surface (Fig. 22.5). Stripes of equal thickness have the form of concentric circles with the center of contact between the lens and the plate.

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

where X/2 is related to the "loss" of the half-wave on reflection from the plate.

Let's use the last two equations. Therefore, in reflected light, the radii of dark rings

value T= 0 corresponds to the minimum of the dark spot in the center of the pattern.

Similarly, we obtain that the radii of light rings are defined as

These formulas for the radii of the rings are valid only in the case of an 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 interference minima in reflected light and vice versa.

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

Typically, optics are enlightened for the middle (yellow-green) region of the visible spectrum. As a result, lenses appear magenta in reflected light due to the mixing of red and violet. 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 metal-oxide-semiconductor structural elements.

When a light wave falls on a thin transparent plate (or film), reflection occurs from both surfaces of the plate. As a result, two light waves arise, 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 throws 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 beam). When entering the plate and when leaving it, the second beam undergoes refraction. In addition to these two beams, the plate will throw upward beams resulting from 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 equal to

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

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

Making the substitution and taking into account that

it is easy to bring the formula for to the form

When calculating the phase difference between 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 and a medium that is optically denser. Therefore, the phase of the wave undergoes a change by . At the point O, the reflection occurs from the interface between the optically denser medium and the optically less dense medium, so that no phase jump occurs. As a result, an additional phase difference arises between beams 1 and 2, equal to It can be taken into account by adding to (or subtracting from) half the wavelength in vacuum. As a result, we get

So, when a plane wave falls on a 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 will be coherent and will be able to 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 of 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 to (see formula (120.9)). Therefore, the condition

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

(twice the plate thickness must be less than the coherence length).

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

Now consider the conditions for observing spatial coherence. Let's put on the path of the reflected beams, the screen E (Fig. 122.2). The rays arriving at the point P and are separated in the incident beam by a distance . If this distance does not exceed the radius of coherence rkg 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 other points of the screen. Thus, the screen will be evenly lit (in the particular case, when the screen is dark). When the beam tilt changes (i.e., when the angle changes), the screen illumination will change.

From fig. 122.1 it can be seen that the distance between the incident beams 1 and 2 is

If we accept then for it turns out and for

For a normal fall at any .

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

(122.7)

(cf. (122.5)). For an angle of incidence on 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 coherences, interference when the plate is illuminated by sunlight is observed only if the thickness of the plate does not exceed a few hundredths of a millimeter. When illuminated with light with a higher degree of coherence, interference is also observed upon reflection 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). The illumination at this point depends on the value of the quantity (122.3). At , maxima are obtained, at - intensity minima ( - an integer). The intensity maximum condition has the form

Let a thin plane-parallel plate be illuminated by scattered monochromatic light (see Fig. 122.3). Let us place a lens parallel to the plate, in the focal plane of which we place the screen. Scattered light contains rays of various directions.

Rays parallel to the plane of the figure 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 traveling in other planes, but falling on the plate at the same angle, will be collected by the lens at other points that are 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 along a circle centered at O. Similarly, rays falling at a different angle Ф "will create a set on the screen in the same way (but differently, since D is different) As a result, a system of alternating light and dark circular stripes with a common center at point O will appear on the screen. 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 bands of equal slope will be different.

Each point of the interference pattern is caused by rays that form a parallel beam before passing through the lens. Therefore, when observing fringes of equal inclination, the screen must be located in the focal plane of the lens, i.e., as it is located to obtain an image of infinitely distant objects on it. In accordance with this, the bands of different slopes are said to be localized at infinity. The role of the lens can be played by the lens, and the screen can be played by the retina. In this case, to observe bands of equal inclination, the eye must be accommodated in the same way as when viewing very distant objects.

According to formula (122.8), the position of the maxima depends on the wavelength. Therefore, in white light, a set of bands shifted relative to each other, formed by rays of different colors, is obtained, and the interference pattern acquires a rainbow color. The possibility of observing an interference pattern in white light is determined by the ability of the eye to distinguish shades of light of close wavelengths. Rays that differ in wavelength by less than 20 A are perceived by the average eye as having the same color. Therefore, to estimate the conditions under which interference from plates in white light can be observed, it should be set equal to 20 A. It is this value that we took when estimating the thickness of the plate (see (122.5)).

2. Plate of variable thickness. Take a plate in the form of a wedge with an angle at the top (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. Before falling on the plate, two practically merging rays (in Fig. 122.4 they are shown as one straight line, denoted by the number ) intersect after reflection at the point Q. Two practically merging rays 1 "intersect at the point It can be shown that the points Q, Q" and other points similar to them lie in the same plane passing through the top of the wedge O. The ray V reflected from the lower surface of the wedge and the ray 2 reflected from the upper surface will intersect at the point R, located closer to the wedge than Q. The analogous rays Г and 3 will intersect at the point Р, further from the surface of the wedge than

The directions of propagation of the waves reflected from the upper and lower surfaces of the klia do not coincide. Temporal coherence will be observed only for parts of the waves reflected from the places of the wedge, for which the thickness satisfies the condition (122.4). Let us assume that this condition is satisfied 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 will be observed in the form of fringes parallel to the top of the wedge O (see the last three paragraphs of § 119). This is the case in particular when the wedge is illuminated with light emitted by a laser.

With limited spatial coherence, the region of localization of the interference pattern (i.e., the region of space in which the screen can be placed on it to observe the interference pattern) is also limited. If you arrange the screen 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 (beams intersect at the points of the screen, which before falling on the wedge matched).

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 point 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 the dotted curve in Fig. 122.4 showing the illumination of the screen E). Each of these bands arises as a result of reflection from sections of the wedge with the same thickness, as a result of which they are called bands of equal thickness.

When the screen is displaced from position E in the direction away 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 screen E 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 the narrower, the lower the degree of spatial coherence of the incident wave. From fig. 122.4 it can be seen that as we approach 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 there is a uniform illumination.

In practice, stripes of equal thickness are observed by placing a lens near the wedge and behind it a screen (Fig. 122.5). The role of the lens can be played by the lens, and the role of the screen can be played by the retina. If the screen behind the lens is located in a plane conjugated with the plane indicated in Fig. 122.4 through E (correspondingly, the eye is accommodated on this plane), the picture will be the most clear. When the screen on which the image is projected is moved (or when the lens is moved), the picture will worsen and disappear completely when the plane associated with the screen goes beyond the localization area of ​​the interference pattern observed without the lens.

When observed in white light, the bands will be colored, so that the surface of the plate or film appears to have an iridescent color. For example, thin films of oil or oil spread on the surface of water, as well as soap films, have such a color. The tint colors that appear on the surface of steel products during their hardening are also due to interference from a film of transparent oxides.

Let us compare the two considered cases of interference upon reflection from thin films. Stripes of equal slope are obtained by illuminating a plate of constant thickness - diffused light, which contains rays of different directions, varies more or less widely). Bands of equal slope at infinity are localized. Stripes of equal thickness are observed when a plate of variable thickness is illuminated (it changes) with a parallel beam of light). Bands 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 but also in transmitted light.

Newton's rings. A classic example of bands of equal thickness are Newton's rings. 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 that are 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 lens, interference fringes do not appear due to reflections from other surfaces). 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 where the plate and the lens touch. At this point, a minimum of intensity is observed, due to the change in phase on when the light wave is reflected from the plate.

Illumination of optics. Interference upon reflection from thin films is the basis of optics antireflection. The passage of light through each refractive surface of the lens is accompanied by the reflection of approximately 4% of the incident light. In complex lenses, such reflections occur many times, and the total loss of the light flux reaches a noticeable value. In addition, reflections from lens surfaces lead to flare. 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 chosen 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, one can observe the interference of light waves reflected from the upper and lower surfaces of the plate (Fig. 26.4). Consider a plane-parallel plate of thickness / with a refractive index P ) on which a plane monochromatic wave is incident at an angle a with a wavelength x. Assume for definiteness that a beam is incident on a plate from air with a refractive index

and the plate lies on a substrate with a refractive index

Rice. 26.4

Such a situation occurs, for example, in the case of interference in a thin plate or film surrounded by air.

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 converging lens (or eye) located further away only reduces the two interfering beams into one. In this case, it should be taken into account that, in accordance with experience, reflection from an optically denser medium at a point A leads to a phase change X/2(to the opposite), and reflection from an optically less dense medium at a point V does not change the phase of the wave. Thus, the optical path difference of the interfering beams 2 and 3, equal to

From aABO it follows that

From aACD taking into account the law of refraction-= P 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 equal to

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

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 in reflected light may look bright or dark. If the plate is illuminated with white light, then the maximum and minimum conditions can be satisfied 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 during their quenching (tint colors).

If monochromatic light is incident on a plate of variable thickness, then the maximum and minimum conditions are determined by the thickness /. Therefore, the plate looks covered with light and dark stripes. At the same time, 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 illumination of the optics. Calculations show that the reflection of light leads to a decrease in the intensity of the transmitted light by several percent, even with almost normal light incidence on the lens. Given that modern optical devices contain a sufficiently large number of lenses, mirrors, beam-splitting elements, the loss of light wave intensity 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 p i. The idea of ​​reducing the intensity of reflected light from the surface of optical parts is 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 coefficient of reflection of light at the boundary of the media is determined by the relative refractive index of the media. So if Rice. 26.5

light passes from air into a lens with a refractive index p y then the condition of equality of the relative refractive indices at the entrance to the film and exit from it reduces to the relation

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