Explain the phenomenon of interference of light in thin films. Light interference in thin films. Strips of equal slope and equal thickness. Newton's rings. The practical application of interference. Examples of solving problems

It is difficult to create two coherent light sources in practice (this is achieved, in particular, using optical quantum generators - lasers). However, there is a relatively simple way to effect interference. It is about splitting one light beam, or rather - each train of a light wave, into two using reflections from mirrors, and then bringing them together at one point. At the same time, the split train interferes “with itself” (being itself coherent)! Figure 7.6 shows a schematic diagram of such an experiment. At the point ABOUTat the boundary of two media with refractive indices of "1 and n 2  the wave train splits into two parts. Using two mirrors R  and P 2  both rays are directed to a point Min which they interfere. The propagation velocities of two rays in two different media are Oi \u003d s / n and u 2 \u003d s / n 2.  At the point M  two parts of the train converge with a shift

Fig. 7.6. Passage of parts of a wave train in two media with n x  and n 2. R  and R 2 -   mirrors

in time equal to where \u003d

= OR x Mand S 2 \u003d OR 2 M -  total geometric paths of light rays from a point ABOUTto the point M  in different environments. Oscillations of vectors of electric field strength at a point Mwill be E t  cos ω (G - Si / v x)  and E 02  cos co (/ - S 2 / v 2)respectively. The square of the amplitude of the resulting oscillation at a point M  will be

Since co \u003d 2p / T (T -  period of oscillations), and and \u003d s / n,  then the expression in square brackets is Der \u003d ( 2n / cT) (S 2 n 2  - 5, l,) \u003d (2n / 0) (S 2 n 2 -- 5i «i), where I / C) is the wavelength of the light in vacuum. Product of path length S  on indicator p  refraction of the medium in which light propagates (Sn)  are called optical path length  and the difference in optical path lengths is denoted by the symbol D and is called optical difference of the wave path.  Bearing in mind that cT \u003d X 0,  can write

This expression relates the phase difference Der of the oscillations and the optical difference of the path D of the rays of the two parts of the “split train”. It is Der who determines the interference effects. Indeed, cos Der \u003d 1 corresponds to the highest intensity, i.e. Der \u003d (2l Do) D \u003d 2 L t  It follows light amplification condition for interference

where t -  any whole (m \u003d 0,  1,2, ...) number.

The greatest attenuation of light corresponds  cos Af \u003d -1, i.e. Df \u003d (2t +  1) 7g. Then (2t  + 1) l \u003d (2lDo) D, or

also with integer t = 0, 1,2,....

It is easy to see that the previously described addition of waves with a four-fold increase in intensity corresponds to the displacement of two "parts" of the split train of light waves relative to each other by an integer number of wavelengths (or, accordingly, a change in the phase difference Df by an even number l), while the total mutual cancellation waves with the equality of their intensities (“light + light” gives darkness!) is observed when two parts of the train are shifted by half the wavelength (by an odd number of half waves, i.e., respectively, with Df \u003d (2t  + 1) l and any whole t  The conclusion made determines the interference effects in all possible cases.

Fig. 7.7.

Let us consider as an example the interference of light reflected from a thin film (or from a thin plane-parallel transparent plate) with a thickness d  (Fig. 7.7). The direction of the incident beam on the film is indicated by an arrow in the figure. The splitting of the trains occurs in this case with a partial reflection of each part of the train on the top (point A)  and bottom (dot IN)  film surface. We assume that the light beam comes from the air and leaves after the point IN  also into air (a medium with a refractive index of unity), and the film material has a refractive index p  \u003e 1. Each train falling at an angle a  beam at point A  splits into two parts: one of them is reflected (beam 1 in the diagram), the other is refracted (lu fAQ).At the point IN  each train of a refracted beam is split a second time: it is partially reflected from the lower surface of the film, and partially refracted (dotted) and goes beyond it. At point C, the train again splits into two, but we will be interested only in that part (beam 2) that leaves the film at the same angle a as beam 1. Rays 1 and 2 reflected from the upper surface of the film are collected by one lens a point (not shown) on the screen or in the lens of the observer’s eye (the same lens). Being parts of the same primary train, rays 1 and 2 are coherent and can participate in interference, moreover, the amplification or attenuation of the light intensity depends on their optical path difference (or the phase oscillation difference).

The phase difference between the oscillations in waves 1 and 2 is created at the path lengths AD  (in the air) and ABC  (in film). The optical path difference is

Bearing in mind that

sin a \u003d p  sin p (law of refraction), you can get D \u003d (2dn /  cos P) (1 - sin 2 p) or D \u003d 2 dn  cos r. Due to the fact that the conditions of the problem are usually specified not by the angle of refraction p, but by the angle of incidence a, it is more convenient to represent the quantity D in the form

In determining the conditions for the maximum or minimum of the light intensity, it would be necessary to equate the quantity Д to an integer or half-integer number of wavelengths (conditions (7.6) and (7.7)). However, in addition to evaluating the optical path difference D, one should also bear in mind the possibility of “losing” (or, equivalently, “acquiring”) half the wavelength of the beam when reflected from an optically denser medium. The implementation of this feature depends on the specific task, or rather, on what environment surrounds the film. If the film is with p  \u003e 1 is surrounded by air with n \u003d  1, half wavelength loss occurs only at the point A  (see fig. 7.7). And if the film lies on the surface of the body (another medium) with a refractive index p  greater than for the film material, the loss of half the wavelength occurs at two points A to B;  but, since an entire wavelength is “incident”, this effect can be ignored - the phase conditions of the interfering waves are preserved. It can be seen that the task requires an individual approach. The main principle of its solution is to first find the optical path difference of the interfering rays, considering the possible loss of half the wavelength at different reflection points (if necessary, add or subtract it in D), and equate to a whole number of wavelengths when determining the conditions for amplification of light intensity or to a half-integer number of wavelengths (an odd number of half-waves) - when finding a minimum of illumination (attenuation due to interference). In the case of the film in the air shown in Fig. 7.7, the condition of the interference maximum has the form

Due to the fact that the refractive index depends on the wavelength (see Subsection 7.5), the conditions for amplification and attenuation of intensity for light

Fig. 7.8.

different wavelengths will be different. Therefore, the film will decompose the incident white light into the spectrum, i.e. in reflected white light a thin film is seen painted in different colors. Each of us met with examples of this, observing multi-colored soap bubbles and oil stains on the surface of the water.

Consider now an example of a thin air wedge (Fig. 7.8). A plate with a well-machined surface lies on another same plate. At a certain place between the two plates there is an object (for example, a thin wire) so that an air wedge is formed with an angle of 5. Consider a ray of light that normally falls on the plates. We assume that the divergence of the trains of light waves at the reflection and refraction points upon reflection from the surfaces of the air wedge is negligible, therefore, the interfering rays are collected at one observation point (they can be collected using an auxiliary lens just like in the previous example). Suppose that at some point A  along the length of the plates, the optical path difference D is equal to an integer t  wavelengths plus Xo / 2 (due to reflection from the optically denser medium of the bottom plate). Such a point will always be found. It turns out that at the point IN  on distance AB \u003d d  counted along the plates and equal ) ^ o/(2   tg 8) (the factor 2 arises due to the fact that the beam passes the space between the plates twice, in one and the other direction), the interference pattern will be repeated for t  ± 1 (phase conditions during wave addition are repeated). Measuring distance d  between these two points, it is easy to relate the wavelength to angle b

Fig. 7.9.

If you look at this picture from above, you can see the geometrical place of the points at which, for certain integers t  light (or dark) bands formed, horizontal and parallel to the base of the wedge (i.e., conditions of interference maxima or minima arose). Along this band, conditions (7.6) or (7.7), as well as (7.10), i.e. along it, the air gap has the same thickness. These bands are called strips of equal thickness.  Provided that the plates are carefully made, strips of equal thickness appear as parallel straight lines. If there are flaws in the plates, the nature of the strips changes noticeably, the position and shape of the flaws appear clearly. On this interference effect, in particular, a method for controlling the quality of surface treatment is based.

Figure 7.9 shows strips of equal thickness: in the middle of the air wedge, a narrow stream of warm air is created, the density of which and, accordingly, the refractive index differ from the values \u200b\u200bfor cold air. You can see the curvature of lines of constant thickness in the flow area.

If the convex lens lies on a flat transparent plate, then with a certain radius ratio R  lens curvature and wavelength X light can be observed the so-called Newton's rings.

They represent the same stripes of equal thickness in the form of concentric circles.

Consider such an interference experiment, leading to the formation of Newton's rings first in the reflected - point M  observations from above (Fig. 7.10, a),  and then in transmitted light (Fig. 7.10, b)  - point M  located below the lens L) and a transparent plate P. Define the radii g t  Newton's light and dark rings (the observed pattern K in the figures), depending on the length /. waves of light and radius R  the curvature of the lens used in the experiment.

The experimental design represents an optical system consisting of a lens A, which is flat on one side and convex on the other! small curvature lying on the glass plate P, of arbitrary thickness.

On the lens A (a plane wavefront of light from a monochromatic source falls, (length to  light waves) which as a result of interference of reflections arising in the air gap between the lens and the plate forms an image K, which can be observed above the lens - the point M  (see Fig. 7.10, a), or below it (see Fig. 7.10, b)  For convenience of observing the image in rays diverging due to non-parallelism of the reflecting planes, an auxiliary collecting lens L 2 is used (at small observation distances, its presence is not necessary). You can observe directly or record an image using an optically sensitive detector (such as a photocell).

Consider the course of two closely spaced rays 1 and 2 (Fig. 7.10, a). These rays before reaching the point of observation M  (observer’s eye in the figure) undergo multiple reflections at the propagation and refraction site “down” at the air-lens A interface, the lens-air gap of thickness d \u003d AB,  and on the "up" section, respectively. But in the formation of the interference pattern that interests us, their behavior in the air gap region is essential d = AB.  It is here that the optical path difference D of the rays 1 and 2 is formed, due to which the conditions for observing interference in the experiment with Newton's rings are created. If the reflection (rotation) of beam 1 occurs at point A, and the reflection (rotation) of beam 2 occurs at point IN  (when beam 2 is reflected at the same point as beam 1, i.e., at A,  there will be no difference in travel D, and beam 2 will simply be "equivalent" to beam 1), then we are interested in the optical difference of the course

those. twice the thickness of the air gap (with a small curvature of the lens and closely spaced rays 1 and 2 AB + VA » 2d) plus or minus half the wavelength (/./2) that is lost (or acquired) when light is reflected from an optically denser (glass refractive index l st \u003d n 2 \u003d  1.5 more air refractive index p TT \u003d P \u003d  1) environment at a point A  (change of the oscillation phase by ± l), where beam 1 is reflected from the glass plate P and returns to the air gap. Losses (acquisitions) of a half-wave by a beam 2 propagating in the glass upon reflection from the interface at a point IN,does not occur (glass-air interface and reflection from air - an optically less dense medium - here n st \u003d P  \u003e "2 \u003d / g air). On the site "up" from the point IN  to the observation point M  the reflected paths 1 "and 2" have the same optical paths (there is no optical path difference).

Fig. 7.10.

From consideration of the experimental design under the assumption that the air gap is small d (d “R  and r m) between the lens A! and plate P, i.e., setting d 2 ~  0, you can write:

this implies that for the optical path difference Δ of the rays in question, we have

Leaving for the “+” sign in the last expression (“-” will result in numbers t  of the same rings differing by one) and taking into account the conditions of the interference maximum, D \u003d tx  and the minimum D \u003d (2 + 1) l / 2, where / and \u003d 0, 1, 2, 3, are integers, we get:

For maximum (light rings)

For a minimum (dark rings)

The results can be combined by one condition

defining t  - as even for maximum (light rings) and odd for minimum (dark rings).

From the result it follows that in the center of the interference pattern, i.e. at t \u003d  0 observed in reflected light will be dark (g ttssh1  \u003d 0) a ring (more precisely, a spot).

A similar consideration can be made for the experiment in transmitted light (Fig. 7.10, b  - point M  observations below). From consideration of the enlarged fragment of the figure it is seen that the difference from the previous experiment in transmitted light is the air gap between A | and plate P passes through beam 1 three times (down, up and down again) and twice it is reflected from an optically denser medium (glass) - at points A  and IN.  In this case, beam 2 passes the air gap between the lens and the plate once (reflections and refractions of this beam at other points at the boundaries are not influenced by the observed picture and are not taken into account) and reflections from an optically denser medium do not occur. Therefore, the optical path difference of rays 1 and 2 in the case under consideration will be

or simply

since the change in the optical path difference by the wavelength X in one direction or another (or an integer number of wavelengths) does not lead to a significant change in phase relationships for interference in interfering waves (rays) - the phase difference between rays 1 and 2 is preserved in this case. Maximum and minimum conditions (D \u003d tx  and D \u003d (2t + 1) X / 2  respectively), as well as

geometric condition for radii g t  matching rings

for experience in transmitted light remain the same, so we get:

For highs

For the lows

at t \u003d  0,1,2,3, ... - that is, the conditions opposite to those considered for experience in reflected light. Redefining again t  as even and odd, we can write a generalized formula for this case in the form

where already for the odd t  we get the maximum (light ring), and for even - the minimum (dark ring). Thus, in transmitted light, compared to reflected light and dark rings are interchanged gt g t  (in the center, at t \u003d  0 it turns out a bright spot g "tsv = 0).

Fig. 7.11.

The phenomena of interference are widely used in engineering and industry. They are also used in interferometry when determining the refractive indices of substances in all three of its states - solid, liquid and gaseous. There are a large number of varieties of interferometers that differ in their purpose (one of them is the Michelson interferometer, which we previously considered when discussing the hypothesis of the world ether (see Fig. 1.39).

We illustrate the determination of the refractive index of a substance using the example of the Jamen interferometer, designed to measure the refractive indices of liquids and gases (Fig. 7.11). Two identical plane-parallel and translucent mirror plates A  and IN  installed parallel to each other. Ray of light from the source S  falls onto the surface of the plate A  at an angle a close to 45 °. As a result of reflection from the outer and inner surfaces of the plate A  two parallel beams 1 and 2 come out. Having passed through two identical glass cuvettes Ki and K2, these rays fall on the plate IN,  again reflected from both of its surfaces and collected using a lens L  at the observation point R.  At this point, they interfere, and interference fringes are examined using an eyepiece, which is not shown in the figure. If one of the cells (for example, K |) is filled with a substance with a known absolute refractive index P,and the second - by a substance whose refractive index "2 is measured, then the optical path difference between the interfering rays will be 6 \u003d (n - n 2) 1, where / is the length of the cell in the path of light. In this case, a shift of interference fringes relative to their position is observed with empty cuvettes. The offset S is proportional to the difference ("! -" 2), which allows one of the refractive indices to be determined, knowing the other. With relatively low requirements for the accuracy of measuring the position of the bands, the accuracy in determining the refractive index can reach 10 ~ * -10 -7 (i.e. 10 -4 - 10 _5%). This accuracy ensures the observation of small impurities in gases and liquids, measuring the dependence of refractive indices on temperature, pressure, humidity, etc.

There are many other designs of interferometers designed for various physical and technical measurements. As already mentioned, using a specially designed interferometer A.A. Michelson and E.V. Morley in 1881 investigated the dependence of the speed of light on the speed of motion of a source emitting it. The fact of the constancy of the speed of light established in this experiment was laid by A. Einstein at the basis of the special theory of relativity.

•   D is measured in units of length (in SI it is meters), and D
•   Generally speaking, the requirement of monochromaticity is not necessary, but in the case of a polychromatic (white) light source, the observed pattern will be an overlap of rings of different colors and make it difficult to isolate the effect of interest to us.

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 )  to which a plane monochromatic wave with a wavelength falls at an angle a X.  For definiteness, suppose that a ray hits a plate of air with a refractive index

and the plate lies on a substrate with a refractive index

Fig. 26.4

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

Find the optical path difference of the interfering rays 2 and 3 between the point A  and plane CD  It is this difference that determines the interference pattern, since the further collecting lens (or eye) only brings two interfering rays into one. It should be borne in mind that, in accordance with experiment, reflection from an optically denser medium at a point A  leads to a phase change to X / 2  (to the opposite), and the reflection from an optically less dense medium at a point IN does not lead to a change in phase of the wave. Thus, the optical path difference of the interfering rays 2 and 3, equal to

Of aAVO follows that

Of aACD   taking into account the law of refraction p  we have

  J  sin p

AD \u003d AC  sina \u003d 2/10 sina \u003d 2 / tgPsina \u003d 2w / tgpsinp \u003d 2rc / sin 2 p / cosp.

Then the optical path difference is

It is more convenient to analyze this formula if, from the law of refraction, the angle of refraction is expressed in terms of 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 light or dark. If the plate is illuminated with white light, then the maximum and minimum conditions can be met for individual wavelengths and the plate looks colored. This effect can be observed on the walls of soap bubbles, on the films of oil and oil, on the wings of insects and birds, on the surface of metals during their hardening (tint colors).

If monochromatic light falls 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. Moreover, in the wedge, these are parallel lines, and in the air gap between the lens and the plate, there are rings (Newton's rings).

Directly related to interference in thin films is enlightenment optics.  As calculations show, the reflection of light leads to a decrease in the intensity of transmitted light by several percent, even with almost normal incidence of light 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, a coating of optical parts with a film with a specially selected thickness / and refractive index is used n and.  The idea of \u200b\u200breducing the intensity of reflected light from the surface of optical parts is the interference suppression of the wave reflected from the outer surface of the film, the wave reflected from the inner surface of the film (Fig. 26.5). To achieve this, it is desirable that the amplitudes of both waves are equal, and the phases differ by 180 °. The light reflection coefficient at the boundary of the media is determined by the relative refractive index of the media. So if Fig. 26.5

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

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

The rainbow color of soap bubbles or gasoline films on water occurs as a result of interference of sunlight reflected by two surfaces of the film.

Let on a plane-parallel transparent film with a refractive index pand thick dat an angle a plane monochromatic wave with a length of (Fig. 4.8).

Fig. 4.8. Thin film light interference

The interference pattern in reflected light occurs due to the superposition of two waves reflected from the upper and lower surfaces of the film. Consider the addition of waves emerging from a point FROM. A plane wave can be represented as a beam of parallel rays. One of the rays of the beam (2) directly hits the point FROMand reflected (2 ") in it upward at an angle equal to the angle of incidence. Another beam (1) hits the point FROM  in a more complicated way: at first it is refracted at a point A  and propagates in the film, then it is reflected from its lower surface at the point 0 and finally comes out, refracted, out (1 ") at the point FROM  at an angle equal to the angle of incidence. So at the point FROM  the film throws up two parallel beams, of which one was formed due to reflection from the lower surface of the film, the second due to reflection from the upper surface of the film. (Beams resulting from multiple reflections from film surfaces are not considered due to their low intensity.)

The optical path difference acquired by beams 1 and 2 before they converge at a point FROMis equal to

Assuming the refractive index of air and taking into account the ratio

We use the law of refraction of light

Thus,

In addition to the optical travel difference , should take into account the phase change of the wave upon reflection. At the point FROM  at the air interface – film "is reflected from optically denser medium, that is, a medium with a high refractive index. At not too large angles of incidence in this case, the phase undergoes a change in . (The same phase jump occurs when a wave traveling along a string is reflected from its fixed end.) At the point 0 at the film – air interface, light is reflected from an optically less dense medium, so that a phase jump does not occur.

As a result, between the rays 1 "and 2" there is an additional phase difference, which can be taken into account if decrease or increase by half the wavelength in a vacuum.

Therefore, when the relation

it turns out maximum interference in reflected light, and in the case of

in reflected light is observed minimum.

Thus, when light falls on a gasoline film in water, depending on the angle of view and film thickness, a rainbow color of the film is observed, indicating the amplification of light waves with certain lengths l.Interference in thin films can be observed not only in reflected, but also in transmitted light.

As already noted, for the appearance of the observed interference pattern, the optical path difference of the interfering waves should not exceed the coherence length, which imposes a limitation on the film thickness.

Example.On a soap film ( n \u003d 1.3), which is in the air, a normal beam of white light falls. Determine at which smallest thickness dfilm reflected light with a wavelength μm  will be maximized as a result of interference.

From the condition of the interference maximum (4.28) we find for the film thickness the expression

(angle of incidence ). Minimum value d  It turns out at:

Light waves from two point light sources. However, we often have to deal with extended light sources in the presence of interference phenomena observed under natural conditions when the light source is a part of the sky, i.e. diffused daylight. The most common and very important case of this kind occurs when illuminating thin transparent films, when the splitting of the light wave necessary for the appearance of two coherent beams occurs due to the reflection of light by the front and back surfaces of the film.

This phenomenon, known as thin film colorseasily observed on soap bubbles, on the thinnest films of oil or oil floating on the surface of the water, etc.

Let a plane light wave fall onto a transparent plane-parallel plate, which can be considered as a parallel wave beam.

The plate reflects 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 represented by only one beam).

Figure 2. Interference in thin films.

At the entrance to the plate and at the exit from it, the second beam undergoes refraction. In addition to these two beams, the plate reflects the beams resulting from three -, five -, etc. multiple reflection from the surface of the plate. However, due to their low intensity, we will not take these beams into account. The path difference acquired by rays 1 and 2 before they converge at point C is (8) where S 1  - the length of the segment of the aircraft; S 2  - the total length of the segments AO and OS; n -refractive index of the plate.

The refractive index of the medium surrounding the plate is set equal to unity,   b  - plate thickness. The figure shows that:

;

substituting these values \u200b\u200bin expression (8) and performing simple calculations, it is easy to bring formula (9) for the path difference Δ to the form

. (9)

However, when calculating the phase difference between the oscillations in beams 1 and 2, it is necessary, in addition to the optical path difference Δ, to take into account the possibility of changing the phase of the wave at point C, where reflection occurs from the interface of an optically less dense medium. Therefore, the phase of the wave undergoes a change in π. As a result, an additional phase difference of π arises between 1 and 2. It can be taken into account by adding to Δ (or subtracting from it) half the wavelength in vacuum. As a result, we get

(10)

The intensity depends on the magnitude of the optical path difference (10). Accordingly, from conditions (5) and (6), when maxima are obtained, and when - minima of intensity ( m  is an integer).

Then the condition for maximum intensity has the form:

, (11)

and for a minimum of illumination we have

. (12)

When illuminated with light of a plane-parallel plate ( b\u003d const) the results of interference depend only on the angles of incidence on the film. The interference pattern has the form of alternating curved dark and light stripes. Each of these bands corresponds to a certain value of the angle of incidence. Therefore they are called stripes or lines of equal slope.  If the optical axis of the lens L is perpendicular to the film surface, the strips of equal inclination should have the form of concentric rings centered on the main focus of the lens. This phenomenon is used in practice for very precise control of the degree of plane parallelism of thin transparent plates; a change in the thickness of the plates by an amount of the order of 10 −8 m can already be detected by distorting the shape of the rings of equal inclination.

The interference strips on the surface of the wedge-shaped film have equal illumination at all points on the surface corresponding to the same film thicknesses. The interference bands are parallel to the edge of the wedge. They are called interference fringes of equal thickness.

Formula (10) is derived for the case of observing interference in reflected light. If interference fringes of equal slope are observed in thin plates or films in the air in the light (in transmitted light), then wave reflection will not occur and the path difference Δ will be determined by formula (9). Therefore, the optical path differences for transmitted and reflected light differ by λ / 2, i.e. the maxima of interference in reflected light correspond to the minima in transmitted light, and vice versa.

Newton's rings.

Stripes of equal thickness can be obtained by placing a plane-convex lens with a large radius of curvature R on a plane-convex plate. An air wedge also forms between them. In this case, strips of equal thickness will have the form of rings, which are called newton's rings  ; the difference in the path of the interfering rays, as in the previous case, will be determined by the formula (10).

Define the radius of the kth Newton ring: from triangle ABC we have , whence, neglecting b 2, since R \u003e\u003e b, we obtain.

Figure 3. Newton's rings

We substitute this expression into the formula (10):

If this path difference is an integer number of wavelengths (the condition for maximum interference), then for the radius of the kth light Newton’s ring in reflected light or dark in transmitted light, we have:

. (14)

Having made similar simple calculations, we obtain a formula for determining the radii of dark rings in reflected light (or light in transmitted):

ex. 1K QC

When light passes through lenses or prisms on each surface, the light flux is partially reflected. In complex optical systems, where there are many lenses and prisms, the transmitted light flux is significantly reduced, and glare also appears. Thus, it was found that up to 50% of the light entering them is reflected in the periscopes of submarines. To eliminate these defects, a technique called enlightenment optics.  The essence of this technique is that the optical surfaces are covered with thin films that create interference phenomena. The purpose of the film is to suppress reflected light.

Questions for self-control

1) What is called the interference and interference of plane waves?

2) What waves are called coherent?

3) Explain the concept of temporal and spatial coherence.

4) What constitutes interference in thin films.

5) Explain what is the multipath interference.

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T.2: Electrical and electromagnetic phenomena: a textbook. - 518 p.

T. 3. Optics. Atomic physics: textbook - 656 p.

the “film – air” borders go backward, are again reflected from the “air – film” border and only after that they go outside (Fig. 19.13). (Of course, there are rays that will experience several pairs of reflections, but their share in the overall “balance” will not be so great, because part of the light waves will go back, that is, to where they came from.)

The interference will take place between the beam (more correctly, of course, a light wave) 1 ¢ and beam 2 ¢. The geometric difference in the path of these rays (the difference in the lengths of the paths traveled) is equal to D s = 2h. Optical travel difference D \u003d pD s = 2ph.

Maximum condition

Minimum condition

. (19.9)

If in the formula (19.9) we put k  \u003d 0, we get, it is with this length that the first minimum of illumination in transmitted light occurs.

Interference in reflected light.Consider the same film from the opposite side (Fig. 19.14). In this case, we will observe interference due to the interaction of rays 1 ¢ and 2 ¢: beam 1 ¢ reflected from the air – film boundary, and the beam 2 ¢ - from the “film – air” boundary (Fig. 19.15).

Fig. 19.14 Fig. 19.15

Reader: In my opinion, here is the situation exactly the same, as with transmitted light: D s = 2h; D \u003d pD s = 2nh, and for h  max and h  min, formulas (19.8) and (19.9) are valid.

Reader: Yes.

Author: And a minimum in passing? It turns out that the light will go  into the film, and out will not work, since both the front and the back are a minimum of illumination. Where did the light energy go if the film does not absorb light?

Reader: Yes, this is really impossible. But where is the mistake?

Author: Here you need to know one experimental fact. If a light wave is reflected from the boundary of the medium more optically dense with less optically dense (glass – air), then the phase of the reflected wave is equal to the phase of the incident wave (Fig. 19.16, a) But if the reflection passes on the boundary of the medium, optically less dense, with the medium more dense (air – glass), then the phase of the wave decreases by p (Fig. 19.16, b) And that means that optical travel difference  decreases by half the wavelength, i.e. Ray 1 ¢ reflected from the outer surface of the plate (see Fig. 19.15) “loses” half a wave, and due to this, the lag of the second beam from it in the optical path difference decreases by l / 2.

Thus, the optical path difference 2 ¢ and 1 ¢ in fig. 19.15 will be equal

Then the maximum condition is written in the form

(19.10)

minimum condition

Comparing formulas (19.8) and (19.11), (19.9) and (19.10), we see that for the same value h  is achieved minimum illumination in transmitted light  and reflected maximumor maximum in passing and minimum in reflected. In other words, the light either mainly reflects or passes through, depending on the thickness of the film.

Task 19.5. Enlightenment Optics. To reduce the fraction of reflected light from optical glasses (for example, from camera lenses), a thin layer of a transparent substance with a refractive index is applied to their surface p  smaller than that of glass (the so-called optical clearing method). Estimate the thickness of the applied layer, assuming that the rays fall on the optical glass approximately normal (Fig. 19.17).

Fig. 19.17

Decision. To reduce the fraction of reflected light, it is necessary that the rays 1   and 2   (see Fig. 19.17), reflected from the outer and inner surfaces of the film, respectively, “extinguished” each other.

Note that both rays, when reflected from a more optically dense medium, lose half a wave each. Therefore, the optical path difference will be equal to D \u003d 2 nh.

The minimum condition will be

Minimum film thickness h  min corresponding k = 0,

Estimate the value h  min. Take l \u003d 500 nm, p  \u003d 1.5, then

  m \u003d 83 nm.

Note that at any film thickness, only light can be repaid 100% specific wavelength  (assuming no absorption!). Usually they “damp” the light of the middle part of the spectrum (yellow and green). The remaining colors are extinguished much weaker.

Reader: And what explains the rainbow-colored film of gasoline in a puddle?

Author: Here, too, there is interference, as in the enlightenment of optics. Since the film thickness is different in different places, in one place some colors are extinguished, and in others - others. The "outstanding" colors we see on the surface of the puddle.

STOP! Decide on your own: B6, C1 – C5, D1.

Newton's rings

Fig. 19.18

Task 19.6.  Consider the experience we have already described in detail (Fig. 19.18): on a flat glass plate lies a plano-convex lens with a radius R. From above, light with a wavelength l is incident on the lens. Light is monochromatic, i.e. the wavelength is rigidly fixed and does not change with time. When viewed from above, an interference pattern of concentric light and dark rings (Newton's rings) is visible. Moreover, as they move away from the center, the rings become narrower. Find radius required Nof the dark ring (counting from the center).

(Fig. 19.19). It is this segment that determines the geometric difference in the path of the rays 1 ¢ and 2 ¢.

Fig. 19.19

Consider D OVS: (by Pythagorean theorem),

h \u003d AC \u003d OA - OS \u003d . (1)

Let us try to simplify expression (1) a little, given that r<< R . Indeed, experiments show that if R  ~ 1 m then r  ~ 1 mm. Multiply and divide expression (1) by the conjugate expression, we obtain

We write the minimum condition for reflected light: the geometric difference in the path of the rays 1 ¢ and 2 ¢ is 2 hbut the beam 2 ¢ loses half-wave due to reflection from an optically denser medium - glass, therefore optical travel difference  it turns out half a wave less than the geometric stroke difference:

We are interested in radius Nof the dark ring. More correctly, we are talking about the radius circumferencewhich is achieved NThe second from the center is the minimum of illumination. If r N  Is the desired radius, then the minimum condition has the form:

where N = 0, 1, 2…

Remember:

. (19.12)

By the way, with N = –1 r  0 \u003d 0. This means that there will be a dark spot in the center.

Answer:

Note that, knowing r N, R  and N, you can experimentally determine the wavelength of light!

Reader: And if we were interested in the radius Nlight ring?

Fig. 19.20

Reader: Is it possible to observe Newton's rings in transmitted light?

STOP! Decide on your own: A7, B7, C6 – C9, D2, D3.

Interference from two slits (Jung's experiment)

The English scientist Thomas Jung (1773–1829) in 1807 set the following experiment. He directed a bright beam of sunlight on the screen with a small hole or narrow slit S  (Fig. 19.21). Light passing through the gap Swalked to the second screen with two narrow holes or slots S  1 and S 2 .

  Fig. 19.21

Cracks S  1 and S 2 are coherent sources, since they had a “common origin” - a gap S. Light from the crevices S  1 and S  2 fell on the remote screen, and on this screen there was an alternation of dark and light areas.

We will deal with this experience in detail. We assume that S  1 and S  2 is a long narrow cracksthat are coherent sources that emit light waves. In fig. 19.21 shows a top view.

Fig. 19.22

The area of \u200b\u200bspace in which these waves overlap is called interference field. In this area, there is an alternation of places with maximum and minimum illumination. If a screen is entered in the interference field, then an interference pattern will appear on it, which has the form of alternating light and dark stripes. In volume, it looks as shown in Fig. 19.22.

Let us be given the wavelength l, the distance between the sources d  and distance to the screen l. Will find x coordinates  min and x  max dark and light stripes. More precisely, the points corresponding to the minimum and maximum illumination. All further constructions will be carried out in the horizontal plane a, which we will “look from above” (Fig. 19.23).

Fig. 19.23

Consider the point R  on screen at a distance x  from point ABOUT  (point ABOUT  - this is the intersection of the screen with a perpendicular restored from the middle of the segment S 1 S  2). At the point R  superimposed beam S 1 Pcoming from the source S  1, and the beam S 2 Pcoming from the source S  2. The geometric difference in the path of these rays is equal to the difference of the segments S 1 P  and S 2 R. Note that since both rays propagate in the air and do not experience any reflections, the geometric path difference is equal to the optical path difference:

D \u003d   S 2 P – S 1 R.

Consider right triangles S 1 AR  and S 2 BP. By the Pythagorean theorem: ,   . Then

.

Multiply and divide the expression this expression by the conjugate expression, we get:

Given that l \u003e\u003e x  and l \u003e\u003e dsimplify the expression

Maximum Condition:

where k = 0, 1, 2, …

Minimum Condition:

, (19.14)

where k = 0, 1, 2, …

The distance between adjacent minima is called interference bandwidth.

Find the distance between ( k  + 1) kminima:

Remember: the width of the interference band does not depend on the serial number of the band and is equal to

STOP! Decide on your own: A9, A10, B8 – B10, C10.

Bilinsa

Task 19.6.  Focal length collecting lens F  \u003d \u003d 10 cm cut in half and halves spaced apart h  \u003d 0.50 mm. Find: 1) the width of the interference fringes; 2) the number of interference fringes on the screen located behind the lens at a distance D \u003d 60 cm, if there is a point source of monochromatic light with a wavelength l \u003d 500 nm in front of the lens, a distance from it a  \u003d 15 cm.

Fig. 19.24

2. First we find the distance b  from lens to image S  1 and S  2. Apply the lens formula:

Then the distance from the sources to the screen:

l \u003d D - b \u003d60 - 30 \u003d 30 cm.

3. Find the distance between the sources. To do this, consider similar triangles SO 1 O  2 and SS 1 S  2. From their likeness follows

4. Now we can well use the formula (19.15) and calculate the width of the interference band:

=   m \u003d 0.10 mm.

5. To determine how many interference fringes will appear on the screen, we will depict interference field, i.e. the region in which waves from coherent sources overlap S  1 and S  2 (Fig. 19.25).

  Fig. 19.25

As can be seen from the figure, the rays from the source S  1 cover area S 1 AA  1, and the rays from the source S  2 cover area S 2 BB  1 . The interference field — the region that is the intersection of these regions — is shown by a darker shading. The size of the interference strip on the screen is a segment AB  1, we denote its length by L.

Consider triangles SO 1 O  2 and SAB  1 . From their likeness follows

If in a section of length L  are contained N  strips, length D x  each then

Answer: D x  \u003d 0.10 mm; N = 25.

STOP! Decide on your own: D4, D5.