Interference fringes of equal inclination. When a thin film is illuminated, a superposition of waves from the same source occurs, reflected from the front and rear surfaces of the film. This may cause light interference. If the light is white, then the interference fringes are colored. Interference in films can be observed on the walls of soap bubbles, on thin films of oil or petroleum floating on the surface of water, on films that appear on the surface of metals or mirrors.
Let us first consider a plane-parallel plate of thickness with a refractive index (Fig. 2.11). Let a plane light wave fall on the plate, which can be considered as a parallel beam of rays. The plate throws upward two parallel beams of light, one of which was formed due to reflection from the upper surface of the plate, the second - due to reflection from the lower surface. Each of these beams is shown in Fig. 2.11 with only one beam.
When entering and exiting the plate, beam 2 undergoes refraction. In addition to two beams and , the plate throws upward beams resulting from three-, five-, etc. multiple reflection from plate surfaces. However, due to their low intensity, they can be ignored.
Let us consider the interference of rays reflected from the plate. Since a plane wave falls on the plate, the front of this wave is a plane perpendicular to rays 1 and 2. In Fig. 2.11 straight line BC represents a section of the wave front by the plane of the drawing. The optical path difference acquired by rays 1 and 2 before they converge at point C will be
| , | (2.13) |
where is the length of the segment BC, and is the total length of the segments AO and OS. The refractive index of the medium surrounding the plate is assumed to be equal to unity. From Fig. 2.11 it is clear that 
, . Substituting these expressions into (2.13) gives . Let's use the law of light refraction: 
; and take into account that , then for the path difference we obtain the following expression: 
.
When calculating the phase difference between oscillations in the rays and it is necessary, in addition to the optical path difference D, to take into account the possibility of a phase change upon reflection at point C. At point C, the wave is reflected from the interface between an optically less dense medium and an optically more dense medium. Therefore, the phase of the wave undergoes a change by p. At a point, reflection occurs from the interface between an optically denser medium and an optically less dense medium, and a phase jump does not occur in this case. Qualitatively, this can be imagined as follows. If the thickness of the plate tends to zero, then the formula we obtained for the optical path difference gives . Therefore, when the rays are superimposed, the oscillations should increase. But this is impossible, since an infinitely thin plate cannot influence the propagation of light at all. Therefore, the waves reflected from the front and back surfaces of the plate must cancel each other out during interference. Their phases must be opposite, that is, the optical path difference D at d→0 should tend to . Therefore, you need to add or subtract to the previous expression for D, where λ 0 is the wavelength in vacuum. The result is:
 .
| (2.14) |
So, when a plane wave falls on a plate, two reflected waves are formed, the path difference of which is determined by formula (2.14). These waves can interfere if the optical path difference does not exceed the coherence length. The last requirement for solar radiation leads to the fact that interference when illuminating the plate is observed only if the thickness of the plate does not exceed a few hundredths of a millimeter.
In practice, interference from a plane-parallel plate is observed by placing a lens in the path of the reflected beams, which collects the beams at one of the points of the screen located in the focal plane of the lens. The illumination at this point depends on the optical path difference. At , we get maxima, and at , we get minima of intensity. Therefore, the condition for intensity maxima has the form:
 ,
| (2.15) |
and the minimums:
 .
| (2.16) |
These relations are obtained for reflected light.
Let a thin plane-parallel plate be illuminated with scattered monochromatic light. Let us place a lens parallel to the plate, in the focal plane of which we place the screen (Fig. 2.12). Scattered light contains rays from a wide variety of directions. Rays parallel to the plane of the pattern and incident on the plate at an angle , after reflection from both surfaces of the plate, will be collected by the lens at a point and create illumination at this point, determined by the value of the optical path difference. Rays coming in other planes, but incident on the plastic at the same angle, will be collected by the lens at other points located at the same distance from the center of the screen as the point . The illumination at all these points will be the same. Thus, rays incident on the plate at the same angle will create on the screen a set of equally illuminated points located in a circle with the center at point O. Similarly, rays incident at a different angle will create on the screen a collection of equally illuminated points located in a circle of a different radius . But the illumination of these points will be different, since they correspond to a different optical path difference.
As a result, a set of alternating dark and light circular stripes with a common center at point O will appear on the screen. Each stripe is formed by rays incident on the plate at the same angle. Therefore, the resulting interference fringes in this case are called fringes of equal inclination.
According to (2.15), the position of the intensity maxima depends on the wavelength, therefore, in white light, a set of stripes shifted relative to each other, formed by rays of different colors, is obtained, and the interference pattern will acquire a rainbow color.
To observe fringes of equal inclination, the screen must be located in the focal plane of the lens, just as it is positioned to obtain objects at infinity. Therefore, they say that bands of equal inclination are localized at infinity. The role of the lens can be played by the lens of the eye, and the role of the screen can be played by the retina.
Interference fringes of equal thickness. Let us now take a wedge-shaped plate. Let a parallel beam of rays fall on it (Fig. 2.13). But now the rays, reflected from different surfaces of the plate, will not be parallel. 
Two almost merging beams before falling on the plate after reflection from the upper and lower surfaces of the wedge intersect at point . After reflection, two practically merging rays intersect at point . It can be shown that the points and lie in the same plane passing through the vertex of the wedge ABOUT.
If you position the screen E so that it passes through the points and, an interference pattern will appear on the screen. At a small angle of the wedge, the difference in the path of the rays reflected from its upper and lower surfaces can be calculated with a sufficient degree of accuracy using the formula 
obtained for a plane-parallel plate, taking as the thickness of the wedge at the point where the rays fall on it. Since the difference in the path of the rays reflected from different parts of the wedge is now unequal, the illumination will be uneven - light and dark stripes will appear on the screen. Each of these stripes arises as a result of reflection from sections of the wedge with the same thickness, as a result of which they are called stripes of equal thickness.
Thus, the interference pattern resulting from the reflection of a plane wave from a wedge turns out to be localized in a certain region near the surface of the wedge. As you move away from the top of the wedge, the optical path difference increases, and the interference pattern becomes less and less distinct.
 Rice. 2.14 |
When observed in white light, the stripes will be colored, so that the surface of the plate will have a rainbow color. In real conditions, when observing, for example, rainbow colors on a soap film, both the angle of incidence of the rays and the thickness of the film change. In this case, bands of a mixed type are observed.
Stripes of equal thickness can easily be observed on a flat wire frame that has been dipped in a soap solution. The soap film that covers it is covered with horizontal interference fringes, resulting from the interference of waves reflected from different surfaces of the film (Fig. 2.14). Over time, the soap solution drains and the interference fringes move down.
If you follow the behavior of a spherical soap bubble, you will easily find that its surface is covered with colored rings, slowly sliding towards its base. The displacement of the rings indicates a gradual thinning of the walls of the bubble.
Newton's rings
A classic example of strips of equal thickness are Newton's rings. They are observed when light is reflected from a plane-parallel glass plate and a plane-convex lens with a large radius of curvature in contact with each other (Fig. 2.15). The role of a thin film, from the surface of which waves are reflected, is played by the air gap between the plate and the lens (due to the large thickness of the plate and lens, interference fringes do not arise due to reflections from other surfaces). With normal incidence of light, stripes of equal thickness look like circles; with inclined light, they look like ellipses.
Let's find the radii of Newton's rings obtained when light is incident normally to the plate. In this case and . From Fig. 2.15 it is clear that , where is the radius of curvature of the lens, is the radius of the circle, all points of which correspond to the same gap. The value can be neglected, then . To take into account the change in phase by p that occurs during reflection from the plate, you need to add to the path difference: , that is, at the point of contact between the plate and the lens, a minimum intensity is observed due to the change in phase by p when the light wave is reflected from the plate.
 Rice. 2.16 |
In Fig. Figure 2.16 shows a view of Newton's interference rings in red and green light. Since the wavelength of red light is longer than that of green light, the radii of the rings in red light are larger than the radii of rings with the same number in green light.
If, in Newton’s installation, the lens is moved upward parallel to itself, then due to the increase in the thickness of the air gap, each circle corresponding to a constant path difference will be contracted towards the center of the picture. Having reached the center, the interference ring turns into a circle, disappearing as the lens moves further. Thus, the center of the picture will alternately become light and dark. At the same time, new interference rings will appear at the periphery of the field of view and move towards the center until each of them disappears in the center of the picture. As the lens moves continuously upward, the rings of the lowest orders of interference disappear and rings of higher orders appear.
Example
Optics coating
Coating of optics is done to reduce the reflectance of the surfaces of optical parts by applying one or more non-absorbing films to them. Without antireflective films, light reflection losses can be very large. In systems with a large number of surfaces, such as complex lenses, light loss can reach 70% or more, which degrades the quality of images generated by such optical systems. This can be eliminated by clearing the optics, which is one of the most important applications of interference in thin films.
When light is reflected from the front and rear surfaces of a film deposited on an optical part, the reflected light will produce a minimum intensity as a result of interference, and therefore, the transmitted light will have a maximum intensity for that wavelength. At normal incidence of light, the effect will be maximum if the thickness of the thin film is equal to an odd number of quarters of the wavelength of light in the film material. Indeed, in this case, the loss of half the wavelength upon reflection does not occur, since on both the upper and lower surfaces of the film the wave is reflected from the interface between an optically less dense and an optically more dense medium. Therefore, the condition for maximum intensity will take the form 
. From here we get 
.
By changing the thickness of the antireflection film, you can shift the minimum reflection to different parts of the spectrum.
In practice, it is difficult to create two coherent light sources (this is achieved, in particular, by using optical quantum generators - lasers). However, there is a relatively simple way to perform interference. We are talking about splitting one light ray, or more precisely, each light wave train, into two using reflections from mirrors, and then bringing them together at one point. In this case, the split train interferes “with itself” (being coherent with itself)! Figure 7.6 shows a schematic diagram of such an experiment. At the point ABOUT at the boundary of two media with refractive indices “1 and n 2 the wave train splits into two parts. Using two mirrors R And R 2 both rays are directed to a point M, in which they interfere. The propagation speeds of two rays in two different media are equal to Oi = s/p and and 2 = s/p 2. At the point M two parts of the train will converge with a shear
Rice. 7.6. Passage of parts of a wave train in two media with p x And n 2. R And R 2 - mirrors
in time equal to where =
= OR x M And S 2 = OR 2 M - total geometric paths of light rays from a point ABOUT to the point M in different environments. Oscillations of electric field strength vectors at a point M will E t cos co (G - Si/v x) And E 02 cos co(/ - S 2 /v 2) respectively. Squared amplitude of the resulting oscillation at a point M will
Since co = 2p/T(T - period of oscillation), and u = s/n, then the expression in square brackets is equal to Der = ( 2n/cT)(S 2 n 2- 5, l,) = (2n/ 0)(S 2 n 2 -- 5i«i), where / H) is the wavelength of light in a vacuum. Product of path length S per indicator P refraction of the medium in which light travels (Sn), called optical path length, and the difference in optical path lengths is denoted by the symbol D and is called optical difference in wave path. Bearing in mind that сТ=Х 0, can be written down
This expression relates the phase difference D of oscillations and the optical path difference D of the rays of the two parts of the “split train”. It is Der who determines the interference effects. Indeed, the highest intensity corresponds to cos Der = 1, i.e. Der = (2lDo)D = = 2 l T. It follows from this condition for amplification of light during interference
Where T - any whole (t = 0, 1,2,...) number.
The greatest attenuation of light corresponds to cos Af = -1, i.e. Df = (2t + 1)7g. Then (2t+ 1)l= (2lDo)D, or
also for integers T = 0, 1,2,....
It is easy to see that the previously described addition of waves with a fourfold increase in intensity corresponds to a 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 Δf by an even number n), while complete mutual cancellation waves with equal intensities (“light + light” gives darkness!) is observed when two parts of the train are displaced by half the wavelength (by an odd number of half-waves, i.e., respectively, with Df = (2t+ 1)l and any whole T. The conclusion drawn determines the interference effects in all possible cases.
Rice. 7.7.
Let us consider, as an example, the interference of light when reflected from a thin film (or from a thin plane-parallel transparent plate) of thickness d(Fig. 7.7). The direction of the beam incident on the film is indicated by an arrow in the figure. The splitting of the trains occurs in this case with partial reflection of each part of the train on the top (point A) and lower (point IN) film surface. We will assume that the light beam comes from the air and leaves after the point IN also into air (a medium with a refractive index equal to one), and the film material has a refractive index P> 1. Each train falling at an angle A beam at a point A splits into two parts: one of them is reflected (ray 1 in the diagram), the other is refracted (ray hAV). At the point IN each train of the refracted ray is split a second time: it is partially reflected from the lower surface of the film, and partially refracted (dotted line) and goes beyond its limits. At point C, the train again splits into two, but we will be interested only in that part of it (ray 2) that exits the film at the same angle a as ray 1. Rays 1 and 2 reflected from the upper surface of the film are collected by the lens into one a point (not shown in the figure) on the screen or in the lens of the observer’s eye (the same lens). Being parts of the same primary train, beams 1 and 2 are coherent and can participate in interference, and the increase or decrease in light intensity depends on their optical path difference (or oscillation phase difference).
The phase difference between oscillations in waves 1 and 2 is created at the path lengths AD(in the air) and ABC(on film). The optical path difference is then
Bearing in mind that
sin a = P sin р (law of refraction), you can get D = (2dn/ cos P)(1 - sin 2 p) or D = 2 dn cos p. 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 value D in the form
When determining the conditions for maximum or minimum light intensity, it would be necessary to equate the value of D to an integer or half-integer number of wavelengths (conditions (7.6) and (7.7)). However, in addition to assessing the optical path difference D, one should also keep in mind the possibility of “loss” (or, what is the same, “gain”) of half the wavelength of the beam when reflected from an optically denser medium. The implementation of this feature depends on the specific task, more precisely on the environment surrounding the film. If the film is with P> 1 surrounded by air with n = 1, the loss of half the wavelength 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 film material, the loss of half the wavelength occurs at two points A to B; but, since in this case a whole wavelength “runs up”, this effect can be ignored - the phase conditions of the interfering waves are preserved. It is clear that the task requires an individual approach. The basic 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 points of reflection (if necessary, add or subtract it in D), and equate it to an integer number of wavelengths when determining the conditions for increasing the light intensity or to a half-integer number of wavelengths (an odd number of half-waves) - when finding the minimum illumination (attenuation due to interference). In the case of a film in the air, shown in Fig. 7.7, the condition for 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 increasing and decreasing the intensity of light
Rice. 7.8.
different wavelengths will be different. Therefore, the film will decompose the incident white light into a spectrum, i.e. In reflected white light, the thin film appears colored in different colors. Each of us has encountered examples of this more than once, observing multi-colored soap bubbles and oil stains on the surface of the water.
Let us now consider an example of a thin air wedge (Fig. 7.8). A plate with a well-treated surface lies on another similar plate. At a certain place between two plates there is an object (for example, a thin wire) so that an air wedge with an angle of 5 is formed. Consider a ray of light falling normally on the plates. We will assume that the divergence of trains of light waves at the points of reflection and refraction when reflected from the surfaces of the air wedge is negligible, so the interfering rays are collected at one observation point (they, as in the previous example, can be collected using an auxiliary lens). Let us assume 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 lower plate). There will always be such a point. It turns out that at the point IN on distance AB = d, measured 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 direction and the other), the interference pattern will be repeated for T± 1 (phase conditions will be repeated when waves are added). Measuring distance d between these two points, it is easy to relate the wavelength to the angle b
Rice. 7.9.
If you look at this picture from above, you can see the locus of points at which, for certain integers T light (or dark) stripes formed, horizontal and parallel to the base of the wedge (i.e., conditions of interference maxima or minima arose). Along this strip, conditions (7.6) or (7.7), as well as (7.10), are satisfied, i.e. along it the air gap has the same thickness. These stripes are called strips of equal thickness. Provided that the plates are carefully manufactured, strips of equal thickness appear as parallel straight lines. If there are flaws in the plates, the nature of the stripes changes noticeably, the position and shape of the flaws appear clearly. In particular, the method for controlling the quality of surface treatment is based on this interference effect.
Figure 7.9 shows stripes of equal thickness: in the middle of the air wedge, a narrow flow of warm air is created, the density of which and, accordingly, the refractive index differ from the values for cold air. One can see the curvature of lines of constant thickness in the flow region.
If a convex lens lies on a flat transparent plate, then at a certain radius ratio R lens curvature and wavelength X of light, one can observe the so-called Newton's rings.
They are the same strips of equal thickness in the form of concentric circles.
Let us consider such an interference experiment, leading to the formation of Newton's rings first in the reflected - point M observation from above (Fig. 7.10, A), and then in transmitted light (Fig. 7.10, b)- dot M located below under the lens L) and the transparent plate P. Let us determine the radii g t light and dark Newton's rings (observed pattern K in the figures) depending on the length /. waves of light and radius R curvature of the lens used in the experiment.
The experimental diagram represents an optical system consisting of a lens L that is flat on one side and convex on the other! small curvature lying on a glass plate P, of arbitrary thickness.
A plane wave front of light from a monochromatic source falls on the lens L (length To waves of light) which, as a result of the interference of reflections arising in the air gap between the lens and the plate, forms an image K, which can be observed from above the lens - point M(see Fig. 7.10, a), or below it (see Fig. 7.10, b). For the convenience of observing the image in rays diverging due to the non-parallelism of the reflecting planes, an auxiliary converging lens L 2 is used (at short observation distances its presence is not necessary). You can observe directly or record the image using an optically sensitive detector (for example, a photocell).
Let's consider the course of two closely spaced rays 1 and 2 (Fig. 7.10, a). These rays before hitting the observation point M(the eye of the observer in the figure) experience multiple reflections in the section of propagation and refraction “down” at the interfaces between the air and the lens L, the lens and the air gap of thickness d = AB, and in the “up” section, respectively. But in the formation of the interference pattern we are interested in, their behavior in the air gap region is essential d = AB. It is here that the optical path difference D of 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 ray 1 occurs at point A, and the reflection (rotation) of ray 2 occurs at point IN(when beam 2 is reflected at the same point as beam 1, i.e. at the point A, there will be no path difference D, and beam 2 will simply be “equivalent” to beam 1), then the optical path difference we are interested in is
those. double the thickness of the air gap (with low lens curvature and closely spaced beams 1 and 2 AB + BA » 2d) plus or minus half the wavelength (/./2), which is lost (or gained) when light is reflected from an optically denser material (refractive index of glass l st = n 2 = 1.5 greater than the refractive index of air p tt = P= 1) environment at a point A(change in oscillation phase by ±l), where beam 1 is reflected from glass plate P and returns to the air gap. Loss (gain) of a half-wave by beam 2 propagating in glass when reflected from the interface at a point IN, does not occur (the glass-air interface and reflection from air - an optically less dense medium - here p st = P> “2 = /g air). On the section “up” from the point IN to the observation point M The reflected rays 1" and 2" have the same optical paths (there is no optical path difference).
Rice. 7.10.
From consideration of the experimental design under the assumption of a small air gap d(d « R and r m) between lens L! and plate P, i.e., assuming d 2 ~ 0, you can write:
it follows from here that for the optical path difference D of the rays under consideration we have 
Leaving the “+” sign in the last expression (“-” will result in the numbers T the same rings, differing by one) and taking into account the conditions of the interference maximum D = TX and minimum D = (2m+1) l/2, where fm = O, 1, 2, 3, integers, we get:
For maximum (light rings)
For minimum (dark rings)
The results obtained can be combined by one condition
having determined T- as even for the maximum (light rings) and odd for the minimum (dark rings).
From the obtained result it follows that in the center of the interference pattern, i.e. at t = 0 observed in reflected light will be dark (g ttsh1= 0) ring (or rather a spot).
A similar consideration can be carried out for experiments in transmitted light (Fig. 7.10, b- dot M observations below). From an examination of an enlarged fragment of the figure, it is clear that, unlike the previous experiment in transmitted light, the air gap between L | and plate P is passed by beam 1 three times (down, up and down again) and it is reflected twice from an optically denser medium (glass) - at points A And IN. In this case, beam 2 passes through the air gap between the lens and the plate once (reflections and refractions of this beam at other points on the boundaries of the beam do not affect the observed picture and are not taken into account) and reflections from an optically denser medium do not occur. Therefore, the optical difference in the path of rays 1 and 2 in the case under consideration will be
or simply
since the change in optical path difference per wavelength X in one direction or another (or by an integer number of wavelengths) does not lead to a significant change in the phase relationships in the interfering waves (rays) for interference - the phase difference between beams 1 and 2 is preserved in this case. Maximum and minimum conditions (D = TX and D = (2t + 1) X/2 respectively), and also
geometric condition for radii g t corresponding rings
for the experiment in transmitted light remain the same, so we get:
For the highs 
For the lows 
at t = 0,1,2,3,... - i.e. conditions opposite to those considered for the experiment in reflected light. Overriding again T as even and odd, we can write a generalized formula for this case in the form
where already for odd T we get the maximum (light ring), and for even numbers we get the minimum (dark ring). Thus, in transmitted light, compared to reflected light, the light and dark rings change places gt g t(in the center, at t = 0 turns out to be a light spot g" tsv = 0).
Rice. 7.11.
Interference phenomena are widely used in technology and industry. They are also used in interferometry to determine the refractive indices of substances in all three of its states - solid, liquid and gaseous. There are a large number of varieties of interferometers, differing 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)).
Let us illustrate the determination of the refractive index of a substance using the example of the Jamyon interferometer, designed for measuring 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. Beam of light from a source S falls on 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 rays 1 and 2 emanate. Having passed through two identical glass cells Ki and K2, these rays hit the plate IN, are again reflected from both its surfaces and collected using a lens L at the observation point R. At this point they interfere, and the interference fringes are viewed using an eyepiece, which is not shown in the figure. If one of the cuvettes (for example K|) is filled with a substance with a known absolute refractive index P, and the second - a substance whose refractive index "2 is measured, then the optical path difference between the interfering rays will be 6 = (n - n 2)1, where / is the length of the cuvettes in the light path. In this case, a shift of the interference fringes is observed relative to their position when the cuvettes are empty. The displacement S is proportional to the difference (“! - “ 2), which makes it possible to determine one of the refractive indices knowing the other. With relatively low requirements for the accuracy of measuring the position of the stripes, 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, measurement of the dependence of refractive indices on temperature, pressure, humidity, etc.
There are many other interferometer designs available for a variety of physical and technical measurements. As already mentioned, with the help of a specially designed interferometer A.A. Michelson and E.V. Morley in 1881 studied the dependence of the speed of light on the speed of movement of the source emitting it. The fact of the constancy of the speed of light, established in this experiment, was used by A. Einstein as the basis for the special theory of relativity.
- D is measured in units of length (in SI these are meters), and D
- Generally speaking, the requirement of monochromaticity is not mandatory, but in the case of a polychromatic (white) light source, the observed pattern will be an overlay of rings of different colors and will make it difficult to isolate the effect of interest to us.
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.
Stripes of equal inclination are localized at infinity, therefore, to observe the interference pattern, the screen is placed in the focal plane of a collecting lens (as for obtaining images of objects at infinity) (Fig. 22.3).
Rice. 22.3.
The radial symmetry of the lens leads to the fact that the interference pattern on the screen will have the form of concentric rings centered at the focal point of the lens.
Let from air (i, ~ 1) onto a plane-parallel transparent plate with refractive index i 2 and thickness d a plane monochromatic light wave with wavelength is incident at an angle O X(Fig. 22.3).
At the point A light beam S.A. partly reflected and partly refracted.
Reflected beam 1 and reflected at the point IN Ray 2 coherent and parallel. If you bring them to a point with a collecting lens 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 dielectric constant (and lower refractive index) to the interface between two media: when the wave is reflected from an optically denser medium ( n 2> i,) its phase changes by l, which is equivalent to the so-called “half-wave loss” (±A/2) upon reflection, i.e. the optical path difference A changes to X/2.
Therefore, the optical path difference of the interfering rays is defined as
Using the law of refraction (sin 0 = "2 sind"), and also 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 difference in the wave path A 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 stripes is determined by the following conditions:
So for the data X, d And n 2 Each inclination of 0 rays relative to the plate corresponds to its own interference fringe.
Stripes of equal thickness. Let a flat 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 rays reflected from the top
on the thickness of the wedge at a given point (d And d" for rays 1 And 2 respectively).
Rice. 22.4. Observation of stripes on the equal and lower surfaces of the wedge depends
Coherent pairs of rays (G And G, 2 And 2") intersect near the surface of the wedge (points O and O, respectively") 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 - strips of equal thickness, each of which occurs when reflected from sections of the wedge 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 almost normally on a transparent wedge, 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 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 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- 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 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 is incident normally on the flat surface of a lens (Fig. 22.5). Strips of equal thickness have the form of concentric circles with the center of contact of the lens with the plate.
We obtain the condition for the formation of dark rings. They arise where the optical path difference D of 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 a half-wave upon reflection from the plate.
We use both last equations. Therefore, in reflected light the radii of the dark rings are
Meaning 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 defined 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.
Enlightening optics. 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 scattered light occur, which reduces the efficiency of optical systems. In prismatic binoculars, for example, the total loss of light flux reaches -50%, but at the boundaries of media it is possible to create conditions when the intensity of light passing through the optical system is maximum. For example, thin transparent films are applied to the surface of lenses. dielectric thickness d with refractive index p ъ (Fig. 22.6). At d - NX/4 (N- odd number) interference of rays G And 2, reflected from the top and bottom surfaces of the film will give a minimum intensity of reflected light.
Typically, optics are cleared for the middle (yellow-green) region of the visible spectrum. As a result, in reflected light, lenses appear purple 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 antireflective protective coatings in optoelectronics based on elements of the metal-oxide-semiconductor structure.
Interference of light- this is the spatial redistribution of the energy of light radiation when two or more coherent light beams are superimposed. It is characterized by the formation of a time-constant interference pattern, i.e., regular alternation, in the space of beam superposition, of areas of increased and decreased light intensity.
Coherence(from lat. Cohaerens - in connection) means the mutual consistency of the time course of light oscillations at different points in space, which determines their ability to interfere, i.e., an increase in oscillations at some points in space and a weakening of oscillations at others as a result of the superposition of two or more waves arriving to these points.
To observe the stability of the interference pattern over time, conditions are necessary under which the frequencies, polarization and phase difference of the interfering waves would be constant during the observation time. Such waves are called Coherent(Related).
Let us first consider two strictly monochromatic waves that have the same frequency. Monochromatic wave is a strictly sinusoidal wave with a constant frequency, amplitude and initial phase over time. The amplitude and phase of the oscillations may change from one point to another, but the frequency is the same for the oscillatory process throughout space. The monochromatic oscillation at each point in space lasts indefinitely, having neither beginning nor end in time. Therefore, strictly monochromatic oscillations and waves are coherent.
Light from real physical sources is never strictly monochromatic. Its amplitude and phase fluctuate continuously and so quickly that neither the eye nor an ordinary physical detector can follow their changes. If two light beams originate from the same source, then the fluctuations arising in them are, generally speaking, consistent, and such beams are said to be partially or completely coherent.
There are two methods for producing coherent beams from a single light beam. In one of them, the beam is divided, for example, passing through holes located close to each other. This method is Wavefront division method- Suitable only for fairly small sources. In another method, the beam is divided into one or more reflective, partially transmitting surfaces. This method is Amplitude division method— can be used with extended sources and provides greater illumination of the interference pattern.
The work is devoted to familiarization with the phenomenon of light interference in thin transparent isotropic films and plates. The light beam emanating from the source falls on the film and is divided due to reflection from the front and rear surfaces into several beams, which, when superimposed, form an interference pattern, i.e., coherent beams are obtained by amplitude division.
Let us first consider the idealized case when a plane-parallel plate of a transparent isotropic material is illuminated by a point source of monochromatic light.
From a point source S to any point P Generally speaking, only two rays can hit - one reflected from the upper surface of the plate, and the other reflected from its lower surface (Fig. 1).
Rice. 1 Fig. 2
It follows that in the case of a point monochromatic light source, each point in space is characterized by a completely definite difference in the path of the reflected rays arriving at it. These rays, when interfering, form a time-stable interference pattern, which should be observed in any region of space. The corresponding interference bands are said to be not localized (or localized everywhere). From symmetry considerations it is clear that the strips in planes parallel to the plate have the form of rings with an axis SN, normal to the plate, and at any position P they are perpendicular to the plane SNP.
When the size of the source increases in the direction parallel to the plane SNP, the interference fringes become less clear. An important exception is the case when the point P is located at infinity, and the observation of the interference pattern is carried out either with an eye accommodated at infinity, or in the focal plane of the lens (Fig. 2). Under these conditions, both beams coming from S To P, namely the rays SADP And SABCEP, come from one incident ray, and after passing through the plates are parallel. The optical path difference between them is equal to:
Where N 2 and N 1 - refractive indices of the plate and the environment,
N- the base of the perpendicular dropped from WITH on AD. The focal plane of the lens and the plane parallel to it NC are conjugate, and the lens does not introduce an additional path difference between the beams.
If H is the thickness of the plate, and j1 and j2 are the angles of incidence and refraction on the upper surface, then
, (2)
From (1), (2) and (3), taking into account the law of refraction
We get that
(5)
The corresponding phase difference is:
, (6)
Where l is the wavelength in vacuum.
One should also take into account the change in phase by p, which, according to Fresnel's formulas, occurs with each reflection from a denser medium (we consider only the electrical component of the wave field). Therefore, the total phase difference at the point P is equal to:
(7)
. (8)
Angle j1, the value of which determines the phase difference, is determined only by the position of the point P in the focal plane of the lens, therefore, the phase difference d does not depend on the position of the source S. It follows that when using an extended source, the fringes are as distinct as with a point source. But since this is true only for a certain observation plane, such stripes are said to be localized, and in this case, localized at infinity (or in the focal plane of the lens).
If the intensities of the coherent rays under consideration are denoted accordingly I 1 and I 2, then full intensity I at the point P will be determined by the relation:
How do we find that the light stripes are located at d = 2 M P or
, M = 0, 1, 2, …, (10A)
And dark stripes - at d = (2 M+ 1)p or
, M = 0, 1, 2, … . (10B)
A given interference fringe is characterized by a constant value of j2 (and therefore j1) and, therefore, is created by light incident on the plate at a certain angle. Therefore, such stripes are often called Stripes of equal slope.
If the lens axis is normal to the plate, then when light is reflected close to normal, the stripes have the form of concentric rings with the center at the focus. The interference order is maximum in the center of the picture, where its magnitude is M 0 is determined by the relation:
.
For now we are considering only light reflected from the plate, but similar reasoning applies to light transmitted through the plate. In this case (Fig. 3) to the point P the focal plane of the lens come from the source S two rays: one that passed through without reflection, and the other after two internal reflections.
The optical path difference of these rays is found in the same way as when deriving formula (5), i.e.
This means that the corresponding phase difference is equal to:
. (12)
However, there is no additional phase difference caused by reflection here, since both internal reflections occur under the same conditions. The interference pattern created by an extended source is also localized at infinity in this case.
Comparing (7) and (12), we see that the patterns in transmitted and reflected light will be complementary, i.e., the light stripes of one and the dark stripes of the other will be at the same angular distance relative to the normal to the plate. Moreover, if the reflectivity R surface of the plate is small (for example, at the glass-air interface at normal incidence it is approximately equal to 0.04), then the intensities of the two interfering rays passing through the plate are very different from each other
(I 1/I 2 @ 1/R 2 ~ 600), therefore the difference in the intensity of the maxima and minima (see (9)) turns out to be small, and the contrast (visibility) of the bands is low.
Our previous reasoning was not entirely rigorous. Since we neglected the multiplicity of internal reflections in the plate. In reality the points P reaches not two, as we assumed, but a whole series of beams coming from S(rays 3, 4, etc. in Fig. 1 or 3).
But if the reflectivity on the surface of the plate is small, then our assumption is quite satisfactory, since the beams after the first two reflections have negligible intensity. With significant reflectivity, multiple reflections greatly change the intensity distribution in the bands, but the position of the bands, i.e., maxima and minima, is precisely determined by relation (10).
Let us now assume that the point source S monochromatic light illuminates a transparent plate or film with flat, but not necessarily parallel, reflective surfaces (Fig. 4).
Neglecting multiple reflections, we can say that to each point P, located on the same side of the plate as the source, again only two rays come, emanating from S, namely SAP And SBCDP, therefore, in this region the interference pattern from a point source is not localized.
Optical path difference between two paths from S before P equal to
Where N 1 and N 2 - refractive indices of the plate and the environment, respectively. The exact value of D is difficult to calculate, but if the plate is thin enough, then the points B, A, D are at a very small distance from each other, and therefore
, (14A)
, (14B)
Where AN 1 and AN 2 - perpendiculars to B.C. And CD. From (13) and (14) we have
In addition, if the angle between the surfaces of the plate is small enough, then
Here N 1¢ and N 2¢ - the base of the perpendiculars dropped from E on Sun And CD, and point E— intersection of the upper surface with the normal to the lower surface at the point WITH. But
, (17)
Where H = C.E. — thickness of the plate near the point WITH, measured normal to the bottom surface; j2 is the angle of reflection on the inner surface of the plate. Consequently, for a thin plate that differs little from a plane-parallel one, we can write, using (15), (16) and (17),
, (18)
And the corresponding phase difference at a point P equal to
. (19)
Magnitude D depends on position P, but it is uniquely defined for everyone P, so that the interference fringes, which are the locus of the points for which D Constant, are formed in any plane of the region where both rays from S. We are talking about such bands that they are not localized (or localized everywhere). They are always observed with a point source, and their contrast depends only on the relative intensity of the interfering beams.
In general, for a given point P both parameters H and j2, which determine the phase difference, depend on the position of the source S, and even with a slight increase in the size of the source, the interference fringes become less clear. It can be assumed that such a source consists of incoherent point sources, each of which creates a non-localized interference pattern.
Then at each point the total intensity is equal to the sum of the intensities of such elementary patterns. If at the point P the phase difference of radiation from different points of an extended source is not the same, then the elementary patterns are shifted relative to each other in the vicinity P and visibility of stripes at a point P less than in the case of a point source. The mutual displacement increases as the size of the source increases, but depends on the position P. Thus, although we are dealing with an extended source, the visibility of the stripes at some points P may remain the same (or almost the same) as in the case of a point source, while elsewhere it will drop to almost zero. Such bands are characteristic of an extended source and are called Localized. We can consider the special case when the point P is located in the plate, and observation is carried out using a microscope focused on the plate, or the eye itself is accommodated to it. Then H is almost the same for all pairs of rays from an extended source arriving at a point P, associated with P(Fig. 5), and the difference in values D at the point P caused mainly by differences in values CosJ 2. If the change interval Cos J 2 is small enough, then the range of values D at the point P much less than 2 P even with a source of considerable size, the stripes are clearly visible. It is obvious that they are localized in the film and localization arises as a consequence of the use of an extended source.
Practically, the condition for the smallness of the interval of changes CosJ 2 can be performed when observing in a direction close to normal, or when limiting the entrance pupil to a diagram D, although the pupil of the naked eye itself may be quite small.
Considering the phase change by P when reflected on one of the surfaces of the plate, we obtain from (9) and (19) that at the point P the maximum intensity will be found if the phase difference is a multiple of 2 P, or, equivalently, when the condition is met
, M = 0,1,2… (20A)
And intensity minima - at
, M = 0,1,2…, (20B)
Where is the average value for those points of the source, the light from which reaches P.
Magnitude CosJ 2, present in the last relations, represents the optical thickness of the plate at the point P, and if our approximation remains valid, then the interference effect in P does not depend on the thickness of the plate in other places. It follows that relations (20) remain valid even for non-flat surfaces of the plate, provided that the angle between them remains small. Then, if sufficiently constant, then the interference fringes correspond to a set of film locations where the optical thicknesses are the same. For the same reason, such stripes are called Stripes of equal thickness. Such stripes can be observed in a thin air gap between the reflective surfaces of two transparent plates, when the direction of observation is close to normal, and the minimum condition (20, B) will take the form:
,
That is, dark stripes will pass in those places of the layer whose thickness satisfies the condition
, M = 0, 1, 2, …, (21)
Where is the wavelength in air.
Thus, the stripes outline the contours of layers of equal thickness at l/2. If the thickness of the layer is constant everywhere, the intensity is the same over its entire surface. It is widely used for quality control of optical surfaces.
With a wedge-shaped air gap between flat surfaces, the strips will run parallel to the edge of the wedge at the same distance from each other. The linear distance between adjacent light or dark stripes is l/2 Q, Where Q- angle at the top of the wedge. In this way, it is easy to measure angles of the order of 0.1¢ or less, as well as detect surface defects with an accuracy available to other methods (0.1l or less).
The interference pattern localized in the film is also visible in transmitted light. As in the case of a plane-parallel plate, the patterns in reflected and transmitted light are complementary. That is, the light stripes of one appear in the same places on the film as the dark stripes of the other. When using weakly reflective surfaces, stripes in transmitted light are poorly visible due to significant inequality in the intensities of the interfering beams.
Until now we have assumed that a point source emits monochromatic radiation. Light from a real source can be represented as a set of monochromatic components incoherent with each other, occupying a certain spectral interval from l to l + Dl. Each component forms its own interference pattern, similar to that described above, and the total intensity at any point is equal to the sum of the intensities in such monochromatic patterns. The zero maxima of all monochromatic interference patterns coincide, but in any other place the appearing patterns are shifted relative to each other, since their scale is proportional to the wavelength. Highs M-th order will occupy a certain area in the observation plane. If the width of this region can be neglected in comparison with the average distance between adjacent maxima, then the same stripes appear in the observation plane as in the case of strictly monochromatic light. In another limiting case, interference will not be observed if the maximum M th order for (l + Dl) will coincide with the maximum ( M+ 1)th order for l. In this case, the gap between adjacent maxima will be filled with maxima of indistinguishable wavelengths of our interval. We write the condition for the indistinguishability of the interference pattern as follows: ( M+ 1)l = M(l + Dl), i.e. M= l/Dl.
But in order for the interference pattern to have sufficient contrast at given values of Dl and l, we have to limit ourselves to observing interference fringes whose order is much less than l/Dl, i.e.
M < < L/ D L. (22)
Therefore, the higher the interference order M, which needs to be observed, the narrower the spectral interval Dl must be, allowing interference to be observed in this order, and vice versa.
Interference order M is associated with the path difference of the interfering light beams, which in turn is associated with the thickness of the plate (see (20)). As can be seen from this formula, in order for the stripes to be distinct, the requirements for the monochromaticity of the source must become stricter, the greater the optical thickness of the plate Hn 2. However, it must be borne in mind that the quality of the observed interference pattern depends significantly on Law of Energy Distribution in the used spectral range and from Spectral sensitivity of the radiation receiver used.
We will study interference in thin films using the example of strips of equal thickness, the so-called Newton's rings.
Newton's rings are a classic example of interference fringes of equal thickness. The role of a thin plate of variable thickness, from the surfaces of which coherent waves are reflected, is played by the air gap between the plane-parallel plate and the convex surface of a plano-convex lens with a large radius of curvature in contact with the plate (Fig. 6). To observe many rings, it is necessary to use light of relatively high monochromaticity.
Let the observation be carried out from the side of the lens. From the same side, a beam of monochromatic light falls on the lenses, i.e., observation is carried out in reflected light. Then the light waves reflected from the upper and lower boundaries of the air gap will interfere with each other. For clarity purposes, in Fig. 6, the rays reflected from the air wedge are slightly shifted away from the incident beam.
At normal incidence of light, the interference pattern in reflected light has the following form: in the center there is a dark spot surrounded by a number of concentric light and dark rings of decreasing width. If the light flux falls from the side of the plate, and observation is still carried out from the side of the lens, then the interference pattern in transmitted light remains the same, only in the center the spot will be light, all the light rings will become dark and vice versa, and, as already noted, more The rings will be contrasting in reflected light.
Let us determine the diameters of the dark rings in reflected light. Let
R- radius of curvature of the lens, Hmm — thickness of the air gap at the location M th ring, Rm — radius of this ring, D H- the amount of mutual deformation of the lens and plate that occurs when they are compressed. Let us assume that only a small area of the lens and plate is deformed and near the center of the interference pattern. To calculate the optical difference in wave paths at the point of occurrence M th ring we use the formula (20 B):
With normal incidence of the wave on the lens and due to the small curvature of its surface, we assume cos j 2 = 1. In addition, we take into account that N 2 = 1, and the phase change is P Or an extension of the optical path by l/2 occurs at the wave reflected from the glass plate (the lower surface of the air gap). Then the optical path difference will be equal and in order for a dark ring to appear in this place, the equality must be satisfied:
. (23)
From Fig. 6 it also follows that
Where, if we neglect the terms of the second order of smallness, 
= >
.
Substituting this expression into (23) after simple transformations gives the final formula connecting the radius of the dark ring with its number M, wavelength L and lens radius R.
. (24)
For experimental testing purposes, it is more convenient to use the formula for the diameter of the ring:
. (25)
If you construct a graph plotting the numbers of dark rings on the abscissa axis, and the squares of their diameters on the ordinate axis, then in accordance with formula (25) you should get a straight line, the continuation of which cuts off the segment on the ordinate axis, and
This makes it possible to calculate the mutual deformation D from the found value H, if the radius of curvature of the lens is known:
By the slope of the graph, you can determine the wavelength of the light in which the observation is being made:
, (28)
Where M 1 and M 2 are the corresponding numbers of the rings, and and are their diameters.
In nature, one can observe rainbow coloring of thin films (oil films on water, soap bubbles, oxide films on metals), resulting from the interference of light reflected by two film surfaces.
Let a plane-parallel transparent film with a refractive index P and thickness d at an angle i a plane monochromatic wave falls (consider one beam). We will assume that on both sides of the film there is the same medium (for example, air) and . Part of the incident wave front perpendicular to the drawing plane is depicted as a segment AB(direction of wave propagation, i.e. rays 1 and 2). On the surface of the film in point A, the beam will be divided into two: it will be partially reflected from the upper surface of the film, and partially refracted. The refracted ray, reaching t .D, will be partially refracted into the air, and partially reflected and will go to the so-called. C. Here it will again be partially reflected (we do not consider it due to its low intensity) and refracted, exiting into the air at an angle i.
Refracted wave (ray 1’’ ) is superimposed on the wave directly reflected from the upper surface (ray 2’) . Rays emerging from the film /’, 1'' and 2' coherent if the optical difference in their path is small compared to the coherence length of the incident wave. If a collecting lens is placed on their path, they will converge in one of the so-called. R the focal plane of the lens and will give an interference pattern. 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. The optical path difference that occurs between two interfering beams from the so-called. A to plane Sun, where the term is due to the loss of a half-wave upon reflection of light from the interface.
If n>n 0, then the loss of a half-wave will occur in the so-called A and will have a minus sign if n 
, where is the refracted angle (9.1)
If n>n 0, .
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 condition of maximum reflection 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 The optical path difference for transmitted light differs from that for reflected light by , then the interference maxima 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. Equal slope strips(interference from a plane-parallel plate).
Def. 9.1. Interference fringes resulting from the superposition of rays incident on a plane-parallel plate at equal angles are called stripes of equal slope.
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, which is why they say that stripes of equal inclination are localized at infinity. To observe them, a collecting lens and a screen (E) located in the focal plane are used
The rays /" and /" / will gather in focus F lenses (in the figure, its optical axis is parallel to the rays G and /"), other rays (ray 2), parallel to ray /, will come to the same point - the overall intensity increases. Rays 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 take the form of concentric rings with the center at the focus of the lens.
Task 1. A beam of rays of monochromatic light falls normally onto 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 film, which in turn is less than the refractive index of glass, then in both cases the reflection occurs from a medium optically denser than the medium in which the incident beam travels. Therefore, the phase of the oscillations changes twice to and the result will be the same as if there was 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 the parallel rays / and 2. R Let's look at the rays / / and / // reflected from the upper and lower surfaces of the wedge. At a certain relative position of the wedge and the lens, rays / / and 1" will intersect at some t. A, which is the image of a point IN.
Since the rays / / and / // are coherent, they will interfere. If the source is located far from the wedge surface and the angle A is small enough, then the optical path difference between the rays / / and / // can be calculated using 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 due to beam division 2, of a wedge falling to another point are collected by a lens in the so-called A". The optical path difference is determined by the thickness d". A system of interference fringes appears on the screen. Each of the stripes arises due to reflection from places of the plate that have the same thickness.
Def. 9.2. Interference fringes that arise as a result of interference from places of the same thickness are called. stripes 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 normally on the plate, then stripes of equal thickness are localized on the upper surface of the wedge. If we want to obtain an image of the interference pattern on the screen, then the collecting lens and the screen must be positioned in such a way in relation 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 orders) according to formula 9.2: 
And 
, where . If the distances from the wedge edge to the interference fringes under consideration are equal to and , then 
and , where is the small angle between the wedge faces (the refracting angle of the wedge), i.e. . Due to its smallness, the refractive angle of the wedge should also be very small, because otherwise, stripes of equal thickness will be so closely spaced that they cannot be distinguished.
Task 2. A beam of rays of monochromatic light falls on a glass wedge normal to its edge. 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 wedge face, is reflected from both the upper and lower faces. These beams are coherent, so a stable interference pattern is observed. Because Since interference fringes are observed at small wedge angles, the reflected rays will be almost parallel.
Dark stripes will be observed in those sections of the wedge for which the difference in the traveling rays is equal to an odd number of half-waves: or , Because , That . Let an arbitrary dark stripe of the number correspond to a certain thickness of the wedge in this place, and let the dark stripe of the number correspond to the thickness of the wedge in this place,. According to the condition, 10 stripes fit into , then, because , That 
.
Newton's rings.
Newton's rings are an example of strips of equal thickness. Observed when light is reflected from an air gap formed by a plane-parallel plate and a plane-convex lens with a large radius of curvature in contact with it. A parallel beam of light falls on a 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 oscillation phase by and no additional path difference arises. When reflected rays are superimposed, stripes of equal thickness appear, which, under normal light incidence, have the form of concentric circles.
In reflected light, the optical path difference ati = 0: R) determine and, conversely, find from the known R..
For both strips of equal slope and strips of equal thickness the position of the maxima depends on. A system of light and dark stripes is obtained only when illuminated with monochromatic light. When observed in white light, a set of stripes shifted relative to each other is obtained, formed by rays of different wavelengths, and the interference pattern acquires a rainbow color. All reasoning was carried out for reflected light. Interference can also be observed in passing 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, i.e. Interference maxima in reflected light correspond to minima in transmitted light, and vice versa.