Operating principle of optical interferometers. Michelson, Jamin, Fabry-Perot interferometers

Application of the interference phenomenon.

The phenomenon of interference is due to the wave nature of light; its quantitative patterns depend on the wavelength l 0 . Therefore, this phenomenon is used to confirm the wave nature of light and to measure wavelengths (interference spectroscopy).

The phenomenon of interference is also used to improve the quality of optical instruments ( optics clearing) and obtaining highly reflective coatings. The passage of light through each refractive surface of the lens, for example through the glass-air interface, is accompanied by reflection of »4% of the incident flux (with a refractive index of glass »1.5). Since modern lenses contain a large number of lenses, the number of reflections in them is large, and therefore the loss of light flux is large. Thus, the intensity of the transmitted light is weakened and the aperture ratio of the optical device decreases. In addition, reflections from lens surfaces lead to glare, which often (for example, in military equipment) reveals the position of the device.

To eliminate these shortcomings, the so-called enlightenment of optics. To do this, thin films with a refractive index lower than that of the lens material are applied to the free surfaces of the lenses.

The phenomenon of interference is also used in very precise measuring instruments called interferometers. All interferometers are based on the same principle and differ only in design.

Russian physicist V.P. Linnik (1889-1984) used the operating principle of the Michelson interferometer to create microinterferometer(a combination of an interferometer and a microscope), which serves to control the cleanliness of surface treatment.

Interferometers are very sensitive optical instruments that make it possible to determine minor changes in the refractive index of transparent bodies (gases, liquids and solids) depending on pressure, temperature, impurities, etc. Such interferometers are called interference refractometers.

MICHAELSON INTERFEROMETER - two-beam interferometer, optical the circuit of which allows for various types of interference, widely used in physics. research and various tech. will measure, instruments for measuring lengths, displacements, for studying the quality of optical fibers. parts, systems, etc. With the help of I.M., the wavelength of light was determined for the first time and Michelson. IM is also used as a spectral device with high aperture and high resolution, which has a number of other advantages.

With a smooth change in the path difference of the interfering beams by λ 0 /2 the interference pattern will shift so much that the maxima will be replaced by minima. Therefore, the phenomenon of interference is used in interferometers for measuring the length of bodies, the wavelength of light, changes in the length of a body with changes in temperature, comparable to λ 0

IN Michelson interferometer monochromatic beam from source S is divided into translucent plate P x two beams 1" And 2", which, reflected from the mirrors M 1 and M 2, again using P 1 are combined into one beam, in which the rays 1" And 2" form an interference pattern. Compensation plate P 2 placed in the beam path 2, so that it, like beam 1, passes through the plate twice. The resulting interference pattern is extremely sensitive to any change in the path difference of the rays (for example, to the displacement of one of the mirrors).

The basis of the Michelson interferometer device is the phenomenon of interference of light waves in thin films. In the device under consideration, this phenomenon is realized by dividing the amplitude of the light wave.

The interferometer contains a plane-parallel plate ($A$), which is coated with silver or aluminum. This plate is fixed on a pedestal at an angle of $45()^\circ $ to the direction of the rays. In addition, there are two flat mirrors ($C\ and\ D$), located perpendicularly (Fig. 1).

To compensate for the difference in the path of the rays, the device uses a plate $B$. Light waves come from the source $S$. These waves experience partial reflection from the plate$\A$, some of them overcome this plate, thus producing two coherent light waves. Waves passing through plate $A$ undergo reflection from mirrors $C\ and\ D$ and return to it. Some of these waves again pass through the plate $A, and some are reflected from it. The resulting waves are capable of interfering on the segment $AK$. Interference results from dividing the amplitude on plate $A$. The interference pattern is observed through a telescope.

Let's rotate the arm $DA$ by an angle $90()^\circ $ (Fig. 1). In this case, the mirror will be located in a position designated as $D"$ in Fig. 1. A small gap appears between the mirrors $D"$ and $C$, which can be likened to a thin film. If the mirrors are positioned strictly normal to each other, then as a result of interference we will obtain stripes of equal inclination in the form of concentric rings. To observe the interference pattern in this case, the telescope should be adjusted to infinity. If the angle between the mirrors is not exactly $90()^\circ $, then the gap between them will be a wedge. The result of such interference will be straight stripes of equal thickness. To examine such an interference pattern, the telescope is directed to the face of the plate $A,$ which is coated with silver.

Interference of monochromatic waves in the direction of the interferometer axis

If the light waves travel clearly along the axis of the interferometer, then the optical difference in their path ($\Delta$) arises as the difference in the lengths of the arms ($p_1\ and\p_2\$) of the interferometer:

\[\Delta =2\left(p_1-\ p_2\right)\left(1\right).\]

In this case, the stroke difference is:

\[\delta =\frac(2\pi \Delta )(\lambda )\left(2\right).\]

Note that in the case under consideration we will not take into account the change in the phase of the wave, which occurs when it is reflected from the mirrors and refracted in plate A, since the interference pattern does not change from this.

Let us assume that when a wave falls on plate A, its energy flux density is divided into two parts. Let's define the waves that go towards the telescope using equations:

where $E_0$ is the amplitude of the incident wave; $\delta =(\varphi )_2-(\varphi )_1$. The intensity of the received wave is:

where $I_0=\frac(1)(2)(E_0)^2$ is the intensity of the source wave.

Following expression (3) with:

\[\delta =\left(2m+1\right)\pi ,\ \left(m=0,\pm 1,\pm 2,\dots \right)\left(4\right),\] \

In case if:

\[\delta =2m\pi ,\ \left(m=0,\pm 1,\pm 2,\dots \right)\left(6\right),\] \

When condition (6) is met, all the energy of the source comes to the “screen”. There is no energy flow returning to the light source.

Examples of problems with solutions

Example 1

Exercise. How can a Michelson interferometer be used in optical research?

Solution. The ability to move the interferometer mirror (for example, mirror D) can change the path difference of the interfering beams. This determines all the possibilities for using this interferometer as an optical device. It can be used to measure wavelengths of light. It should be taken into account that the mirror is moved so that its reflecting surface is parallel to itself.

A Michelson interferometer can measure changes in the refractive index of light. An additional plate of thickness $d$ and refractive index $n"$ is introduced into one of the equal arms of the interferometer, then a path difference will appear between the interfering rays:

\[\Delta =2d\left(n"-n\right)\left(1.1\right),\]

where $n=1$ is the refractive index of air. To restore the interference pattern in the field of view of the pipe, the other length of the interferometer arm should be increased by an amount equal to:

\[\Delta p=\frac(\Delta )(2)=d\left(n"-1\right)\left(1.2\right).\]

Michelson used the device to test the relationship between the direction of propagation of a light beam relative to the Earth and the speed of light.

Using the Michelson interferometer, for the first time, a systematic study of the fine structure of spectral lines was carried out and a reference meter was compared with the wavelength of light. At the moment, the Michelson interferometer is outdated as an instrument for scientific research.

Example 2

Exercise. How much should mirror D be shifted parallel to itself (Fig. 2) in order for the interference pattern to shift by $k$ fringes? The wavelength of light is $\lambda$. \textit()

Solution. As a basis for solving the problem, we use the condition for obtaining interference maxima

\[\Delta =m\lambda \ \left(m=0,\pm 1,\pm 2,\dots \right)\left(2.1\right).\]

On the other hand, we know that for the interferometer in the first position of the mirrors:

\[(\Delta )_1=2\left(p_2-\ p_1\right)=m_1\lambda \left(2.2\right).\]

In the interferometer state, when one mirror is shifted by a distance $\Delta p$ (the desired distance):

\[(\Delta )_2=2\left(p_2+\Delta p-\ p_1\right)=m_2\lambda \left(2.3\right).\]

Let us find the difference between equations (2.2) and (2.3), we have:

\[(\Delta )_2-(\Delta )_1=m_2\lambda -m_1\lambda =2\left(p_2+\Delta p-\ p_1\right)-2\left(p_2-\ p_1\right)\left (2.4\right).\]

According to the problem:

transforming expression (2.4), we obtain:

Answer.$\Delta p=\frac(k\lambda )(2)$

Target: familiarization with the optical design and operation of the interferometer; determination of the wavelength of light, measurement of small deformations.

Introduction

When two coherent light waves are added, the light intensity at some arbitrary point M will depend on the difference in the phases of the oscillations arriving at this point.

Let at the point ABOUT the wave is divided into two coherent waves, which superimpose each other at the point M. The phase difference at this point of coherent waves depends on the time of propagation of waves from the point ABOUT exactly M. For the first wave this time is equal, for the second
, Where ,- path and speed of propagation of the first wave from a point ABOUT exactly M; ,- for the second wave. As is known,

,
, (1)

Where With- speed of light in vacuum; n 1 and n 2 - refractive indices of the first and second medium, respectively.

Then the phase difference of the two waves at the point M can be represented in the form

, (2)

where  is the optical difference between the paths of two waves;
And
- optical lengths of the first and second waves.

From formula (2) it is clear that if the path difference is equal to an integer number of wavelengths in vacuum

,k= 0, 1, 2, (3)

then the phase difference turns out to be a multiple of 2 and oscillations excited at the point M both waves will occur with the same phase. Thus (3) is the condition for the interference maximum.

Optical measuring instruments based on the interference of light are called interferometers. In this work, a Michelson interferometer is used, the schematic diagram of which is shown in Fig. 1.

Its main elements are: light source I, dividing cube K and two mirrors - movable Z1 and fixed Z2. A beam of light from source I falls on a cube K, glued together from two halves along a large diagonal plane. The latter plays the role of a translucent layer that divides the original beam into two - 1 and 2. After reflection from the mirror and combination, rays 1 and 2 fall on the screen E, where the interference pattern is observed. The type of interference pattern is determined by the configuration of the wave surfaces of the interfering waves. If the wave surfaces are flat (a collimated beam comes from the source), then a system of parallel alternating light and dark stripes will appear on the screen (see § 2 section 2), and the distance between the dark and light stripes is determined by the relation

, (4)

Where - wavelength of light; - angle between wave vectors And interfering waves.

Angle size and, therefore, the width of the stripes, convenient for observation, can be set by changing the inclination of mirrors Z1 and Z2 and cube K.

In the case when the folded waves are spherical (see § 6 section 2), the interference pattern has the form of rings with the distances between the stripes being greater, the less the radii of curvature of the wave surfaces differ.

The distances from the dividing cube to the mirrors are usually called interferometer arms, which in general are not equal to each other. The doubled difference in arm lengths is the optical difference in the path of interfering waves . Changing the length of any arm by an amount leads to a change in the optical path difference by and, accordingly, to a shift of the interference pattern on the screen by one band. Thus, the interferometer can serve as a sensitive device for measuring very small displacements.

You can change the optical path difference between two beams in various ways. You can move one of the mirrors, and the optical path difference will change by twice the amount of movement of the mirror. You can change the optical path length of one of the rays by changing the refractive index of the medium in a certain area, and the change in the path difference of the interfering rays will be equal to twice the optical path length of the light in this medium. The work used methods that make it possible to measure various physical quantities.

Glass plate. Let a glass plate of thickness stand in the path of one of the rays d with refractive index n. When turning the plate at an angle from a position perpendicular to the incident beam of light, an additional path difference arises:

. (5)

If, during rotation, the interference pattern shifts by m stripes, then
and you can find the refractive index. For small corners
approximately from (5)

IN Michelson interferometer The phenomenon of interference in thin films is used. The phenomenon of interference in this device is carried out by dividing the amplitude of the wave.

What is this device? On a massive pedestal there is a plane-parallel plate ($A$) lightly coated with silver, located at an angle of $45^0$ to the direction of propagation of the rays and two mutually perpendicular flat mirrors $C$ and $D$ (Fig. 1).

Picture 1.

Plate B (Fig. 1) serves as an auxiliary plate; it compensates for the difference in the path of the rays. Light waves travel from ($S$). Some of them are reflected from the silver surface of the plate $A$, and some pass through this plate. This is how the process of splitting a wave of light into two coherent waves occurs. The waves that pass through the plate are reflected from the mirrors $C$ and $D$. The reflected waves are again partially reflected and partially transmitted through the silver plate $A$. These waves can interfere in the $AK$ region. This interference pattern is observed through a telescope. Thus, on the plate $A$ the amplitude is divided; the wave front on it is preserved; only the direction of its movement changes.

If, hypothetically, arm $DA$ is rotated by $90^0$, then mirror $D$ will end up in position $D"$. A gap appears between $D"$ and $C$, which may be similar to a thin film. If the mirrors $C$ and $D$ are strictly perpendicular, then stripes of equal inclination are observed, which look like circles. The telescope in this case should be set to infinity. If the mirrors $C$ and $D$ are not completely perpendicular, then the gap between us becomes like a wedge, then stripes of equal thickness appear in the form of straight stripes. In this case, the telescope is focused on the silvered edge of the plate $A$.

Interference of monochromatic waves that propagate along the axis of the interferometer

In the case of wave propagation strictly along the axis of the interferometer, the optical difference in the path of the rays ($\triangle $) appears due to the difference in the lengths of the arms ($l_1\ and\l_2\\$) of the interferometer:

The resulting phase difference is:

In a rigorous calculation, one should take into account the change in wave phases upon reflection from mirrors and refraction in the plate $A$; here we will not do this, since this is not of fundamental importance for the interference pattern in our case.

where $E_0$ is the amplitude of the wave before it hits the plate $A$. $\delta =(\varphi )_2-(\varphi )_1$. Therefore, for the intensity observed as a result we obtain:

where $I_0=\frac(1)(2)(E_0)^2$ is the intensity of the wave incoming from the light source.

In case if:

intensity (3) is zero. If:

the intensity is equal to $I_0$, which means: all the energy from the source falls on the “screen”, there is no flow of energy that returns in the direction of the light source.

Comment

The Michelson interferometer is used to measure small distances and small changes in refractive indices. Michelson himself used his interferometer for an experiment to test the connection between the speed of light and the direction of motion of the beam in relation to the Earth.

Example 1

Exercise: In order to calculate the refractive index of ammonia, a glass tube with a vacuum inside is placed in one arm of the Michelson interferometer. Its length is $l=15\cm=15\cdot 10^(-2)m$. If this tube is filled with ammonia, the interference pattern for a wavelength equal to $\lambda =589\nm=589\cdot (10)^(-9)m$ shifts by $192$ band. What is the refractive index of ammonia?

Solution:

The difference in the optical path of the wave ($\triangle $) in vacuum and ammonia can be found as:

\[\triangle =ln-ln_v\left(1.1\right),\]

where $n_v$=1 refractive index for vacuum. Let us write down the condition for interference minima:

\[\triangle =m\frac(\lambda )(2)\ \left(m=0,\pm 1,\pm 2,\dots \right)\left(1.2\right).\]

Let us equate the right sides of expressions (1.1) and (1.2), we obtain:

Let us express the refractive index from (1.3):

Let's carry out the calculations:

Answer:$n=1.000377.$

Example 2

Exercise: In a Michelson interferometer, when one of the mirrors moves forward, the interference pattern either disappears or appears. What is the displacement ($\triangle l$) of the mirror between two successive occurrences of a clear interference pattern if the waves $(\lambda )_1$ and $(\lambda )_2$ are used?

Solution:

The reason for the disappearance of the interference pattern can be considered to be that the maxima and minima of the interference pattern of waves of different lengths are shifted relative to each other. With a sufficient difference in wavelength, the maxima in the interference of one wave can fall on the minima of another, then the interference pattern completely disappears.

Let us write down the condition for the transition from one clear picture to another:

\[\left(z+1\right)(\lambda )_1=z(\lambda )_2\left(2.1\right),\]

where $z$ is an integer. The required mirror displacement ($\triangle l$) can be defined as:

Using the system of equations (2.1) and (2.2) we express $\triangle l$:

\[\left(z(\lambda )_1+(\lambda )_1\right)=z(\lambda )_2\to z((\lambda )_2-(\lambda )_1)=(\lambda )_1\to z=\frac((\lambda )_1)(((\lambda )_2-(\lambda )_1)),\] \[\triangle l=\frac((\lambda )_1(\lambda )_2)( 2((\lambda )_2-(\lambda )_1)).\]

Answer:$\triangle l=\frac((\lambda )_1(\lambda )_2)(2((\lambda )_2-(\lambda )_1)).$

MICHAELSON INTERFEROMETER consists of two mirrors M 1 and M 2 and a semi-permeable reflecting partition S, inclined at an angle of 45° (Fig. 1). This partition transmits 50% of the light incident on it and reflects the remaining 50%. Distances to mirrors L 1 and L 2 are the same: L 1 = L 2 = L. Monochromatic light from the source passes halfway through the partition S, is reflected from M 1 and then hits the detector, half reflected from S (ray 1). This path light travels in the direction of the speed of the Earth as it moves in orbit and in the opposite direction, which corresponds to the movement of a swimmer with and against the current. The other part of the light beam is reflected by the partition S to the mirror M 2, and on the way back it passes through the partition, hitting the detector (beam 2). This corresponds to the swimmer moving across the current.

If the interferometer is at rest relative to the ether, then the time spent by the first and second rays of light on their path is the same, and two coherent rays enter the detector in the same phase ( cm. COHERENCE). Consequently, interference occurs and a central bright spot can be observed in the interference pattern ( cm. OSCILLATIONS AND WAVES; OPTICS). If the interferometer moves relative to the ether, then the time spent by the rays on their path turns out to be different. Indeed, let c the speed of light relative to the ether, and v speed of the interferometer relative to the ether. Then the time spent on the first path (downstream and back) is equal to

To calculate time t 2 it should be taken into account that while the light travels from the semi-permeable partition to the mirror M 2, the mirror itself moves with the Earth relative to the ether. Therefore, the path traveled by light to the mirror M 2 is equal to the hypotenuse of the triangle. The speed of light does not change, since light moves perpendicular to the direction of the Earth's speed. From simple geometric considerations

Using approximate formulas:

This time delay corresponds to the difference in paths of two rays of light

Consequently, such a difference in the path of the rays will correspond to the total number of wavelengths of light that fit within this path difference, equal to

Interference maxima and minima alternate when the path difference changes by p/2. Thus, having calculated the value n for specific installation parameters and knowing the speed of the Earth, you can find out how the interference fringes should move. Of course, the effect is very small. To strengthen it, Michelson maximized the base of the interferometer L, causing the light to be reflected repeatedly from additional mirrors. In addition, the experiment was carried out a second time with the device rotated 90°, due to which the rays swapped places and the effect of shifting the interference fringes was doubled.

For monochromatic light corresponding to the sodium line with wavelength l = 590 nm, and at L= 11 m, v/c= 10 8, it turns out that the total shift is approximately 0.37 stripes. However, Michelson and Morley claimed, based on preliminary tests of the instrument, that they were able to clearly detect a shift of 0.01 fringes.

Alexander Berkov