Epigraph:

“I value one experience higher than a thousand opinions born only of imagination.”
M. Lomonosov.

Lesson objectives:

1. Development of abilities.
   The ability to use the studied material to solve calculations and practical problems. Be able to apply mathematical knowledge to physical laws.
2. Formation of values.
   White light has complex structure, knowing which can explain the variety of colors in nature. Using a diffraction grating or prism, white light can be separated into a spectrum, which consists of seven primary colors: red, orange, yellow, green, blue, indigo, violet.
3. Reasonable behavior in the environment.
   Outside of us, there are no colors in nature, there are only waves of different lengths. The eye is a complex optical device capable of detecting differences in color, which correspond to a slight (about 10-6 cm) difference in the length of light waves.

Expected results:

1. Formation of students’ skills in working with studied formulas and skills in performing practical work.
2. Use mathematical knowledge to calculate the result experimental task.
3. Ability and skills of students to work with additional and reference literature.

1. Application of the studied material to perform test task
2. View in/fragment “Fraunhofer Diffraction”, frontal conversation on this material (questions are written on the board).
3. Work on the board. Solution of problem No. 2405 from the collection of problems in physics by G.N. Stepanova.
4. Execution experimental work on the topic “Determination of the wavelength of light (for a specified color) using a diffraction grating.”
5. Working with a reference book on physics and technology by A.S. Enochovich. Comparison of the results obtained with the data from the reference book and generalization of the results of the experiment.
6. Summing up the lesson. Assign differentiated homework.

Lesson objectives:

• Educational : Repeat formulas learned in previous lessons, apply mathematical knowledge to solve calculation problems. Use the studied material when solving problems and performing experimental work to determine the wavelength of light using a diffraction grating.
• Educational: Develop cognitive interest students, the ability to think logically and generalize. Develop motives for learning and interest in physics and mathematics. Develop the ability to see the connection between physics and mathematics. Improve students’ ability to highlight the main thing, analyze the conditions of a task, and develop a culture of oral and written speech.
• Educational To cultivate a love of student work, perseverance in achieving a goal, and the ability to work in pairs. Foster a culture of mathematical calculations. Mutual respect.

Progress of the lesson.

1. Repetition and generalization of the studied material

White light has a complex structure, knowing which can explain the variety of colors in nature. Using a diffraction grating or prism, white light can be separated into a spectrum, which consists of seven primary colors: red, orange, yellow, green, blue, indigo, violet. Outside of us, there are no colors in nature, there are only waves of different lengths. The eye is a complex optical device capable of detecting differences in color, which correspond to a slight (about 10-6 cm) difference in the length of light waves. In previous lessons we learned about the properties of light waves: interference, dispersion, diffraction, polarization.

Today we will summarize the knowledge gained in practice. But first, we will recall the material from the previous lesson, in which we got acquainted with the device and principle of operation optical device– diffraction grating.

2. Presentation on the topic: “Diffraction grating.”

The diffraction grating is based on the phenomenon of diffraction, which is a set of large number very narrow slits separated by opaque spaces. ( Appendix 1, slide 2)

The width of the transparent slits is equal to A, and the width of the opaque slits is equal to b.

a +b =d,d – period of the diffraction grating.

Let's consider elementary theory diffraction grating. Let a plane monochromatic wave of length λ be incident on the grating. (Appendix 1, slide 3).
Secondary sources in the slits create light waves that travel in all directions.

Let us find the condition under which the waves coming from the slits reinforce each other. For this purpose, let us consider waves propagating in the direction determined by the angle φ.
The path difference between the waves from the edges of adjacent slits is equal to the length of the segment AC . If this segment contains an integer number of wavelengths, then the waves from all the slits, adding up, will reinforce each other. From a triangle ABC you can find the length of the leg AC:
AC=ABsinφ.

Maximums will be observed at an angle φ , determined by the condition

d*sinφ =k * λ

It must be kept in mind that when performing this condition The waves coming from all other points of the slits are amplified. Each point in the first slit corresponds to a point in the second slit, located at a distance d from the first point. Therefore, the difference in the path of the secondary waves emitted by these points is equal to k * λ, and these waves are mutually amplified.
A collecting lens is placed behind the grating and behind it a screen on focal length from the lens. The lens focuses rays running parallel to one point. At this point, the waves combine and their mutual amplification occurs. Angles φ , satisfying the condition, determine the position of the maxima on the screen.

Since the position of the maxima (except for the central one, corresponding k = 0) depends on the wavelength, the grating decomposes white light into a spectrum (the spectra of the second and third orders overlap). The more λ , the further a particular maximum corresponding to a given wavelength is located from the central maximum. Each value has its own spectrum. Between the maxima there are minima of illumination. How larger number gaps, the more sharply defined the maxima are and the wider the minima they are separated by. (Appendix 1, slide 4) The light energy falling on the grating is redistributed by it so that most of it falls on the maxima, and a small part of the energy falls on the minima.
Using a diffraction grating, very precise wavelength measurements can be made. If the grating period is known, then determining the wavelength is reduced to measuring the angle φ , corresponding to the direction to the maximum. (Appendix 1, slide 5)

d * sin φ =k * λ

λ = , because the angles are small, then sin φ = tan φ

tan φ = , then λ = ,

Examples of diffraction gratings include: our eyelashes, with the spaces between them, are a rough diffraction grating. (Appendix 1, slide 6) Therefore, if you squint at a bright light source, you can see rainbow colors. White light is decomposed into a spectrum by diffraction around the eyelashes. A laser disk with grooves running close together is similar to a reflective diffraction grating. If you look at the light reflected by it from an electric light bulb, you will find the decomposition of light into a spectrum. Several spectra can be observed corresponding to different meanings k. The picture will be very clear if the light from the light bulb hits the plate at a large angle.

3. Performing a test task.

Option I.

1. 
   A.ν 1 = ν 2
   B. Δφ = 0
   IN.Δφ = const
   G.ν 1 = ν 2 , Δφ = const
2. λ ℓ 1 And ℓ 2 from point M. ( Figure 1) At point M the following is observed:
   A. Maximum;
   B. Minimum;
   IN. The answer is ambiguous;
   G.
3. n 1 n 2. What is the relationship between n 1 And n 2?
   A. n 1< n 2
   B. n 1 = n 2
   IN. n 1 > n 2
   G. the answer is ambiguous
4. d λ φ , under which the first main maximum is observed?
   A. sinφ =λ/d
   B. sinφ =d/λ
   IN. cos φ= λ/d
   WITH. cos φ= d/λ
5. 
   A.
   B. Diffraction sound waves, because . λsound>> λlight
   IN. λsound<< λсв .
   G.
6. 
   A. A
   B. b
   IN. or a or b depending on the size of the disk.

II option.

1. Light waves are coherent if:
   A.ν1 = ν2 , Δφ = const B.ν1 = ν2 IN. Δφ = 0 G. Δφ = const
2. Two coherent sources with wavelength λ located at different distances ℓ1 And ℓ2 from point M.( Figure 2) At point M the following is observed: A. Maximum; B. Minimum; IN. The answer is ambiguous; G. There is no correct answer among the answers A-B.
3. To “clarify” optics onto a glass surface with a refractive index n1 apply a thin transparent film with a refractive index n2. What is the relationship between n1 And n2?
   A. n1 = n2 B. n1 > n2 IN. n1< n2 G. the answer is ambiguous
4. Diffraction grating with period d illuminated by a normally incident light beam with a wavelength λ . Which of the following expressions defines the angle φ , under which the second main maximum is observed? A. sinφ = 2λ/d B. sinφ =d/2λ IN. cos φ= 2λ/d WITH. cos φ= d/2λ
5. What's in everyday life Is it easier to observe: diffraction of sound or light waves?
   A. Diffraction of light waves, because λsound<< λсв .
   B. Diffraction of light waves, due to the peculiarity of the visual organism - the eye.
   IN. Diffraction of sound waves, because they are longitudinal, and light waves are transverse.
   G. Diffraction of sound waves, because . λsound>> λlight
   
6. When a small disk is illuminated with monochromatic white light, a diffraction pattern is observed on the screen. In the center of the diffraction pattern the following is observed: A. white spot; b. dark spot.
   A. A B. b IN. or a or b depending on the radius of the hole.

View in/fragment “Fraunhofer Diffraction”.

Questions for this material:

1. What is a diffraction grating?
   Answer: A diffraction grating is a collection of a large number of very narrow slits separated by opaque spaces.
2. How do the spectra produced by a prism differ from diffraction spectra?
   Answer: Diffraction grating and prism - spectral devices - spectrum analyzers. The spectrum obtained using a prism is more stretched in the short-wavelength part, and compressed in the long-wavelength part, because The prism deflects violet rays more strongly. The diffraction grating deflects red rays more strongly, the spectrum is almost uniform.
3. What determines the angular distance between the maxima in the diffraction spectrum?
   Answer: The angular distance between the maxima in the diffraction spectrum depends on the constant of the diffraction grating. The smaller the diffraction grating constant, the greater the angular distance between the spectra.
4. What determines the resolving power of a device?
   Answer: The sharpness of the spectral lines increases with the number of slits; the greater the number of slits, the wider the spectrum; this determines the resolving power of the device.
5. What kind of gratings are called reflective?
   Answer: Since the end of the last century, reflective grilles have become widespread. In such gratings there are up to several thousand lines per 1 mm. The more lines per 1 mm, the greater the angular width of the spectrum.
6. What types of gratings do you know?
   Answer: Michelson echelon - diffraction at the edges of steps;
   Concave spherical lattice– serves as a focusing mirror without a lens;
   Crossed diffraction gratings - form a 2-dimensional diffraction structure that decomposes the spectrum into two coordinates;
   Disordered structure (dusty window) – forms rainbow rings;
   Human eyelashes, with the spaces between them, form a rough diffraction grating.
7. Name the optical instruments that use diffraction gratings and in what areas of science are they used?
   Answer: Diffraction gratings are used in spectroscopes, spectrographs, special microscopes, in astronomy, physics, chemistry, biology, technology, to study the absorption and reflection spectra of substances, to study optical properties various materials, in production for express analysis of various substances.

Many narrow slits at a short distance from each other form a wonderful optical device - a diffraction grating. The grating turns the light into a spectrum and allows you to very accurately measure the wavelength of light.

Before moving on to the experimental work, we will solve the problem of determining the wavelength using a diffraction grating and repeat the formula to determine the condition under which the waves coming from the slits reinforce each other.

Solving the problem. Work on the board.

No. 2405 – S.

Using a diffraction grating with a period of 0.02 mm, the first diffraction image was obtained at a distance of 3.6 cm from the central maximum and at a distance of 1.8 m from the grating. Find the wavelength of the light.

4. Completing the experimental task. Work in groups.

Subject: « Determination of the wavelength of light using a diffraction grating."

Experimental task: using the setup shown in Figure 3, determine the wavelength (of the indicated color).

Pay attention to the figure (Appendix 1, slide 7). The grid is installed in holder 2, which is attached to the end of ruler 1. On the ruler there is a black screen 3 with a narrow vertical slit in the middle. There are millimeter scales on the screen and ruler. The entire setup is mounted on a tripod.

Work order:

1. Move the scale with the aiming slit to the maximum possible distance from the diffraction grating. ( Appendix 2).
2. Point the axis of the device at a lamp with a straight filament. (in this case, the lamp filament should be visible through the narrow sighting filament of the shield. Look carefully first to the left and then to the right of the slit. In this case, diffraction patterns (spectra) will be visible to the right and left of the slit, on a black background above the scale).
3. Without moving the device, use the scale to determine the position of the centers of the color bands in the first-order spectra. Record the results in the table.
4. From the measurement data, calculate the wavelength. Compare it with the wavelength value for this color of light given in the reference book. Draw a conclusion.

d * sin φ =k * λ

λ = d * sin φ/ k, because the angles are small, then sin φ = tan φ

tan φ = , then λ =

Results table:

So, in today's lesson, we once again repeated the properties of light waves, carried out practical definition light wavelength using an optical device - a diffraction grating, compared the data obtained with reference results,

All this allowed us to conclude that a diffraction grating allows us to determine the wavelength of light with great accuracy.

Literature used.

1. Physics: Textbook. For 11th grade. general education institutions / G.Ya. Myakishev, B.B. Bukhovtsev. – 12th ed. – M: Education, 2004.
2. Physics: Textbook. For 11th grade. general education institutions / N.M.Shakhmaev, S.N.Shakhmaev, D.Sh.Shodiev – M: Education, 2000.
3. Wave optics: textbook. - M.: Bustard, 2003.
4. School course physics: tests and assignments. – M.: Shkola-Press, 1996.
5. Handbook of physics and technology: Textbook. A manual for students - M.: Education, 1989.
6. Collection of problems in physics for grades 10-11, author. G.N. Stepanova - M.: Education, 2001.

DETERMINING THE WAVELENGTH OF LIGHTBY USING

DIFFRACTION GRATING

PURPOSE OF THE WORK: Determine the wavelength of red and violet light.

EQUIPMENT: 1. A device for determining the wavelength of light,

2. light source, 3. diffraction grating.

THEORY: A parallel beam of light, passing through a diffraction grating, due to diffraction behind the grating, propagates in all possible directions and interferes. An interference pattern can be observed on a screen placed in the path of interfering light. Light maxima are observed at points on the screen for which the following condition is met:  =n, where D is the wave path difference,n– maximum number,l- light wavelength. The central maximum is called zero; for it  = 0. To the left and right of it are maxima of higher orders.

Diffraction Screen

lattice

The condition for the occurrence of a maximum can be written differently:

n = dsin

Whered– period of the diffraction grating,j– the angle at which the light maximum is visible (diffraction angle).

Since diffraction angles are, as a rule, small, for them we can take

sin  = tan ,Atan  = a/b

Therefore n×l = d×a/b

White light is complex in composition. The zero maximum for it is a white stripe, and the maximum of higher orders is a set of seven colored stripes, the totality of which is called the spectrum, respectively 1 th , 2 th , ... order, and the longer the wavelength, the further the maximum is from zero.

The diffraction spectrum can be obtained using a device to determine the wavelength of light.

ORDER OF WORK:

Place the lamp on the demonstration table and turn it on.

Looking through the diffraction grating, point the device at the lamp so that the lamp filament is visible through the window of the device screen.

Install the instrument screen at a distance of 400 mm from the diffraction grating and obtain a clear image of the spectra on it 1 th and 2 th orders of magnitude.

Determine the distance from the zero division “0” of the screen scale to the middle of the purple stripe, as in left side"A l ", and to the right "a n ", for first order spectra and calculate the average value "a sr.f »

A sr.f1 = (a l + a n ) / 2

diffraction grating

screen

Repeat the experiment with a second order spectrum. Determine a for him sr.f2

Perform the same measurements for the red stripes diffraction spectrum.

Calculate the wavelength of violet light, the wavelength of red light (for 1 th and 2 th orders) according to the formula:

= ,

Whered = 10 -5 m – constant (period) lattice,

n – spectrum order,

b– distance from the diffraction grating to the screen, mm

8. Determine the average values:

λ f = ; λ cr =

9. Determine measurement errors:

absolute –Δ λ f = |λ sr.f. - λ tab.f. | ; Whereλ tab.f = 0.4 µm

– Δ λ cr = |λ Wed cr. - λ tab.cr. | ; Whereλ tab.cr = 0.76 µm

relative –δ λ f = %; δ λ cr = %

10. Prepare a report. Enter the results of measurements and calculations into the table.

spectrum

spectrum edge

violet. colors

spectrum edge

red colors

light wavelength

op.

« A l »,

mm

« A n »,

mm

« A Wed »

mm

« A l »,

mm

« A n »,

mm

« A Wed »

mm

f ,

cr ,

11. Draw a conclusion.

TEST QUESTIONS:

1. What is light diffraction?

   What is a diffraction grating?

   At what points on the screen are the 1st, 2nd, 3rd maximums obtained? What do they look like?

   Determine the constant of the diffraction grating if, when illuminated with light with a wavelength of 600 nm, the second-order maximum is visible at an angle of 7

   Determine the wavelength if the first-order maximum is 36 mm from the zero maximum, and a diffraction grating with a constant of 0.01 mm is located at a distance of 500 mm from the screen.

   Determine the wavelength incident on a diffraction grating with 400 lines on each millimeter. The diffraction grating c is located at a distance of 25 cm from the screen, the third-order maximum is 27.4 cm away from the zero maximum.

Laboratory work №6

Determining the wavelength of light

Purpose of the work : Determine the wavelength of light using a diffraction grating.

Equipment:

a diffraction grating with a period indicated on it;

measuring installation;

semiconductor laser (laser pointer).

Work progress

In this work, to determine the wavelength of light, we use diffractionlattice with a period (the period is indicated on the hash mark). It is the main part of the measuring setup shown in Figure 1 .

Before starting laboratory work, place the screen on the bench so that when you turn on the laser with the button, the red dot coincides with the zero division of the screen scale.

Place the frame in the holder diffraction grating and turn on the laser. A pattern of maxima and minima is formed on the screen, coming from different slits of the grating in the same direction. This picture shows a series of bright red dots radiating symmetrically from a central spot - the zero maximum. By changing diffraction gratings, observe how the diffraction pattern changes depending on the number of lines per millimeter.

To) exactly coincides with the entire millimeter division of the screen scale, and measure the distance b from it to the central maximum. Determine the distance A along a ruler on the bench from the screen to the bars.

The wavelength is determined by the formula:
,

Where: d - grating period; To - spectrum order;

- the angle at which the maximum light of the corresponding color is observed;

Since the angles at which the 1st and 2nd order maxima are observed do not exceed 5 0, their tangents can be used instead of the sines of the angles.

From Figure 2 it is clear that
.

Distance counted along a ruler from the grating to the screen, distance b - on the screen scale from the slit to the selected spectrum line.

ABOUT

the final formula for determining the wavelength is:

Prepare a report form with a table to record the results of measurements and calculations.

Collect measuring setup, install the screen at an arbitrary distance from the grille.

After observing a qualitative picture of a series of maximums, move the slider with the grid along the groove of the bench so that any maximum (write down its number) To) exactly coincides with the whole millimeter division of the screen scale, and measure the distance b from it to the central maximum.

Determine the position of the centers of the color bands in the 1st order spectra.

Enter the data into the table.

From the measurement data, calculate the wavelengths

Compare your results with table value wavelength of the visible part of the spectrum.

Carry out the experiment with another diffraction grating and compare the results obtained with each other and the table ones.

To avoid eye damage, it is strictly forbidden to direct the laser beam at a person’s face.

Security Question:

How does the diffraction spectrum differ from the dispersion spectrum?

Laboratory work No. 2 (solutions, answers) in physics, grade 11 - Determination of a light wave using a diffraction grating

2. Install the screen at a distance L ~ 45-50 cm from the diffraction grating. Measure L at least 5 times, calculate the average value . Enter the data into the table.

5. Calculate the averages. Enter the data into the table.

6. Calculate the lattice period d, write its value in the table.

7. By measured distance from the center of the slit in the screen to the position of the red edge of the spectrum and the distance from the diffraction grating to the screen, calculate sin0cr, under which the corresponding spectrum band is observed.

8. Calculate the wavelength corresponding to the red edge of the spectrum perceived by the eye.

9. Determine the wavelength for the violet end of the spectrum.

10. Calculate the absolute errors in measuring distances L and l.

L = 0.0005 m + 0.0005 m = 0.001 m
l = 0.0005 m + 0.0005 m = 0.001 m

11. Calculate the absolute and relative error wavelength measurements.

Answers to security questions

1. Explain the principle of operation of a diffraction grating.

The principle of operation is the same as that of prisms - deflection of transmitted light at a certain angle. The angle depends on the wavelength of the incident light. The longer the wavelength, the larger the angle. It is a system of identical parallel slits in a flat opaque screen.

Click to enlarge

2. Indicate the order of the primary colors in the diffraction spectrum?

In the diffraction spectrum: violet, blue, cyan, green, yellow, orange and red.

3. How will the diffraction spectrum change if you use a grating with a period 2 times greater than in your experiment? 2 times smaller?

Spectrum in general case there is a frequency distribution. Spatial frequency is the quantity inverse period. It is therefore obvious that doubling the period leads to a compression of the spectrum, and decreasing the spectrum will lead to a doubling of the spectrum.

Conclusions: A diffraction grating allows one to very accurately measure the wavelength of light.

National research university"MEI"

(Moscow Energy Institute)

Department of Physics named after. V. A. Fabrikanta

Lab 3

at the rate " General physics»

Determining the wavelength of light using a diffraction grating

Completed:

2nd year student

gr. FM-1-14

Navoev M. M.

Accepted:

senior lecturer

Bamburkina I. A.

Moscow 2015

Purpose of the work: observing the diffraction spectrum of a grating, measuring the wavelengths of light emitted by a spectral lamp, and studying the spectroscopic characteristics of a diffraction grating.

1. Introduction

A flat transparent diffraction grating is a system of equally spaced transparent narrow slits separated by opaque stripes. Width sum b cracks and opaque stripes a called lattice period d(Fig. 1).

Let a plane monochromatic wave fall on the grating perpendicular to its surface. After the wave passes through the grating, the direction of wave propagation changes and diffraction occurs.

Diffraction in parallel rays commonly called Fraunhofer diffraction. To fulfill the conditions for the formation and observation of the diffraction spectrum of the grating, the following scheme is used (Fig. 2). Monochromatic light from source 1 illuminates the crack 2 , located in the focal plane of the collecting lens 3 . After the lens 3 parallel beam of light incident on a diffraction grating 4 . The light wave diffracts when passing through the grating, forming secondary coherent waves. They are collected by the lens 5 on the screen in its focal plane 6 .

Light intensity distribution in diffraction pattern we obtain if we take into account the intensity distribution during diffraction at each slit and the redistribution of energy in space due to the interference of waves from all slits. At small diffraction angles the calculation is easier graphical method addition of amplitudes.

Let a slit whose length be l much larger than its width b (l >> b) a parallel beam of light falls. According to the Huygens-Fresnel principle, each point on the wave surface becomes a source of secondary spherical waves propagating in all directions at diffraction angles q. These waves are coherent and can interfere when superimposed. Let's break the open part wave front in the plane of the slit into narrow strips of equal width, length l, parallel to the edges of the slot (see Fig. 3). Each such strip will play the role of a secondary source of waves. Since the areas of the strips are equal, the vibration amplitudes Δ A i, coming from these sources will be equal to each other, and also equal initial phases these waves, since the plane of the slit coincides with wave surface falling wave. Oscillations from each strip will arrive at the observation point with the same phase lag, which, in turn, depends on the diffraction angle q. This lag can be found from the relationship (Fig. 3).

The phase difference of the rays coming from the edges of the slit, where – geometric difference the course of the extreme rays (Fig. 3).

To find the resulting amplitude of oscillations of waves arriving at observation point P, we proceed as follows. We represent the amplitude of the oscillations sent by each strip in the form of a vector, the lag of these oscillations in phase by the amount g i, we represent it by rotating the vector counterclockwise. Then the sum of vectors will look like a chain of vectors, identical in magnitude and rotated relative to each other by the same angle g i(Fig. 4). The resulting amplitude () is a vector that is a chord of a circular arc of radius R. It's obvious that . Let us denote by A 0 length of the arc consisting of chain links (). Since, then. From these two relations we obtain that . Since the light intensity I ~ A 2, then for the distribution of screen illumination we obtain the formula:

Where . Zero illumination (diffraction minimum) will be observed at points where, i.e. at (At g = 0, all vectors line up along a straight line, and I = I 0 – zero maximum).

From here we obtain the condition for the minima during the diffraction of light by one slit:

, m = 1, 2, 3… (2)

Dependency graph I from sin q is shown in Fig. 5.

The diffraction grating contains N such cracks (up to a thousand or more). When light falls on the grating, each of the slits will give a picture in the plane of the screen as shown in Fig. 5.

When superimposed, these patterns will spatially coincide, since their spatial position is determined not by where the rays came from, but by the angle q at which these rays go (in Fig. 2 it can be seen that the rays emerging from different slits, but at the same angle same angle q, will hit one point on the screen). If the waves coming from the slits were not coherent, then such an overlap would lead to simple increase light intensity on the screen N times compared to the illumination from a single slit. But these waves are coherent and this leads to a new redistribution of energy on the screen, but within each of the maxima from one slit.

To find this new redistribution of energy, consider the rays coming from two corresponding points of adjacent slits, i.e. from points lying at a distance d from each other (Fig. 1). The path difference D of the waves coming from these points at the diffraction angle q is equal to (Fig. 1).

If the interference maximum condition is met – then a light stripe will be located on the screen in the appropriate place.

Thus, the position of the so-called main maxima is determined by the formula:

, n = 0, 1, 2, 3… (3)

Intensity minima during mutual interference occur in cases where the phase difference of waves coming from adjacent slits is equal, etc. For these diffraction angles, the chain of vectors closes into a circle once (Fig. 4a), twice, etc. and the total vector . That is, these diffraction angles correspond to the so-called additional minimums, the position of which can be found using the formula

, k= 1, 2, 3…, but k N, 2N, 3N… (4)

Thus, between the main maxima there is N– 1 additional minimum. Between the additional lows are weak secondary highs. The number of these maxima falling within the interval between adjacent main maxima is equal to N – 2.

Diffraction angles in the direction of which none of the slits send light correspond to major lows, which are determined by formula (2).

The resulting picture of the light intensity distribution on the screen taking into account formulas (1), (2), (3) and (4) is presented in Fig. 6. Here the dotted line repeats the intensity distribution during diffraction by a single slit.

When a grating is illuminated with non-monochromatic light, diffraction is accompanied by the decomposition of light into a spectrum. The central maximum will have the same color as the source, since at q = 0 light waves of any length have zero path difference. To the left and right of it there will be maxima for different wavelengths of the 1st, 2nd, etc. orders of magnitude, and the longer wavelength will correspond to larger angle diffraction q. Thus, a diffraction grating can serve as a spectral device (Fig. 7). The main purpose of such devices is to measure the wavelength of the light being studied.

2. Description of installation and measurement method

The problem of measuring wavelength using a grating with a known constant d reduces to measuring the angles q at which diffraction maxima are observed.

The optical diagram of the installation is shown in Fig. 8.

Light source 1 illuminates the crack 2 , located in the focal plane of the lens 3 collimator. After the collimator, a parallel beam of light falls normally onto the diffraction grating 4 installed on the device table. Diffracted light wave hits the lens 5 telescope 6 and observed through the eyepiece 7 .

Diffraction angles are measured using an optical device - a goniometer (Fig. 9).

Its main parts: spotting scope 1 , her eyepiece 2 , tube focusing screw 3 , reading microscope 4 , table 5 , collimator 6 , micrometric collimator screw 7 , which regulates the size of the collimator slit. The telescope is mounted on a rotating base 8 .

The angles at which diffraction maxima are observed are measured using a reading device. The magnitude of the angle q is determined by the limb, which is viewed through the microscope eyepiece 4 with the lights on. On the surface of the glass dial there is a scale with divisions from 0° to 360°. The divisions are digitized in 1° increments. Each degree is divided into three parts. Therefore, the price of dividing the limb is 20." (With accepted method measurements do not use the reverse image and the scale in the right window of the field of view of the reference microscope.) The field of view of the reference microscope is shown in Fig. 10.

The counting is carried out as follows. In the left window there are images of diametrically opposite sections of the limb and a vertical index for counting degrees. The number of degrees is equal to the visible number closest to the left of the vertical index on the upper scale. The number of minutes is determined with an accuracy of 5" by the position of the vertical index. The reading in the figure is approximately equal to 0°15´.

3. Work order

1. Turn on the light source (spectral lamp) in front of the collimator slit. The lamp lights up within 5-7 minutes.

2. Let's get acquainted with the installation and fill out the table of specifications of measuring instruments.

3. By rotating the telescope, align the eyepiece crosshair with the image of the collimator slit. The image of the slit should be clearly visible and approximately 1 mm wide.

4. By rotating the tube eyepiece frame, we will achieve a clear image of the crosshair in the field of view of the eyepiece.

5. We install a diffraction grating with a known constant on the goniometer table so that its plane is perpendicular to the collimator axis.

6. Turn on the goniometer lighting.

7. By turning the telescope left and right, we observe the lines of the lamp spectrum, located symmetrically from the zero (uncolored) maximum. The telescope should be rotated slowly and smoothly. Let us determine the number of visible orders of the spectrum on each side of the zero maximum. At the same time, we will make sure that the reading on the limb scale when observing spectral lines does not go beyond the angle range from 20° to 270°. Otherwise, release the table screw 5 and turning the nozzle with this screw around vertical axis device, we enter the required section of the limb. Then fasten the screw again. This makes it possible not to cross the zero scale of the dial during measurements and thereby simplifies calculations.

8. Let's measure the angles at which different lines are observed in the spectra ±1, ±2, ±3, etc. orders of magnitude. To do this, we sequentially draw the crosshairs of the telescope eyepiece to each line to the left and right of the central one. We take a reading along the limb using a reading microscope, as described above.

9. We will enter the measurement data into the table. 1. When measuring through α the angular position of the spectrum lines is indicated to the right of the zero maximum, and by β - to the left of the zero maximum.

Table 1

Lattice constant d = 6,03*10 -5

4. Processing of measurement results

1. Calculate the diffraction angle q using the formula

2. For each value of the angle q, we find the wavelength using the formula

(violet),

(green).

3. Calculate the average wavelength for a line of a given color. We write the calculation results in the table. 1.

4. From formula (6) we derive the formula for calculating the error Δλ and calculate the error. Δα = Δβ = 5´.

5. Write down the final result

5. Additional task

The main characteristics of a spectral device are angular dispersion and resolution.

Determination of angular dispersion

Angular dispersion– characteristic of the device’s ability to spatially separate waves of different lengths. If two lines differ in wavelength by δλ and there is a corresponding angle difference δq, then the measure of angular dispersion is .

Let there be two close spectral lines with wavelengths λ 1 and λ 2. The distance between the maxima δq for wavelengths λ 1 and λ 2 is found from the condition of the main intensity maxima. After differentiation in formula (3) we have: d·cos(q)·δq = nδλ. Where

Let's take measurements angular distances for the yellow doublet in all visible spectral orders.

Knowing the difference δλ = λ 1 – λ 2, we calculate the angular dispersion of the diffraction grating in the spectrum of the 1st and 2nd orders (or other orders). Dimension D– min/nm.

The obtained result is comparable to the theoretical one (formula 7).

During laboratory work, measurements of two light waves were made. It was found that they correspond to the table values.