Conclusion to the laboratory measurement of the wavelength of light. in a diamond is...

Topic: "Measurement of the wavelength of light using grating».

Lesson Objectives: experimentally obtain the diffraction spectrum and determine the wavelength of light using a diffraction grating;

to cultivate attentiveness, goodwill, tolerance in the process of working in small groups;

develop an interest in the study of physics.

Lesson type: a lesson in the formation of skills and abilities.

Equipment: wavelengths, OT instructions, lab instructions, computers.

Conducting methods: laboratory work, work in groups.

Interdisciplinary connections: mathematics, informatics ICT.

All knowledge real world

comes from experience and ends with it

AND.Einstein.

During the classes

I. Organizing time.

Message about the topic and purpose of the lesson.

ІІ. 1. Actualization of basic knowledge. Survey of students (Appendix 1).

Performing laboratory work.

Students are encouraged to measure the wavelength of light using a diffraction grating.

Students are united in small groups (4-5 people each) and together they perform laboratory work according to the instructions. With the help of computer Excel programs make calculations and the results of the work are entered in a table (in the Word program).

Evaluation criteria:

The team that completed the task first receives - a score of 5;

the second - a score of 4;

third - score 3

Life safety rules during work.

Work in groups under the guidance of a teacher.

Generalization and systematization by students of the results of work.

The result of the work is entered into a table on the computer (Appendix 2).

ІІІ.

Summarizing. Compare the obtained results with tabular data. Draw conclusions.

Reflection.

Did everything turn out the way I envisioned?

What was done well?

What went wrong?

What was easy to do, and what turned out to be unexpectedly difficult?

Work in small group helped me or created additional difficulties?

VI. Homework.

File a job.

Repeat theoretical material on the topic "Interference and diffraction of light".

Make a crossword puzzle on the topic "Properties of electromagnetic waves."

Addition 1

1. What is light?

2. What it consists of White light?

3. Why is light called visible radiation?

4. How to decompose white light into a color spectrum?

5. What is a diffraction grating?

6. What can be measured with a diffraction grating?

7. Can two different colors light waves, for example, red and green radiation, have same lengths waves?

8. And in one environment?

Supplement 2

Red

10 -7 m

Orange

10 -7 m

Yellow

10 -7 m

Green

10 -7 m

Blue

10 -7 m

Blue

10 -7 m

Purple

10 -7 m

Laboratory work

Theme: Measurement of the length of a light wave.

Objective: measure the wavelength of red and purple flowers, compare the obtained values ​​with the tabular ones.

Equipment: electric bulb with a straight filament, a device for determining light wavelength.

Theoretical part

In this work, a diffraction grating with a period of 1/100 mm or 1/50 mm is used to determine the length of a light wave (the period is indicated on the grating). It is the main part of the measuring setup shown in the figure. The grate 1 is installed in the holder 2, which is attached to the end of the line 3. On the line there is a black screen 4 with a narrow vertical slot 5 in the middle. The screen can move along the ruler, which allows you to change the distance between it and the diffraction grating. There are millimeter scales on the screen and ruler. The entire setup is mounted on a tripod 6.

If you look through the grating and the slot at the light source (an incandescent lamp or a candle), then on the black background of the screen it is possible to observe diffraction spectra of the 1st, 2nd, etc. orders on both sides of the slit.

Rice. 1

Wavelengthλ is determined by the formulaλ = dsinφ/k , whered - grating period; k - the order of the spectrum;φ - the angle at which the maximum light of the corresponding color is observed.

Since the angles at which the maxima of the 1st and 2nd orders are observed do not exceed 5°, one can use their tangents instead of the sines of the angles. It can be seen from the figure thattgφ = b/a . Distancea counted along the ruler from the grate to the screen, the distanceb - on the scale of the screen from the slit to the selected line of the spectrum.

Rice. 2

The final formula for determining wavelength has the formλ = db/ka

In this work, the measurement error of wavelengths is not estimated due to some uncertainty in the choice of the middle part of the spectrum of a given color.

Work can be done using instructions #2 or #2

Instruction #1

Progress

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

2. Assemble the measuring setup, install the screen at a distance of 50 cm from the grid.

3. Looking through the diffraction grating and the slit in the screen at the light source and moving the grating in the holder, set it so that the diffraction spectra are parallel to the screen scale.

4. Calculate the wavelength of the red color in the 1st order spectrum to the right and left of the slit in the screen, determine the average value of the measurement results.

5. Do the same forothersColourov.

6. Compare your results withtabularwavelengths.

Instruction number 2

Progress

Measure the distance b to the corresponding color in the spectrum first in line to the left and right of the central maximum. Measure the distance from the diffraction grating to the screen (see Figure 2).

Determine or calculate the grating period d.

Calculate the length of light for each of the seven colors in the spectrum.

Record the results of measurements and calculations in the table:

Colour

b ,left,m

b ,right,m

b , average, m

a ,m

Order

spectrumk

Lattice period

d , m

measuredλ , nm

fipurple

Synuy

Blue

Greenth

Yellow

Orangeth

Red

4. Calculate relative error experiment for each color according to the formula

Lesson-study

Self-control table

		Multimedia	Pages of history	Trust but check	Terms. Formulas.	Additionally
student

Testing

Lesson-study

on the topic "Determination of the wavelength of light"

Self-control table

F. I. student ___________________________

	Testing ( level A,B,C )	Multimedia	Pages of history	Trust but check	Terms. Formulas.	Additionally
student

Testing

"Lesson Development"

Lesson - study

(Grade 11)

Length determination

light wave

Teacher: Radchenko M.I.

Theme: Determining the wavelength of light. Laboratory work "Measuring the length of a light wave."

Lesson - research. ( Appendix.)

Goals:

Generalize, systematize knowledge about the nature of light, experimentally investigate the dependence of the wavelength of light on others physical quantities, to teach to see the manifestations of the studied patterns in surrounding life, to form the skills of teamwork in combination with the independence of students, the education of motives for learning.

Without a doubt, all our knowledge begins with experience.

Kant Immanuel

(German philosopher, 1724-1804)

Decor - portraits of scientists curriculum vitae, achievements in science. Main links scientific creativity Keywords: initial facts, hypothesis, consequences, experiment, initial facts.

During the classes

Org. moment.

Introduction by the teacher. The topic of the lesson and goals are made in Power Point, projected over the network on monitor screens and interactive whiteboard.

The teacher reads out and explains the words of the epigraph and the main links of scientific creativity

Knowledge update. Repetition, generalization of the studied material about the nature of light. Problem solving. Students present their results theoretical research prepared in the form of presentations in Power Point (dispersion, interference, light diffraction, diffraction grating. Applications).

Performing laboratory work"Measuring the wavelength of light".(Application, textbook material.) Analysis of the obtained results, conclusions.

Computer testing. Tasks are prepared in four levels of difficulty. The result is entered in the "Table of self-control". ( Appendix).

Summarizing.

Students fill out self-control tables with marks on various types activities.

The teacher analyzes the results of the work together with the students.

View document content
"Light phenomena level A"

LIGHT PHENOMENA

Level A

A. TV.

B. Mirror.

G. Sun.

2. In order to find out the speed of light in an unknown transparent substance, it is enough to determine ...

A. Density.

B. Temperature.

B. Elasticity.

G. Pressure.

D. Index of refraction.

3. A light wave is characterized by its wavelength, frequency and propagation speed. When moving from one environment to another, it does not change ...

A. Speed.

B. Temperature.

B. Wavelength.

D. Frequency only.

D. Index of refraction.

4. Optical system The eye builds an image of distant objects behind the retina. What is this vision defect and what lenses are needed for glasses?

B. Myopia, collecting.

B. There is no visual defect.

5. If the refractive index of diamond is 2.4, then the speed of light (s = 3 * 10 8 m / s)

in a diamond is...

A. 200,000 km/s.

B. 720,000 km/s.

V. 125,000 km/s.

D. 725,000 km/s.

D. 300,000 km/s.

B. The wavelength changes.

D. Only the frequency is the same.

7. A person approaches a flat mirror at a speed of 2 m/s. The speed at which he approaches his image is ...

A. Lightning.

B. Shine precious stones.

V. Rainbow.

G. The shadow of the tree.

9. During operation, the light should fall ...

A. Right.

B. From above.

G. Front.

10.

AND. flat mirror.

B. Glass plate.

B. Converging lens.

D. Diverging lens.

11. An image on the retina of the eye ...

View document content
"Light Phenomena Level B"

LIGHT PHENOMENA

Level B

1. In order to find out the speed of light in an unknown transparent substance, it is enough to determine ...

A. Density.

B. Temperature.

B. Elasticity.

G. Pressure.

D. Index of refraction.

2. A light wave is characterized by its wavelength, frequency and propagation speed. When moving from one environment to another, it does not change ...

A. Speed.

B. Temperature.

B. Wavelength.

D. Frequency only.

D. Index of refraction.

3. The optical system of the eye builds an image of distant objects behind the retina. What is this vision defect and what lenses are needed for glasses?

A. Farsightedness, collecting.

B. Myopia, collecting.

B. There is no visual defect.

G. Myopia, scattering.

D. Hyperopia, scattering.

4. If the refractive index of a diamond is 2.4, then the speed of light (c \u003d 3 * 10 8 m / s)

in a diamond is...

A. 200,000 km/s.

B. 720,000 km/s.

V. 125,000 km/s.

D. 725,000 km/s.

D. 300,000 km/s.

5. Determine the wavelength if its speed is 1500 m/s and the oscillation frequency is 500 Hz.

B. 7.5 * 10 5 m.

D. 0.75 * 10 5 m.

6. A reflected wave occurs if ...

A. The wave falls on the interface between media with different densities.

B. The wave falls on the interface between media with the same density.

B. The wavelength changes.

D. Only the frequency is the same.

D. The refractive index is the same.

7. A person approaches a flat mirror at a speed of 2 m/s. The speed at which he approaches his image is ...

8. Which of the following phenomena is explained rectilinear propagation Sveta?

A. Lightning.

B. Shine of precious stones.

V. Rainbow.

G. The shadow of the tree.

9. What optical device can give an enlarged and real image of an object?

A. Flat mirror.

B. Glass plate.

B. Converging lens.

D. Diverging lens.

10. An image on the retina of the eye ...

A. Increased, direct, real.

B. Reduced, inverted (reverse), real.

B. Reduced, direct, imaginary.

G. Enlarged, inverted (reverse), imaginary.

11. Find the period of the grating if the first-order diffraction image is obtained at a distance of 2.43 cm from the central one, and the distance from the grating to the screen is 1 m. The grating was illuminated with light with a wavelength of 486 nm.

View document content
"Light phenomena level D"

LIGHT PHENOMENA

Level D

1. From the bodies listed below, select the body that is natural source Sveta.

A. TV.

B. Mirror.

G. Sun.

2. The angle of incidence of the light beam is 30º. The angle of reflection of the light beam is equal to:

3. When solar eclipse on the Earth, a shadow and penumbra from the Moon are formed (see fig.). What does the person in the shadow at point A see?

4. 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 length of the light wave.

5. The focal length of a biconvex lens is 40 cm. To get the image of an object in full size, it must be placed from the lens at a distance equal to ...

6. First diffraction maximum for light with a wavelength of 0.5 μm, it is observed at an angle of 30 degrees to the normal. At 1 mm, the diffraction grating contains strokes ...

7. When photographing from a distance of 200 m, the height of the tree on the negative turned out to be 5 mm. If the focal length of the lens is 50 mm, then the actual height of the tree is ...

8. In order to find out the speed of light in an unknown transparent substance, it is enough to determine ...

A. Density.

B. Temperature.

B. Elasticity.

G. Pressure.

D. Index of refraction.

9. A light wave is characterized by wavelength, frequency and propagation speed. When moving from one environment to another, it does not change ...

A. Speed.

B. Temperature.

B. Wavelength.

D. Frequency only.

D. Index of refraction.

10. The optical system of the eye builds an image of distant objects behind the retina. What is this vision defect and what lenses are needed for glasses?

A. Farsightedness, collecting.

B. Myopia, collecting.

B. There is no visual defect.

G. Myopia, scattering.

D. Hyperopia, scattering.

11. Determine the wavelength if its speed is 1500 m/s and the oscillation frequency is 500 Hz.

B. 7.5 * 10 5 m.

D. 0.75 * 10 5 m.

12. If the refractive index of a diamond is 2.4, then the speed of light (c \u003d 3 * 10 8 m / s)

in a diamond is...

A. 200,000 km/s.

B. 720,000 km/s.

V. 125,000 km/s.

D. 725,000 km/s.

D. 300,000 km/s.

13. A reflected wave occurs if ...

A. The wave falls on the interface between media with different densities.

B. The wave falls on the interface between media with the same density.

B. The wavelength changes.

D. Only the frequency is the same.

D. The refractive index is the same.

14. A person approaches a flat mirror at a speed of 2 m/s. The speed at which he approaches his image is ...

15. Find the period of the grating if the first-order diffraction image was obtained at a distance of 2.43 cm from the central one, and the distance from the grating to the screen was 1 m. The grating was illuminated with light with a wavelength of 486 nm.

16. The optical system of the eye adapts to the perception of objects located at different distances due to ...

A. Changes in the curvature of the lens.

B. Additional lighting.

B. Approximation and removal of objects.

G. Light stimulation.

1 7. Which of the following phenomena is explained by the rectilinear propagation of light?

A. Lightning.

B. Shine of precious stones.

V. Rainbow.

G. The shadow of the tree.

18. What optical device can give an enlarged and real image of an object?

A. Flat mirror.

B. Glass plate.

B. Converging lens.

D. Diverging lens.

19. During operation, the light should fall ...

A. Right.

B. From above.

G. Front.

20. An image on the retina of the eye ...

A. Increased, direct, real.

B. Reduced, inverted (reverse), real.

B. Reduced, direct, imaginary.

G. Enlarged, inverted (reverse), imaginary.

"Diffraction grating."

Diffraction grating

The device of a remarkable optical device, the diffraction grating, is based on the phenomenon of diffraction.

Determining the wavelength of light

AC=AB*sin φ=D*sin φ

Where k=0,1,2...

View presentation content
"Diffraction"

Diffraction

straightness deviation

wave propagation, wave bending around obstacles

Diffraction

mechanical waves

Diffraction

Experience cabin boy

Fresnel theory

Yung Thomas (1773-1829) English scientist

Fresnel Augustin (1788 - 1821) French physicist

View presentation content
"Interference"

Interference

Addition in the space of waves, in which a time-constant distribution of the amplitudes of the resulting oscillations is formed

Discovery of interference

Newton observed the phenomenon of interference

Discovery and term interference belong to Jung

Maximum condition

• The amplitude of oscillations of the medium at a given point is maximum if the difference between the paths of two waves that excite oscillations at this point is equal to an integer number of wavelengths

∆ d=k λ

Minima condition

• The amplitude of oscillations of the medium at a given point is minimal if the difference between the paths of two waves that excite oscillations at this point is equal to an odd number of half-waves.

∆ d=(2k+1) λ /2

“A soap bubble floating in the air… lights up with all the shades of colors inherent in the surrounding objects. The soap bubble is perhaps the most exquisite miracle of nature.

Mark Twain

Interference in thin films

• The difference in color is due to the difference in wavelength. Light beams of different colors correspond to waves of different lengths. Mutual amplification of waves requires different film thicknesses. Therefore, if the film has an unequal thickness, then when it is illuminated with white light, different colors should appear.

• A simple interference pattern occurs in a thin layer of air between a glass plate and a plano-convex lens placed on it, the spherical surface of which has a large radius of curvature.

• Waves 1 and 2 are coherent. If the second wave lags behind the first by an integer number of wavelengths, then, adding up, the waves amplify each other. The vibrations they cause occur in one phase.
• If the second wave lags behind the first by odd number half-waves, then the oscillations caused by them will occur in opposite phases and the waves cancel each other

• Checking the quality of surface treatment.
• It is necessary to create a thin wedge-shaped layer of air between the surface of the sample and a very smooth reference plate. Then the irregularities will cause noticeable curvature of the interference fringes.

• Illumination of optics. Part of the beam after multiple reflections from internal surfaces still passes through the optical device, but is scattered and no longer participates in creating a clear image. To eliminate these consequences, optical enlightenment is used. A thin film is applied to the surface of the optical glass. If the amplitudes of the reflected waves are the same or very close to each other, then the extinction of the light will be complete. Lens reflection cancellation means that all light passes through the lens.

View presentation content
"Determination of the wavelength of light l p"

Formula:

λ =( d sin φ ) /k ,

where d - grating period, k – spectrum order, φ is the angle at which the maximum light is observed

Distance a is measured along the ruler from the grating to the screen, distance b is measured along the screen scale from the slit to the selected spectrum line

Maximum light

Final Formula

λ = db/ka

light wave

Interference experiments allow you to measure the wavelength of light: it is very small - from 4 * 10 -7 to 8 * 10 -7 m

Diffraction grating

Objective

Using a diffraction grating, obtain a spectrum, study it. Determine the wavelength of violet, green and red rays

Theoretical part of the work

A parallel beam of light, passing through a diffraction grating, due to diffraction behind the grating, propagates throughout possible directions and interfere. An interference pattern can be observed on a screen placed in the path of the interfering light. At point O of the screen placed behind the grating, the difference in the path of rays of any color will be equal to zero, here there will be a central zero maximum - white stripe. At the screen point, for which the path difference of the violet rays will be equal to the wavelength of these rays, the rays will have the same phases; there will be a maximum - a purple stripe - F. At the point of the screen, for which the difference in the path of the red rays will be equal to their wavelength, there will be a maximum for the red light rays - K. Between the points F and K there will be maxima of all other components white color in ascending order of wavelength. A diffraction spectrum is formed. Immediately behind the first spectrum is the spectrum of the second order. The wavelength can be determined by the formula:

Where λ is the wavelength, m

φ is the angle at which the maximum is observed for a given wavelength,

d is the period of the diffraction grating d= 10 -5 m,

k is the order of the spectrum.

Since the angles at which the maxima of the first and second orders are observed do not exceed 5 0, it is possible to use their tangents instead of the sines of the angles:

where a is the distance from the center of the window to the middle of the rays of the spectrum, m;

ℓ - distance from the diffraction grating to the screen, m

Then the wavelength can be determined by the formula:

Equipment

A device for determining the length of a light wave, a diffraction grating, an incandescent lamp.

Progress

1. Install the screen at a distance of 40-50 cm from the grille (ℓ).

2. Looking through the grating and the slit in the screen at the light source, ensure that the diffraction spectra are clearly visible on both sides of the slit.

3. On the scale on the screen, determine the distance from the center of the window to the middle of the violet, green and red rays (a), calculate the wavelength of the light by the formula: ,

4. Changing the distance from the grating to the screen (ℓ), repeat the experiment for the second-order spectrum for rays of the same color.

5. Find the average wavelength for each of the monochromatic rays and compare with the tabular data.

Table Wavelength values ​​for some colors of the spectrum

Table Results of measurements and calculations

Computing

1. For the spectrum of the first order: k=1 , d= , ℓ 1 =

a f1 = , a h1 = , and kr1 =

Wavelength for the first order spectrum:

- purple: , λ f1 =

- Green colour: , λ c1 =

- Red: , λ cr1 =

2. For the spectrum of the second order: k=2 , d= , ℓ 2 =

a φ2 = , a z2 = , and kr2 =

Wavelength for second order spectrum:

- purple: , λ f2 =

- Green colour: , λ z2 =

- Red: , λ cr2 =

3. Average value of wavelengths:

- purple: , λ fsr =

- Green colour: , λ sav =

- Red: , λ rsr =

Conclusion

Record responses to questions complete sentences

1. What is called light diffraction?

2. What is called a diffraction grating?

3. What is called the lattice period?

4. Write down the lattice period formula and comments on it

Lab #6

"Measuring the length of a light wave with a diffraction grating"

Belyan L.F.,

Physics teacher

MBOU "Secondary School No. 46"

city ​​of Bratsk

Objective:

Continue the formation of ideas about the phenomenon of diffraction.

Learn how to determine the length of a light wave using a diffraction grating with a known period.

k=-3 k=-2 k=-1 k=0 k=1 k=2 k=3

Equipment:

1. Ruler

2.Diffraction grating

3. Screen with a narrow vertical slot in the middle

4. Light source - laser (monochromatic light source)

Diffraction grating

The diffraction grating is a set a large number very narrow gaps separated by opaque gaps.

a - the width of the transparent stripes

b - width of opaque stripes

d = a + b

d- grating period

Derivation of the working formula:

Maximum

Sveta

a

Lattice

Screen

d sin φ = k λ

because angles are small

sin φ = tg φ , then

Measurement table

Spectrum order

in

a

m

d

m

m



 10 -9 m

 Wed

 10 -9 m

CALCULATIONS:

1 .  =

2.  =

3.  =

 cf =

Table values:

λ kr = 760 nm

In the output, compare the measured values ​​of the wavelength and the tabular ones.

Test questions:

1. How does the distance between the maxima of the diffraction pattern change as the screen moves away from the grating?

2. How many orders of the spectrum can be obtained from the diffraction gratings used in the work?

RESOURCES:

Physics. Grade 11. Myakishev G.Ya., Bukhovtsev B.B., Charugin V.M.

Textbook for educational institutions.

Basic and profile levels.

http://ege-study.ru/difrakciya-sveta/

http://kaf-fiz-1586.narod.ru/11bf/dop_uchebnik/in_dif.htm

http://www.physics.ru/courses/op25part2/content/chapter3/section/paragraph10/theory.html#.WGEjg1WLTIU

Determination of the wavelength of light using a diffraction grating

Objective: determination using a diffraction grating of the wavelength of light in various parts visible spectrum.

Instruments and accessories: diffraction grating; a flat scale with a slit and an incandescent lamp with a matte screen mounted on an optical bench; millimeter ruler.

1. THEORY OF THE METHOD

Wave diffraction is the wave bending around obstacles. Obstacles are understood as various inhomogeneities that waves, in particular light waves, can go around, deviating from rectilinear propagation and entering the region of a geometric shadow. Diffraction is also observed when waves pass through holes, bending around their edges. Diffraction is noticeably pronounced if the dimensions of the obstacles or holes are of the order of the wavelength, and also at large distances from them compared to their dimensions.

Light diffraction finds practical application in diffraction gratings. A diffraction grating is any periodic structure that affects the propagation of waves of one nature or another. The simplest optical diffraction grating is a series of identical parallel very narrow slits separated by identical opaque stripes. In addition to such transparent gratings, there are also reflective diffraction gratings, in which light is reflected from parallel irregularities. Transparent diffraction gratings are usually a glass plate on which stripes (strokes) are drawn with a diamond using a special dividing machine. These strokes are almost completely opaque gaps between the intact parts of the glass plate - the slits. The number of strokes per unit length is indicated on the grid. The period of the (constant) lattice d is the total width of one opaque stroke plus the width of one transparent slit, as shown in Fig. 1, where it is understood that the strokes and stripes are located perpendicular to the plane of the pattern.

Let a parallel beam of light fall on the grating (GR) perpendicular to its plane, Fig. 1. Since the slits are very narrow, the diffraction phenomenon will be strongly pronounced, and the light waves from each slit will go in different directions. In what follows, rectilinearly propagating waves will be identified with the concept of rays. From the entire set of rays propagating from each slit, we select a beam of parallel rays going at a certain angle  (diffraction angle) to the normal drawn to the grating plane. From these rays, consider two rays, 1 and 2, which come from two corresponding points A and C neighboring slots, as shown in Fig. 1. Draw a common perpendicular to these rays AB. At points A and C the phases of the oscillations are the same, but on the segment CB between the rays there is a path difference  equal to

 = d sin. (1)

After straight AB the path difference  between beams 1 and 2 remains unchanged. As can be seen from fig. 1, the same path difference will exist between rays traveling at the same angle  from the corresponding points of all adjacent slots.

Rice. Fig. 1. Passage of light through a diffraction grating DR: L is a converging lens, E is a screen for observing the diffraction pattern, M is the convergence point of parallel rays

If now all these rays, i.e., waves, are reduced to one point, then they will either strengthen or weaken each other due to the phenomenon of interference. The maximum amplification, when the amplitudes of the waves are added, occurs if the path difference between them is equal to an integer number of wavelengths:  = k, where k is an integer or zero,  is the wavelength. Therefore, in directions satisfying the condition

d sin = k , (2)

maxima of light intensity with wavelength  will be observed.

To bring rays going at the same angle  to one point ( M) a converging lens L is used, which has the property of collecting a parallel beam of rays at one of the points of its focal plane, where the screen E is placed. The focal plane passes through the focus of the lens and is parallel to the plane of the lens; distance f between these planes is equal to the focal length of the lens, Fig. 1. It is important that the lens does not change the path difference , and formula (2) remains valid. The role of the lens in this laboratory work is played by the lens of the observer's eye.

In directions for which the value of the diffraction angle  does not satisfy relation (2), there will be a partial or complete attenuation of light. In particular, light waves arriving at the meeting point in opposite phases will completely cancel each other out, and illumination minima will be observed at the corresponding points on the screen. In addition, each slit, due to diffraction, sends rays of different intensity in different directions. As a result, the picture that appears on the screen will have a rather complex form: between the main maxima determined by condition (2), there are additional, or side maxima, separated by very dark areas - diffraction minima. However, practically only the main maxima will be visible on the screen, since the light intensity in the secondary maxima, not to mention the minima, is very small.

If the light incident on the grating contains waves of different lengths  1 ,  2 ,  3 , ..., then by formula (2) it is possible to calculate for each combination k and  their values ​​of the diffraction angle , for which the main maxima of the light intensity will be observed.

At k= 0 for any value of , it turns out  = 0, i.e., in the direction strictly perpendicular to the plane of the grating, waves of all lengths are amplified. This is the so-called spectrum. zero order. Generally, the number k can take values k= 0, 1, 2, etc. Two signs, , for all values k 0 correspond to two systems of diffraction spectra located symmetrically with respect to the zero-order spectrum, to the left and to the right of it. At k= 1 the spectrum is called the spectrum of the first order, when k= 2 the spectrum of the second order is obtained, etc.

Because always |sin|  1, then from relation (2) it follows that for given d and  value k cannot be arbitrarily large. Maximum possible k, i.e., the limiting number of spectra k max , for a specific diffraction grating can be obtained from the condition that follows from (2) taking into account that |sin|  1:

therefore k max is equal to the maximum integer not exceeding the ratio d/. As mentioned above, each slit sends rays of different intensity in different directions, and it turns out that at large values ​​of the diffraction angle , the intensity of the rays sent is weak. Therefore, spectra with large values |k|, which should be observed at large angles , will not be practically visible.

The picture that appears on the screen in the case of monochromatic light, i.e., light characterized by one specific wavelength , is shown in fig. 2a. On a dark background, you can see a system of separate bright lines of the same color, each of which corresponds to its own value. k.

Rice. 2. View of the picture obtained using a diffraction grating: a) the case of monochromatic light, b) the case of white light

If non-monochromatic light falls on the grating, containing a set of waves of different lengths (for example, white light), then for a given k 0 waves with different lengths  will be amplified at different angles , and the light will be decomposed into a spectrum when each value k corresponds to the entire set of spectral lines, Fig. 2b. The ability of a diffraction grating to decompose light into a spectrum is used in practice to obtain and study spectra.

The main characteristics of a diffraction grating are its resolution R and variance D. If there are two waves with close wavelengths  1 and  2 in the light beam, then two closely spaced diffraction maxima will appear. With a small difference in wavelengths  =  1   2 these maxima merge into one and will not be seen separately. According to the Rayleigh condition, two monochromatic spectral lines are still visible separately in the case when the maximum for a line with a wavelength of  1 falls into place of the nearest minimum for a line with a wavelength of  2 and vice versa, as shown in Fig. 3.

Rice. 3. Scheme explaining the Rayleigh condition: I– light intensity in relative units

Usually, to characterize a diffraction grating (and other spectral instruments), not minimum value, when the lines are seen separately, and the dimensionless quantity

called resolution. In the case of a diffraction grating, using the Rayleigh condition, one can prove the formula

R = kN, (5)

where N – total number lattice strokes, which can be found by knowing the width of the lattice L and period d:

Angular dispersion D is determined by the angular distance  between two spectral lines, referred to the difference in their wavelengths :

It shows the rate of change in the diffraction angle  of rays depending on the change in wavelength .

The ratio / included in (7) can be found by replacing it with the derivative d/d, which can be calculated using relation (2), which gives

. (8)

For the case of small angles , when cos  1, from (8) we obtain

Along with the angular dispersion D also use linear dispersion D l, which is determined by the linear distance  l between the spectral lines on the screen, referred to the difference in their wavelengths :

where D is the angular dispersion, f is the focal length of the lens (see Fig. 1). The second formula (10) is valid for small angles  and is obtained if we take into account that for such angles  l  f .

The more resolution R and variance D, the better any spectral device containing, in particular, a diffraction grating. Formulas (5) and (9) show that a good diffraction grating should contain a large number of grooves N and have a short period d. In addition, it is desirable to use spectra of higher orders (with large values k). However, as noted above, such spectra are poorly visible.

The goal of this lab is to determine the wavelength of light in various fields spectrum using a diffraction grating. The setup diagram is shown in fig. 4. The role of the light source is played by a rectangular hole (slit) AND in Shk scale, illuminated by an incandescent lamp with a matte screen S. The observer's eye D, located behind the DR diffraction grating, observes a virtual image of the slit in those directions in which the light waves coming from different grating slits mutually amplify, i.e., in the directions of the main maxima.

Rice. 4. Scheme of the laboratory setup

We study spectra not higher than the third order, for which, in the case of the diffraction grating used, the diffraction angles  are small, and therefore their sines can be replaced by tangents. In turn, the tangent of the angle , as can be seen from Fig. 4, equal to the ratio y/x, where y- distance from hole A to virtual image spectral line on the scale, and x is the distance from the scale to the grating. In this way,

. (11)

Then instead of formula (2) we will have , whence

2. ORDER OF PERFORMANCE OF WORK

1. Install as shown in fig. 4, scale with hole AND at one end of the optical bench near the incandescent lamp S, and the diffraction grating at its other end. Turn on the lamp in front of which there is a matte screen.

2. Moving the grid along the bench, make sure that the red border of the right spectrum of the first order ( k= 1) coincided with any integer division on the Shk scale; write down its value y in table. 1.

3. Using a ruler, measure the distance x for this case and also enter its value in the table. 1.

4. Do the same operations for the purple border of the right spectrum of the first order and for the middle of the green section located in the middle part of the spectrum (hereinafter, this middle will be called the green line for brevity); values x and y for these cases, also enter in the table. 1.

5. Perform similar measurements for the left spectrum of the first order ( k= 1), entering the measurement results in Table. 1.

Note that for left spectra of any order k y.

6. Do the same operations for the red and violet borders and for the green line of the second-order spectra; record the measurement data in the same table.

7. Enter in the table. 3 grating width L and the value of the lattice period d that are indicated on it.

Table 1

lamp spectrum incandescent		x, cm	y, cm	 i, nm		 i =  i, nm
purple

3. PROCESSING OF EXPERIMENTAL DATA

Using formula (12), calculate the wavelengths  i for all measurements taken

(d = 0.01 cm). Enter their values ​​in the table. 1.

2. Find the average wavelengths separately for the red and violet boundaries of the continuous spectrum and the green line under study, as well as the average arithmetic errors in determining  using the formulas

where n= 4 is the number of measurements for each part of the spectrum. Enter the values ​​\u200b\u200band in table. 1.

3. Present the measurement results in the form of a table. 2, where write down the boundaries of the visible spectrum and the wavelength of the observed green line, expressed in nanometers and angstroms, taking as  the average values ​​of the obtained wavelengths from Table. 1.

table 2

4. Using formula (6), determine the total number of lattice strokes N, and then using formulas (5) and (9) calculate the resolution R and the angular dispersion of the lattice D for the second-order spectrum ( k = 2).

5. Using formula (3) and its explanation, determine the maximum number of spectra k max , which can be obtained using a given diffraction grating, using as  the average wavelength of the observed green line.

6. Calculate the frequency  of the observed green line using the formula  = c/, where With is the speed of light, taking as  also the value .

All calculated in paragraphs. 4-6 values ​​enter in the table. 3.

Table 3

4. CONTROL QUESTIONS

1. What is the phenomenon of diffraction and when is diffraction most pronounced?

Wave diffraction is the wave bending around obstacles. Light diffraction is a set of phenomena observed when light propagates through small holes, near the boundaries of opaque bodies, etc. and due to the wave nature of light. The phenomenon of diffraction, common to all wave processes, has features for light, namely here, as a rule, the wavelength λ is much smaller than the dimensions d of the barriers (or holes). Therefore, diffraction can only be observed at sufficiently large distances. l from the barrier ( l> d2/λ).

2. What is a diffraction grating and what are such gratings used for?

A diffraction grating is any periodic structure that affects the propagation of waves of one nature or another. The diffraction grating carries out multibeam interference of coherent diffracted light beams coming from all slits.

3. What is usually a transparent diffraction grating?

Transparent diffraction gratings are usually a glass plate on which stripes (strokes) are drawn with a diamond using a special dividing machine. These strokes are almost completely opaque gaps between the intact parts of the glass plate - the slits.

4. What is the purpose of a lens used with a diffraction grating? What is the lens in this work?

To bring rays coming at the same angle φ into one point, a converging lens is used, which has the property of collecting a parallel beam of rays at one of the points of its focal plane where the screen is placed. The role of the lens in this work is played by the lens of the observer's eye.

5. Why does a white band appear in the central part of the diffraction pattern when illuminated with white light?

White light is non-monochromatic light containing a set of different wavelengths. In the central part of the diffraction pattern k = 0, a zero-order central maximum is formed; therefore, a white band appears.

6. Define the resolution and angular dispersion of a diffraction grating.

The main characteristics of a diffraction grating are its resolution R and dispersion D.

Usually, to characterize a diffraction grating, not the minimum value of Δλ is used, when the lines are seen separately, but a dimensionless value

The angular dispersion D is determined by the angular distance δφ between two spectral lines, divided by the difference in their wavelengths δλ:

It shows the rate of change of the diffraction angle φ of rays depending on the change in wavelength λ.

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Calculation formula for calculating lengths light waves at help diffractive gratings. Measurement length waves boils down to definition deflection angle...