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Експериментальна робота №7 "Вимірювання довжини світлової хвилі"

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Актуалізація опорних знань та вмінь: що таке дифракційна гратка, період дифракційної гратки, інтерференціний максимум першого порядку та способи розрахунку довжини світлової хвилі.

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Обладнання та підготовка до експерименту

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Експеримент: Дивлячись крізь дифракційну ґратку і щілину на лампу розжарювання, спостерігайте на екрані приладу різкі дифракційні спектри, лінії яких паралельні штрихам на шкалі. Дані експерименту занесіть в таблицю.

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фізика11

Дифракція

Принцип Гюйґенса – Френеля

Дифракційна ґратка

стала ґратки

Формула дифракційної ґратки

00:00:02

lesson 66,

00:00:04

experimental work number 7,

00:00:07

measuring the length of light wavelength,

00:00:14

the topic of measuring the length of light wavelength,

00:00:19

the goal of the work is to learn how to measure the length of a

00:00:23

light wave using a

00:00:26

diffraction grating, let

00:00:29

's remember what a diffraction grating is, the

00:00:32

period of a diffraction grating, the

00:00:35

interference maximum of the first

00:00:37

order, and methods of calculating the length of a

00:00:40

light wave using the example of a

00:00:42

solution solving the following problem

00:00:49

to measure the wavelength of light, a

00:00:53

diffraction grating with

00:00:56

1000 lines per 1 mm of the

00:00:59

first order maximum on the screen is used.

00:01:02

Obtained at a distance of 24 cm

00:01:07

from the central maximum,

00:01:10

determine the wavelength if the distance from the

00:01:14

diffraction grating

00:01:17

to the screen is

00:01:20

1 meter.

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measurement of

00:01:32

light wavelength lambda is small, a

00:01:35

diffraction grating is used, which has

00:01:38

1000

00:01:40

n, large strokes on the length of the grating L is

00:01:43

small, 1 mm is 10 in the minus third of a

00:01:46

meter,

00:01:47

the maximum of the first order, that is, when k is

00:01:51

equal to one on the screen Obtained at a

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distance of

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24 cm from the central maximum,

00:01:59

we denote this distance by x

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determine the wavelength if the distance

00:02:05

from the diffraction grating to the screen L is

00:02:08

large, let it be equal to 1 m, a

00:02:13

diffraction grating is an optical

00:02:15

device whose action is based on the phenomenon of

00:02:18

light diffraction and which is a

00:02:20

collection of a large number of parallel

00:02:23

paths drawn on a certain surface at the

00:02:26

same distance from each other

00:02:31

if on a flat light wave falls on the grating,

00:02:35

then each slit becomes a source of

00:02:39

secondary waves that are

00:02:44

coherent and propagate in all

00:02:47

directions,

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since they are coherent, they are capable of

00:02:52

forming an interference pattern on the screen,

00:02:58

and

00:03:00

interference maxima will be

00:03:02

formed if the difference in the course of

00:03:05

two rays coming from two

00:03:08

adjacent slits

00:03:09

Delta d, which is equal to D by sin 6 DD, the

00:03:13

period of our grating is equal to the total number of

00:03:19

waves,

00:03:23

this ratio is called the

00:03:26

diffraction grating formula, and we will use

00:03:30

it to develop our problem,

00:03:35

since we have

00:03:39

where the sine of phi is equal to K lambda,

00:03:42

from it we can determine the

00:03:44

wavelength of lambda But in we are left with an

00:03:48

unknown angle phi for

00:03:50

the maximum of the first order K = 1 and

00:03:54

unknown. We have a

00:03:56

parameter D, which is called the constant

00:04:00

Bratka or lattice period. If we have a

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known number of strokes N is large on a

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lattice segment of length L, then the

00:04:15

lattice period can be calculated using the formula D

00:04:18

small is equal to l it is small to divide by

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N the number of strokes,

00:04:24

since we have this information, we can

00:04:28

substitute the value for calculating the

00:04:30

period of the grating into the diffraction grating formula,

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and we will have the need to

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find out at what angle the

00:04:41

first-order maximum is visible,

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for this we will make a

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scheme

00:04:51

that is shown in the figure,

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we previously depicted rays

00:04:57

that propagated from neighboring slits at an

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angle to the left, but we know that each

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slit is a source of secondary waves that

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spread in all directions, so we can

00:05:13

make a schematic drawing in which

00:05:17

the beam will spread, including to the

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right. At the same time, the

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angle phi under which is visible

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under which a maximum of a certain order is visible

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in the picture will be shown in this

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form

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Thus, if we have a maximum of the first

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order visible under some designated

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frames, one at a certain angle, then in

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this case, the distance from the

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central maximum, which lies on a

00:05:50

perpendicular drawn to the screen, to

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this maximum will be equal to X

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and it is known to us, we also know the

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distance from the

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diffraction grating to the screen

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by looking at the picture We see that

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we are dealing with a right triangle in

00:06:15

which we know the legs, but the

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angle phi and the hypotenuse are unknown,

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since for the Formula of the diffraction grating

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We need the sine of the angle phi, we can

00:06:27

immediately write it down the sine of the angle phi, this is

00:06:30

the ratio of the opposite leg, i.e. x in

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our notation to the hypotenuse with

00:06:38

X, we know the hypotenuse C, we can

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immediately determine from the Pythagorean theorem,

00:06:44

according to the Pythagorean theorem, the square of

00:06:47

the hypotenuse is equal to the sum of the squares

00:06:49

of the legs,

00:06:51

so

00:06:53

it will accordingly be Equal

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square root of

00:06:58

l² + x squared, the

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final value of sine phi will look

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as follows x divide by the

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square root of l²+x²

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we will substitute this value for sines and for the

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formula for calculating the grating period in the

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diffraction grating formula we will have L

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small divide by n * x /

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√x² and this all has be equal to the calendar

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for the case of

00:07:32

maxima,

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hence lambda will be equal to LX

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divide by NK multiply by the

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square root of l²+x²

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let's proceed to the calculations

00:07:53

we check the units of measurement L and x

00:07:56

is measured in meters

00:08:00

n dimensionless k dimensionless square root

00:08:04

of meters squared plus

00:08:06

meters squared meters squared plus

00:08:09

square meters will give simply

00:08:11

square meters after extracting the root we will get meters

00:08:13

in 1 m we shorten we will get that the

00:08:15

wavelength is measured in meters What is

00:08:19

correct we substitute numerical values

00:08:20

lambda is equal to l small 10 in the minus

00:08:25

third of a meter multiply by X 24 cm is

00:08:29

0.24 m divide by n1000 strokes

00:08:33

multiply by K the maximum of the first order

00:08:37

multiply by the square root of

00:08:41

L is large one

00:08:44

meter squared plus x 0.24 squared

00:08:48

After doing the calculations we get that it

00:08:51

will be approximated Equal to 230 times 10 to the minus ninth power of a

00:08:54

meter or taking into account

00:08:57

that 10 to the minus nine so it is prefixed with no,

00:09:00

we will have that the wavelength will be equal to 230

00:09:04

nanometers,

00:09:07

since in this problem we are considering

00:09:10

the maximum of the first order, it can be

00:09:13

solved in a simpler way, allowing for

00:09:16

some approximations.

00:09:20

So, we have a diffraction grating formula

00:09:23

and a method for calculating the grating period D,

00:09:27

the tangent of the angle phi as it is known from mathematics that is

00:09:32

equal to the ratio of the sine of the angle all to the

00:09:35

cosine of the angle phi at the maximum of the first

00:09:40

order, the

00:09:41

angle phi will be small, it will be different from

00:09:46

zero, but small enough so that the

00:09:50

cosine of the given angle phi is

00:09:55

approximately equal to one, because as you

00:09:59

know, the cosine of the angle phi is equal to one, the

00:10:01

derivative is zero and also at small

00:10:05

angles phi And in the case of a maximum of the first

00:10:09

order, we will have the smallest of the angles

00:10:13

at which the possible diffraction maximum

00:10:15

can be put That the cosine of the angle phi for the

00:10:21

maximum of the first order will be approximately

00:10:24

equal to unity, then the sine of phi will

00:10:28

also be approximately equal to tg,

00:10:32

and the tangent of the angle phi from our

00:10:35

rectangular of the triangle will be equal to the

00:10:38

ratio of the opposite leg, that is, x

00:10:41

to the

00:10:42

adjacent leg, that is, to L, the distance

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from the diffraction grating to the screen.

00:10:49

And from this it follows that the sines will be equal to the

00:10:52

ratio of x to L. Substituting this

00:10:55

value into the formula of the diffraction grating and

00:10:58

also the value for calculating D, we

00:11:02

will get that L is small divide by N

00:11:05

multiply by x and divide by large is

00:11:07

equal to the lambda

00:11:11

from this formula the lambda will be Equal to LX

00:11:14

divide by N multiply by L

00:11:18

substituting the value We will get that the

00:11:22

wavelength in our case will be

00:11:26

equal to 240 nanometers which is only 10

00:11:32

nanometers different from the

00:11:36

method of calculating a more accurate which we

00:11:40

used in the previous version of

00:11:42

the solution of the problem for the maximum of the first

00:11:46

order, this method is also quite

00:11:49

admissible. But it is clear that for

00:11:51

maxima of a higher order, when the angle

00:11:54

will already be significantly greater than zero, this

00:11:58

simplification will give too large an error for

00:12:02

us to use it.

00:12:05

Finally, consider the video which

00:12:09

shows how

00:12:13

maxima are formed when the diffraction

00:12:17

grating is illuminated by monochromatic light, similar

00:12:21

to how it happened in our problem

00:12:25

So, here we have the central maximum And

00:12:27

here we have the maximum of the first order when the

00:12:31

distance between the grating and the screen decreases, the

00:12:33

distance between the central maximum and the

00:12:38

maximum of the first order decreases as the

00:12:41

distance increases, the

00:12:43

distance between the maximum of the first order and the

00:12:46

central maximum increases. It is

00:12:53

also worth noting that the angle

00:12:59

phi at which the

00:13:01

interference maximum is observed depends on the

00:13:04

wavelength,

00:13:05

therefore diffraction gratings decompose

00:13:09

non-monochromatic white light into a spectrum,

00:13:14

such a spectrum is called a diffraction spectrum, the

00:13:18

wavelength of the red color is greater

00:13:22

than the wavelength of the violet colors,

00:13:24

therefore, in the diffraction spectrum, the red

00:13:27

lines

00:13:29

are located further from the central

00:13:32

maximum than the violet ones,

00:13:35

and

00:13:37

therefore each of the diffraction maxima, in

00:13:41

turn, will be decomposed into a

00:13:44

diffraction spectrum

00:13:47

for the central maximum, the difference in the course of

00:13:51

waves of any length is zero,

00:13:53

therefore, it always has the color light that

00:13:57

illuminates the grating if the grating is illuminated by

00:14:00

white light, then, accordingly, the central

00:14:03

maximum will also have a white color

00:14:06

by measuring the angle phi at which the

00:14:09

interference

00:14:10

maximum of the order of magnitude is observed and knowing

00:14:14

the period of the diffraction grating, where you can

00:14:17

measure the length of the light wave

00:14:19

falling on the grating using the formula

00:14:22

lambda is equal to 10

00:14:25

divided by K

00:14:33

equipment

00:14:35

lamp with with a straight incandescent filament, a

00:14:40

device for determining the length of a light

00:14:44

wave, a

00:14:45

tripod with a coupling, a

00:14:48

diffraction grating,

00:14:57

preparation for the experiment, first

00:15:01

determine the period D small of the diffraction

00:15:05

grating,

00:15:06

usually the number of N strokes per 1 mm is indicated on the grating,

00:15:15

in this case, in order to determine the period of the

00:15:18

grating,

00:15:23

it is necessary to

00:15:25

divide 10 into minus the third

00:15:29

on N,

00:15:31

i.e. on the number of

00:15:34

strokes per 1 mm, the

00:15:40

second

00:15:41

assemble the installation

00:15:43

shown in the figure,

00:15:50

attention when preparing and conducting

00:15:54

the experiment strictly Follow the

00:15:57

safety instructions, the

00:16:07

results of measurements and calculations are

00:16:09

immediately entered in the next table, the

00:16:13

period of the lattice where is small in meters, the

00:16:17

color of the spectrum is violet and red, the

00:16:21

distance from the center slits to the border A1 in

00:16:26

meters

00:16:27

A2 in meters the

00:16:30

average value of the distance from the center of the

00:16:32

slit to the border and the average in meters the

00:16:36

distance from the grating to the screen L

00:16:40

small in meters the

00:16:43

measured wavelength of lambda in Nano

00:16:48

meters

00:16:49

tabular values of wavelengths in

00:16:52

nanometers for violet and red

00:16:56

colors

00:17:03

experiment

00:17:05

first

00:17:07

looking through the diffraction grating and the

00:17:11

slit to the incandescent lamp,

00:17:14

sharp diffraction spectra can be observed on the screen of the device,

00:17:17

the lines of which are

00:17:20

parallel to the strokes on the

00:17:31

second

00:17:32

scale on the screen,

00:17:35

first determine the

00:17:38

distance A1 from the center of the

00:17:43

slit to the limit of the violet

00:17:48

color of the spectrum of the first order

00:17:50

located to the right of the slit, the

00:17:54

distance A2 from the center of the slit to the limit of the

00:17:58

violet color of the spectrum of the first

00:18:01

order located To the left of the slit, the

00:18:10

third

00:18:14

repeat the

00:18:16

actions

00:18:18

described in the second point for the limit of the

00:18:21

red color of the

00:18:24

spectrum of the first order,

00:18:30

the fourth, measure the distance L small

00:18:34

from the grid to the screen,

00:18:42

if it is impossible to perform the experiment

00:18:45

yourself, use the following photo

00:18:48

and video to capture the data of the experiment

00:18:52

beforehand we enter into the table the value of the

00:18:55

distance from the grating to the screen L is small

00:19:00

in meters based on the data from the photo

00:19:04

we also calculate the period of the grating D

00:19:08

based on the formula and also enter into the table

00:19:11

with the results of the experiment

00:19:25

wavelength determination setup

00:19:29

light source

00:19:31

amplitude phase holographic

00:19:33

diffraction grating

00:19:37

obtained diffraction spectrum

00:19:40

maximum of the first order from the source of

00:19:45

white light

00:20:00

according to the scale on the screen, we first determine the

00:20:04

distance A1 from the central maximum

00:20:09

to the limit of the violet color of the

00:20:13

spectrum of the first order located to the

00:20:17

right of the slit,

00:20:25

then the distance A2 from the central

00:20:30

maximum to the limit of the violet color of the

00:20:34

spectrum of the first order

00:20:37

located To the left of the slit,

00:20:46

we repeat the described actions for the limit of the

00:20:51

red

00:20:56

spectrum of the first order of

00:21:08

processing the results of the experiment,

00:21:10

first, calculate the average values of the

00:21:14

distances from the slit to the corresponding limits of the

00:21:17

violet and red colors of the

00:21:20

spectrum of the first order according to the formula, and the

00:21:23

average is equal to A1 + A2, divide by 2,

00:21:30

second,

00:21:32

calculate the length of the light wave of

00:21:34

violet color and the light wave of

00:21:37

red color according to the formula lambda

00:21:41

is where the small is multiplied by and the

00:21:44

average is divided by H where where is the small

00:21:48

period of the grating and the average is the average distance

00:21:52

from the center of the slit or the central

00:21:55

maximum to the limit L is the small distance

00:21:59

from the grating to the screen

00:22:03

third estimate the relative error of the

00:22:07

experiment

00:22:09

Epsilon with the subscript lambda

00:22:12

equals modulo one minus

00:22:17

divide the measured wavelength lambda by

00:22:20

the table lambda and multiply by 100%

00:22:30

. analyze the experiment and its

00:22:33

results, formulate a conclusion in your own

00:22:36

words, in which you must indicate the

00:22:39

first, what physical quantity you determined, the

00:22:43

second, what result did you get, the third,

00:22:48

what are the reasons for the possible error of the

00:22:51

experiment,

00:22:56

repeat homework, paragraph 31.

00:23:02

If you liked it, do not forget to

00:23:06

like the subscription and share the video among those to

00:23:09

whom it may also be interesting, thank you

00:23:13

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Фізика 11 клас. Презентація до уроку 66: Експериментальна робота №7 за темою «Вимірювання довжини світлової хвилі». Дистанційне навчання: Фізика - https://www.facebook.com/physics.ukr Мета уроку: Навчальна. У процесі дослідницької діяльності закріпити знання про дифракцію, дифракційну гратку, навчити учнів вимірювати довжину світлової хвилі за допомогою дифракційної гратки; розвивати спостережливість, увагу, пам’ять, уяву, мислення; Розвивальна. Сприяти: розвитку спостережливості, уваги, пам’яті, уяви, мислення; виробленню звички до планування своїх дій; формуванню вміння самостійно контролювати проміжні і кінцеві результати роботи; формуванню вміння організовувати своє робоче місце. Виховна. Виховувати в учнів охайність під час проведення експерименту, дбайливе ставлення до лабораторного обладнання; виховувати учнів працювати в парах та групах. Онлайн урок для дистанційного навчання від проєкту "Фізика онлайн". За книгою "Фізика 11" за редакцією В. Г. Бар'яхтара, С. О. Довгого. Деякі матеріали для даного відео взяті із сайту ФІЗИКА НОВА https://www.fizikanova.com.ua/ Якщо Вам сподобалось не забудьте про лайк, підписку та поширити відео серед тих, кому воно теж може бути цікавим. 00:00 Експериментальна робота №7 "Вимірювання довжини світлової хвилі" 00:48 Актуалізація опорних знань та вмінь: що таке дифракційна гратка, період дифракційної гратки, інтерференціний максимум першого порядку та способи розрахунку довжини світлової хвилі. 14:33 Обладнання та підготовка до експерименту 16:06 Експеримент: Дивлячись крізь дифракційну ґратку і щілину на лампу розжарювання, спостерігайте на екрані приладу різкі дифракційні спектри, лінії яких паралельні штрихам на шкалі. Дані експерименту занесіть в таблицю. 21:08 Опрацювання результатів експерименту

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