Download "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы"

Please wait. We're preparing links for easy ad-free video watching and downloading.

Previous video: ЛР-11-2-01 Определение фокусного расстояния рассеивающей линзы Next video: Кем были бившийся с Челубеем Пересвет и его брат Ослябя?

Кем были бившийся с Челубеем Пересвет и его брат Ослябя?

Кем были бившийся с Челубеем Пересвет и его брат Ослябя?

Channel: Неизвестная История России - Шортс

ЛР-11-2-01 Определение фокусного расстояния рассеивающей линзы

ЛР-11-2-01 Определение фокусного расстояния рассеивающей линзы

Channel: Павел ВИКТОР

Урок 213 (осн). Линейное увеличение линзы

Урок 213 (осн). Линейное увеличение линзы

Channel: Павел ВИКТОР

13 Оптика (8-9 класс)

13 Оптика (8-9 класс)

Channel: Павел ВИКТОР

03 Оптика (8-11 кл)

03 Оптика (8-11 кл)

Channel: Павел ВИКТОР

06 Оптика (8-11 кл)

06 Оптика (8-11 кл)

Channel: Павел ВИКТОР

Желтоқсан ызғары

Желтоқсан ызғары

Channel: BIG100

09 Оптика (8-9 кл)

09 Оптика (8-9 кл)

Channel: Павел ВИКТОР

20 Оптика (8-9 кл)

20 Оптика (8-9 кл)

Channel: Павел ВИКТОР

14 Оптика (8-9 кл)

14 Оптика (8-9 кл)

Channel: Павел ВИКТОР

Формула тонкой линзы

00:00:07

today's lesson we will devote to

00:00:10

determining the focal length of

00:00:13

collecting lenses experimentally from

00:00:15

installation you will have the same ones in

00:00:17

laboratory work in the sixth seventh

00:00:19

lessons so the topic of the lesson is the topic of

00:00:27

experimental determination

00:00:37

experimental determination of focal

00:00:43

length

00:00:54

collecting and lenses

00:01:07

in this lesson we will look at how this is

00:01:10

done and in the next lesson you

00:01:13

do the laboratory work yourself,

00:01:17

determine the focal length, so

00:01:19

homework is to complete the laboratory

00:01:26

work for tomorrow,

00:01:30

but now what is the first thing that comes to mind

00:01:34

when you decide to determine the focal

00:01:36

length of a converging lens, you and I know

00:01:38

that a converging lens allows you

00:01:40

to create a real image

00:01:41

of an object and we know the formula for a thin lens,

00:01:45

which means the first idea first thought

00:01:53

to use the thin lens formula:

00:01:57

one divided by the distance from the object

00:02:00

to the lens

00:02:01

plus one divided by the distance from the

00:02:06

lens to the image

00:02:07

equals one divisor by the focal

00:02:10

length of the lens

00:02:12

d is the distance from the object to the lens, but

00:02:18

imagine that your lens has

00:02:21

this shape

00:02:26

many collecting spectacle lenses have

00:02:29

such a shape where there are distances

00:02:33

where the optical center is, but that’s not so bad,

00:02:44

you can assume that this lens is thin and

00:02:46

somewhere here, and if the lens is

00:02:52

thick enough and by the way you will have to deal

00:02:55

with a rather thick one, here it’s

00:02:57

no less than a centimeter relatively It’s

00:03:03

not clear why we should calculate these distances,

00:03:06

so this method has very low

00:03:09

accuracy and we need to do something differently,

00:03:11

and now we’ll look at a

00:03:14

more advanced method, the one you’ll

00:03:16

use

00:03:17

today in laboratory work, for this

00:03:20

we’ll have to solve an auxiliary

00:03:22

problem and Goldfarb formulated before us,

00:03:26

so we will solve the auxiliary problem

00:03:38

number twenty-six twenty-seven from

00:03:42

Gald Farb,

00:03:51

this can already be erased,

00:04:00

let's get acquainted with the condition of the problem,

00:04:12

here is 2627, let's move it

00:04:17

up a little like this and so the

00:04:22

distance from the illuminated object to the

00:04:25

screen is written here is 100 centimeters a lens placed

00:04:28

between them gives a clear image of an

00:04:30

object on the screen in two positions, the

00:04:33

distance between which is 20 centimeters,

00:04:35

and so from the light source to the screen the

00:04:40

distance is el large

00:04:42

100 centimeters, and now the most important thing is

00:04:46

a lens placed between them gives a clear

00:04:49

image on the screen in two

00:04:51

positions, the

00:04:52

distance between which is 20 centimeters

00:04:55

gel small equals 20 centimeters

00:04:59

we need to find the focal length, it’s

00:05:04

just ideal, the

00:05:07

problem from the problem book is adapted to our task, why is it better

00:05:14

than the task of determining the focal

00:05:17

length, which I described as the first

00:05:19

thought because we don’t need to know where the

00:05:23

optical center of the lens is,

00:05:25

because we need to know to how much do you need

00:05:29

to move the lens in order for it to

00:05:31

create a sharp image

00:05:33

and why do you get two images

00:05:35

why do the lens have two positions

00:05:37

Artem from the reversibility of light rays yes,

00:05:42

but we don’t know how these images will differ

00:05:46

until it comes to mind

00:05:49

let’s see, we have a

00:05:51

setup that we will use it

00:05:53

now, I’ll show it, it’s big,

00:06:04

here it is, we have a light source,

00:06:09

a lens, and there you see an image, and

00:06:12

what an object,

00:06:14

and here’s an object, a small arrow, you see the

00:06:20

image, which is enlarged, and now I

00:06:25

’ll move the lens, the image became

00:06:29

fuzzy, then nothing is visible at all, and

00:06:33

then a tiny, reduced one image

00:06:37

simple . even on a video camera it is not

00:06:42

resolvable, but because it is too bright

00:06:44

and the focusing of the lens is lost, but the

00:06:47

eyes can be fixed, and so in the first

00:06:49

case, the image is sharply enlarged and in the

00:06:52

second case, the image is reduced, also

00:06:54

sharp, and the distance by which you need to

00:06:57

move the lens has

00:06:59

nothing to do with the position of the

00:07:01

optical lens center, so we

00:07:03

won’t have any problems with this,

00:07:12

we can determine this distance

00:07:15

quite accurately, regardless of the

00:07:18

thickness of the lens and what its shape,

00:07:21

let’s then proceed to solve this

00:07:24

problem, there are two positions, let’s

00:07:26

depict both positions in the figure, here is the

00:07:31

main optical axis in the first In this case, here is

00:07:34

the main optical axis, in the second

00:07:36

case, in both cases, we don’t

00:07:41

move the object,

00:07:53

here the object is the same, in the same

00:07:56

place on our installation, this is this

00:08:01

arrow, then here is the screen

00:08:10

on our installation, here is the game, in the first

00:08:13

case, we get a reduced

00:08:16

image, well, for me on the contrary, in the first case

00:08:19

we got an enlarged image, okay,

00:08:21

let's show it here is the

00:08:22

lens, the converging lens, that's how it

00:08:27

is designated and we got an enlarged

00:08:30

image, I won't build the details

00:08:33

of it, here it is in the second position, if you

00:08:39

move the lens, it will be

00:08:42

here, it will be here

00:08:45

and the image will be reduced,

00:08:50

but also clearly and here it is in both cases,

00:08:53

really inverted, here it is

00:08:55

enlarged, here it is reduced, you can

00:08:57

build tricks, but we don’t need this now,

00:08:59

we need something else, here is the distance from the

00:09:06

object to the lens in the first case d1 here is the

00:09:11

distance from the lens to the image in the

00:09:15

second case f1 here the situation

00:09:21

has changed here the distance from the object to

00:09:26

the lens in the second case d 2 and here is the

00:09:30

distance from the lens to the image in the

00:09:35

second case f2 now what is given by the

00:09:39

condition is the distance from the object to the screen

00:09:43

it is indicated by the letter r large

00:09:46

let's show it here here it is

00:09:55

el large and it is also known that in

00:10:01

order to get a clear 2 1 image,

00:10:05

the lens must be moved at a distance of

00:10:08

el small, this is the Elm distance,

00:10:25

we will use the lens formula, for example,

00:10:29

for this case,

00:10:30

one on d1 plus one on f1

00:10:37

equals one divisor by the focal

00:10:40

length

00:10:45

if we know d1 f1 we can from

00:10:49

this formulas to express the focal

00:10:51

length without problems, how can we find it

00:10:54

if we instead of d1 f1 to at the distance

00:10:57

from the object to the screen and the distance by

00:11:00

which we need to move the lens, first

00:11:03

of all, pay attention to this,

00:11:06

according to the principle of reversibility of light

00:11:09

rays that the order just mentioned,

00:11:11

yes we can change in some places the image of

00:11:15

the object in this case the distance that

00:11:18

was previously the

00:11:19

distance from the lens to the image will

00:11:23

now become the distance from the lens to

00:11:26

the object therefore d2 and f1 we have

00:11:31

the same i d 2 equals f1 in the same

00:11:37

way f2 equals which the distance in the

00:11:44

second case from the lens to the image is the same

00:11:46

as the distance in the first case

00:11:49

from the lens to the object, that is, f 2

00:11:52

equals d 1 and now look, the

00:12:00

usual geometry has gone, let's go from this

00:12:05

position

00:12:06

to this along this path,

00:12:10

move first to d1 then noel and

00:12:16

then to f2

00:12:19

in total we will move at a distance

00:12:23

el large means we can write el

00:12:26

large equals d1

00:12:30

plus el small + f2

00:12:35

but as we found out from the principle of

00:12:37

reversibility of light rays f2 coincides

00:12:40

with d1 in magnitude, which means plus d1 we can

00:12:45

find d1 from here since according to the conditions of the

00:12:47

problem el large and small

00:12:49

we are given, we get it l equals

00:12:54

2d one plus el small or 2d one

00:13:01

equals el big minus film all and and

00:13:04

here I will write d 1 equals than

00:13:09

el big minus and small in half is ready

00:13:19

now let's again go from this

00:13:22

position to this in a

00:13:23

different way so that we have

00:13:27

not d1a f1 was involved,

00:13:31

look how you can do this, look

00:13:35

at the second picture from this position we

00:13:39

move to d2

00:13:41

and this is the same as f1 but we need to

00:13:43

get here moving here is not

00:13:46

interesting let's go back to el malo and

00:13:49

back and move to f1 again

00:13:53

forward as a result we will move to

00:13:55

el large means el large

00:13:57

equals again look at el large

00:14:01

equals f1 she is 2

00:14:05

now back noel minus n and forward to

00:14:12

f1

00:14:15

that is, we get l equals 2

00:14:20

f1 minus el mal and then

00:14:24

2 f1 from here equals hotel big plus

00:14:30

or small or I'll write it down here

00:14:35

f1 equals look el big plus and

00:14:39

small in half

00:14:44

hoh we solved the most difficult part of the problem, we solved the most difficult part of the problem

00:14:48

through those quantities that are given according to the

00:14:51

condition we received those quantities that are

00:14:54

included in the lens formula, then it’s a matter of technology,

00:14:57

this can be erased, we substitute the unit for

00:15:07

t1 and one on f1 here

00:15:09

one on t1 reciprocal fraction 2 divided by

00:15:15

p minus or plus one by f12 divided

00:15:23

by l + l equals the inverse focal

00:15:27

length what do you propose to do

00:15:29

next what to do bring Andrey

00:15:35

correctly to a common denominator

00:15:37

common denominator we will have a goal minus

00:15:41

n multiplied by n plus and here we will have

00:15:47

2 multiplied by an additional factor 2

00:15:50

multiplied by l + l add 2

00:15:55

by this additional factor and

00:15:59

minus or what is this inverse and focal

00:16:03

length

00:16:11

now open the brackets 2 el large

00:16:16

this is the numerator of our fractions + 2 l

00:16:21

small + 2n large -2 they don’t have enough

00:16:28

x in the denominator that the difference of

00:16:33

squares is written or a large square minus or a

00:16:38

little squared and all together it’s the

00:16:40

opposite and the focal length is

00:16:42

ready look at the numerator look

00:16:46

here two with a plus here two with a minus

00:16:48

they cancel each other out

00:16:52

and it remains that two large plus 2 el

00:16:57

large this will be 4 large 4l large

00:17:03

divided by r square minus r square

00:17:06

equals the inverse focal length

00:17:09

but or optical power

00:17:11

but we are asked to find the focal length,

00:17:13

then we will reverse this freak and we

00:17:17

will get the focal length equals a

00:17:20

fraction in the numerator will be the difference of

00:17:24

squares in the denominator will be 40 large

00:17:35

here is our working formula since in the

00:17:39

conditions the numbers are given then the answer can be

00:17:42

obtained in the form of a number

00:17:44

we calculate f equals the fractional line in the

00:17:51

denominator 4 l 4 per 100 centimeters in the

00:17:57

numerator the difference of squares you know it’s

00:17:59

probably easier to calculate using the formula

00:18:01

abbreviated multiplication the difference of

00:18:03

squares is n minus

00:18:04

el that is 100 minus 20 multiplied by n +

00:18:13

n 100 plus 20 centimeters square

00:18:18

centimeters square centimeters are

00:18:19

partially reduced equal to the fraction

00:18:26

here we have 80

00:18:29

multiplied by 120 divided by 4 and on the table

00:18:36

let's do this once 2 we reduced a

00:18:40

hundred times the denominator

00:18:41

1 2 we decreased a hundred times the numerator the

00:18:44

result is centimeters equals now 8

00:18:47

divided by 4 will be how much

00:18:49

2 2 multiplied by 12 24 the

00:18:55

focal length of this lens is 24

00:18:58

centimeters now we have done on the board

00:19:02

what you did in the laboratory work with your

00:19:05

own hands on a live

00:19:07

installation I will show you on this large

00:19:09

installation how we will work in what

00:19:11

sequence

00:19:13

and you will have this miniature one,

00:19:15

each of us has 9 of them,

00:19:17

we are exactly 34 people, which means in

00:19:22

groups when we work there will be enough

00:19:24

for everyone and so we proceed to the actual

00:19:29

measurements for That’s probably why I

00:19:32

’ll now remove all the unnecessary stuff and organize a table

00:19:35

that needs to be filled out as we complete the

00:19:38

work. The only thing we need from this task is

00:19:40

this working form,

00:19:42

naturally, in the brief theory of the series I

00:19:45

wanted to ask about this before in the brief theory

00:19:49

you talk about how

00:19:52

this formula is obtained, necessarily because we

00:19:55

define the focal length, then in a

00:19:57

brief theory it will be necessary to describe what the

00:20:00

focal length is and list the

00:20:03

main elements of geometric optics,

00:20:07

that is, what is

00:20:08

the lens, the main optical axis, what is the

00:20:12

collecting lens, which

00:20:15

means that we will measure we

00:20:18

will set several distances from the

00:20:21

object to the screen which

00:20:24

We denote the letter l for several distances, that

00:20:27

is, we will have several experiments,

00:20:29

so the table will look something

00:20:31

like this:

00:20:37

experiment number You and I will now take two

00:20:45

measurements, in laboratory work you will

00:20:47

take 5 measurements 1 2 3 4 5

00:20:55

then we will change the

00:21:00

distances from object to the screen,

00:21:04

it is marked with us el large

00:21:08

centimeters further on the measuring

00:21:15

device on the optical bench, this

00:21:16

device is called an optical bench,

00:21:19

here there is a scale,

00:21:22

this scale allows you to determine the position

00:21:26

I or the coordinate of any object that is

00:21:31

located here, let's designate the two positions of

00:21:34

the lens on this scale as x 1, in

00:21:41

centimeters and

00:21:45

x2, in centimeters

00:21:50

means what it is these are the positions of the lens

00:21:54

at which a sharp

00:21:56

image is obtained, once enlarged

00:21:58

another time reduced, it

00:22:01

turns out that the countdown comes from here,

00:22:03

which means for an enlarged image

00:22:06

x1 will be a smaller value and for a

00:22:09

reduced image x1 will be a larger

00:22:11

value

00:22:12

therefore how to find the distance at which

00:22:14

we move the lens this was here the

00:22:19

coordinate x 1 became here the

00:22:21

coordinate x 2

00:22:22

must be subtracted, which means the next column l

00:22:28

is equal to x 1 minus x 2 centimeters,

00:22:35

in this formula we have everything

00:22:40

necessary to calculate the focal

00:22:42

length l we set x1 x2 we measure

00:22:47

here x1

00:22:50

by subtraction we find or it means we have

00:22:53

everything to calculate the focal length

00:22:56

we write f centimeters and take measurements

00:23:02

5 times

00:23:17

prepare this table for yourself now in your notebooks

00:23:21

and please tell me why I

00:23:23

left this column and

00:23:25

haven’t plundered it yet and and what do you think for the

00:23:28

average value f average centimeters

00:23:33

here our table ends with this is the

00:23:36

structure it has and you will need to take 5

00:23:40

measurements and fill in these five lines

00:23:44

calculate the focal length and find the

00:23:47

average value here is the table but

00:23:52

now let’s try to put into practice

00:23:54

what is planned so that you can clearly

00:24:00

see the camera, take a closer look, now we’ll

00:24:07

probably prepare a mouse so that I don’t run

00:24:09

to the computer all the time, now it

00:24:13

recognizes it,

00:24:14

now it recognizes the computer, let’s remove all the

00:24:18

unnecessary things,

00:24:25

now we can see either

00:24:31

everything in small plans or everything in close-

00:24:37

ups, pointing it towards the world, for example, at an

00:24:42

object

00:24:46

here now we need to determine where

00:24:49

the object is,

00:24:50

let's look at the scale, the position of

00:24:58

the object is against the number 5, which means the object

00:25:03

is at a point with coordinate 5,

00:25:05

let's set the distance from the object

00:25:08

to from to the image, that is, to the screen

00:25:11

70 centimeters, then we move along the

00:25:14

scale here and against 75 we set

00:25:22

the screen here this means that now the distance from

00:25:28

the object to the screen is 70 centimeters,

00:25:34

we write down l70 now the fun begins,

00:25:42

we need to build first an enlarged

00:25:45

image and then a reduced one,

00:25:50

we look at the screen,

00:26:01

I move the lens,

00:26:14

we make sure that the edges of the image

00:26:17

are as clear as possible,

00:26:20

we have built an enlarged image,

00:26:22

now we look at the position of the lens

00:26:27

here is the lens itself, here is the coordinate, let's say the

00:26:33

lens is how much 16.6 16.6 centimeters

00:26:45

we write down x1 16.6

00:26:52

now we move the lens so as to

00:26:56

get a sharp reduced image

00:27:11

look closely

00:27:13

it is very bright so we look where it

00:27:17

is and the lens itself the

00:27:28

position of the lens is

00:27:30

how much 60 2.1

00:27:35

is our x 2 62 and one tenth now

00:27:43

we take a calculator and subtract x we even need

00:27:48

x 2 minus x1

00:27:50

from the larger the smaller you can put it here modulo

00:27:53

like this because it can

00:27:55

be the other way around for you if you put the screen on

00:27:58

the side where the smaller coordinate is the

00:28:01

opposite for you so that means 62 and 162

00:28:05

and one minus 16 and 6

00:28:10

it turns out 45 and a half 45.5

00:28:18

centimeters, and now the most exciting

00:28:21

moment is looking at this formula and using a

00:28:25

calculator we calculate the focal

00:28:28

length and so r-square I will immediately

00:28:33

calculate r in such large blocks

00:28:36

-square 70 squared minus

00:28:43

minus el small square here it is -45

00:28:48

and 5 squared is equal to what I did I

00:28:53

calculated the numerator now we need to

00:28:56

divide this result by 4 and also what we

00:29:00

get by dividing by el large

00:29:03

by 70 it turned out

00:29:15

just tell me please how many

00:29:21

significant ones I have the right to write down 3 digits

00:29:26

because we had length measurements with

00:29:28

three significant figures, which means three

00:29:31

significant figures it will be how much 10 point

00:29:34

one tenth 10 point one tenth

00:29:38

centimeter here is the focal length

00:29:42

further it’s too early to believe it

00:29:46

let’s change the distance between the

00:29:51

object and the screen it was 70 centimeters

00:29:54

let's do 50 for example like this oh Kolya

00:30:01

well done if it says centimeter here and

00:30:04

there is no need to write everything carefully,

00:30:05

look so now once again I remind you that

00:30:10

the object has a coordinate, that is,

00:30:13

its position is specified by the

00:30:15

five is at the position point of

00:30:18

which 5 centimeters the coordinate of the point is 5

00:30:21

centimeters we want to make it so that we

00:30:24

were at a distance of 50 centimeters,

00:30:27

so I move to a point 50

00:30:31

more, here it is 55 and here I place

00:30:38

the screen ready, record

00:30:46

50 centimeters and repeat the measurements

00:30:51

first you look and it’s a habit so first

00:31:01

we determine where the lens is when the

00:31:04

image is enlarged the screen must be

00:31:06

perpendicular to the main optical axis

00:31:09

we will fix let's move the lens like this,

00:31:14

let's let you see what's going on, I have the

00:31:18

camera in sight, so we'll move the

00:31:24

lens, turn on the mouse, retract,

00:31:38

move the lens,

00:31:42

here it is, a sharp image is ready,

00:31:51

measure the position of the lens at which

00:31:54

such a sharp image is

00:31:56

obtained, here's the lens, its position,

00:32:05

how many centimeters 18 3

00:32:12

x 118 and three centimeters, I’m no longer writing

00:32:19

now, please note that when we

00:32:23

moved the screen, the object had to be

00:32:26

moved a little from the lens to this

00:32:28

distance more than this, now

00:32:30

we move the lens so that we get a

00:32:33

reduced image,

00:32:38

it will be a little bit larger, so

00:32:43

we look where it is located where is

00:32:45

the lens at this position of the lens

00:32:53

what is 40.6

00:32:58

centimeters 40.6 centimeters we calculate the

00:33:05

distance by which we moved

00:33:07

the lens

00:33:08

means 46 -18 340. 6 minus 18 3

00:33:15

it turns out 22.3

00:33:22

now I use these values

00:33:24

el large and this value al

00:33:27

small we calculate the focal length and

00:33:30

so r-squared 50 squared well,

00:33:35

I’ll immediately write 2500 minus 22 and 3

00:33:40

squared 22 and 3 squared equals how much

00:33:45

then it turned out there now it needs to be

00:33:47

divided by 4 and by 50 but 4 by 50 is

00:33:53

200 means we divide by 200 let

00:34:00

’s see what happens what happens

00:34:11

10 point one hundredth

00:34:14

if you write down three significant figures and then it

00:34:18

will be that

00:34:20

10.0 class is a wonderful result

00:34:24

10.0 a why do I think that these are

00:34:29

wonderful results, yes, because

00:34:31

on this lens its focal

00:34:38

length is indicated as 100 millimeters, that is, 10

00:34:45

centimeters, see how well

00:34:47

this method works,

00:34:49

how accurate it is, although the lens is quite

00:34:51

thick, so it is impossible to obtain the position of the

00:34:55

optical center with great accuracy,

00:34:58

so using this method

00:35:01

you you will be in the sixth and seventh lessons in

00:35:03

groups doing laboratory work,

00:35:05

now I will give you the distances between the

00:35:11

object and the screen that you will

00:35:13

install as el large

00:35:17

el large equals 45 centimeters 40

00:35:25

centimeters thirty-five centimeters 30

00:35:31

centimeters and

00:35:33

25 centimeters that is, you will get five

00:35:37

lines, calculate focal length of another

00:35:40

lens there will be another five times you will find

00:35:44

the average this is the task we

00:35:47

face in lesson 2 now we are finishing the lesson

00:35:51

we are drinking tea

Description:

Урок физики в Ришельевском лицее

Preparing download options

Popular

HD video

Only sound

All

* — If the video is playing in a new tab, go to it, then right-click on the video and select "Save video as..."

** — Link intended for online playback in specialized players

Questions about downloading video

How can I download "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы" video?

http://unidownloader.com/ website is the best way to download a video or a separate audio track if you want to do without installing programs and extensions.
The UDL Helper extension is a convenient button that is seamlessly integrated into YouTube, Instagram and OK.ru sites for fast content download.
UDL Client program (for Windows) is the most powerful solution that supports more than 900 websites, social networks and video hosting sites, as well as any video quality that is available in the source.
UDL Lite is a really convenient way to access a website from your mobile device. With its help, you can easily download videos directly to your smartphone.

Which format of "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы" video should I choose?

The best quality formats are FullHD (1080p), 2K (1440p), 4K (2160p) and 8K (4320p). The higher the resolution of your screen, the higher the video quality should be. However, there are other factors to consider: download speed, amount of free space, and device performance during playback.

Why does my computer freeze when loading a "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы" video?

The browser/computer should not freeze completely! If this happens, please report it with a link to the video. Sometimes videos cannot be downloaded directly in a suitable format, so we have added the ability to convert the file to the desired format. In some cases, this process may actively use computer resources.

How can I download "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы" video to my phone?

You can download a video to your smartphone using the website or the PWA application UDL Lite. It is also possible to send a download link via QR code using the UDL Helper extension.

How can I download an audio track (music) to MP3 "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы"?

The most convenient way is to use the UDL Client program, which supports converting video to MP3 format. In some cases, MP3 can also be downloaded through the UDL Helper extension.

How can I save a frame from a video "Урок 218 (осн). Экспериментальное определение фокусного расстояния собирающей линзы"?

This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.

What's the price of all this stuff?

It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.