Ege in physics optics solution. Limiting angle of total internal reflection. Relative refractive index

The light beam exits the glass into the air (see figure).

1) increases;

2) decreases;

3) does not change.

Solution.

When a light beam passes from glass to air, the frequency of electromagnetic oscillations in a light wave does not change, since it does not depend on the medium in which the wave propagates. Since glass is an optically denser medium than air, as it leaves the glass, the propagation speed of the light wave increases. In turn, the wavelength is related to the frequency of electromagnetic oscillations and the propagation velocity by the relation In view of the invariance of the frequency and the increase in speed, it follows that the wavelength increases.

Answer: 311.

Answer: 311

The light beam passes from air to glass (see figure).

What happens in this case with the frequency of electromagnetic oscillations in a light wave, the speed of their propagation, the wavelength?

For each value, determine the appropriate nature of the change:

1) increases;

2) decreases;

3) does not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

When a light beam passes from air to glass, the frequency of electromagnetic oscillations in a light wave does not change, since it does not depend on the medium in which the wave propagates. Since glass is an optically denser medium than air, when passing into glass, the speed of propagation of a light wave decreases. In turn, the wavelength is related to the frequency of electromagnetic oscillations and the speed of propagation by the relation In view of the invariance of the frequency and the decrease in speed, it follows that the wavelength decreases.

Answer: 322.

Answer: 322

The object is in front of the converging lens between the focal length and the double focal length. How will the distance from the lens to its image, the linear size of the image of the object and the appearance of the image (imaginary or real) change when the object is moved to a distance greater than the double focal length ()?

Solution.

A converging lens gives a real image of an object if it is at a distance greater than the focal length from the lens. Consequently, when moving an object from a position between the focal and double focal lengths to a distance greater than double focal length, the appearance of the image will not change, it will remain valid (B - 3). According to the thin lens formula, the distance from the object to the lens, the distance from the lens to the image and the focal length are related by the ratio Therefore, as a result of the transfer, the distance from the lens to the image will decrease (A - 2).

It can be seen from the figure that the linear dimensions of the object and image are related to the distances from the object and image to the lens by the ratio Thus, when the object is removed, the linear size of the image will decrease (B - 2).

Answer: 223.

Answer: 223

A small object is located on the main optical axis of a thin converging lens, at twice the focal length from it. How will the following three quantities change when the object is removed from the lens: the size of the image, its distance from the lens, the optical power of the lens?

For each value, determine the appropriate nature of the change:

1) increase;

2) decrease;

3) will not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

Under the initial conditions, the image of the object given by the lens is inverted, and the same size as the original. In accordance with the thin lens formula, the farther an object is from the lens under the same initial conditions, the closer its image will be to it. As for the optical power of the lens, it, like the focal length, is a characteristic of the lens and does not depend on the location of the object and its image.

The optical power of the lens (the reciprocal of the focal length) is a characteristic of the lens itself, therefore, when the object is removed from the lens, it does not change in any way. According to the thin lens formula, the distance from the object to the lens, the distance from the image to the lens and the focal length are related by the ratio

Therefore, as the object moves away from the lens, the image will move closer to the lens. It can be seen from the figure that the linear dimensions of the object and the image are related to the distances from the object and the image to the lens by the ratio. Thus, when the object is removed, the size of the image will decrease.

Answer: 223.

Answer: 223

Establish a correspondence between optical instruments and the types of images they give. For each position of the first column, select the desired position of the second and write down the selected numbers in the table under the corresponding letters.

Explanation.

When completing this task, it is useful to ask yourself a leading question: which rays are convenient to use to build an image in the case of the two devices mentioned above? The answer to it will help solve two other questions:

1) Is the image upside down or upside down?

2) is it real or imaginary?

The answers to them are obvious - provided that you imagine what flat mirror and how the simplest camera works.

Solution.

A flat mirror gives a direct virtual image (A - 1). The lens of the simplest camera is a converging lens that produces a real image on a photographic plate. This results in an inverted image. Therefore, the correct answer among the listed: B - 2.

Answer: 12.

Answer: 12

A beam of light falls on the glass-air interface. How will the following three quantities change with an increase in the refractive index of glass: the wavelength of light in glass, the angle of refraction, and the angle of total internal reflection?

For each value, determine the appropriate nature of the change:

1) increase;

2) decrease;

3) will not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

When a light beam passes from glass to air, the frequency of electromagnetic oscillations in a light wave does not change, since it does not depend on the medium in which the wave propagates. The wavelength is related to the frequency of electromagnetic oscillations and the speed of propagation by the relation As the refractive index increases, the speed of propagation of the light wave in the medium decreases, which means that the wavelength of light in the glass also decreases.

According to Snell's law of refraction, the sines of the angles of incidence and refraction when light exits glass into air are related to the refractive index of glass by the relation Therefore, as the refractive index increases, the angle of refraction will increase.

Finally, the angle of total internal reflection is given by the relation Thus, an increase will lead to a decrease in the angle of total internal reflection.

Answer: 212.

Answer: 212

Establish a correspondence between the varieties of a thin lens and the results of the refraction of parallel rays in it. For each position of the first column, select the desired position of the second and write down the selected numbers in the table under the corresponding letters.

Solution.

Rays parallel to the main optical axis of the converging lens, passing through it, will then pass through its far focus (A - 1). Rays parallel to the main optical axis of a divergent lens, passing through it, will appear to diverge from its focus (B - 3).

Answer: 13.

Answer: 13

Alexey (St. Petersburg)

I don't know, I have to think, it sounds like it. I see two scenarios here:

I'm more inclined towards the second option.

Valery Grigoriev 19.06.2016 06:42

And what is "near focus" and "far focus" in general? Focus is one, no?

Or is it for obfuscation?

Anton

The lens has two focal points - on both sides of it. The one on the side of the incident light is near, on the other side it is far.

A beam of light passes from air to glass. The frequency of the light wave is the speed of light in air - With refractive index of glass relative to air - n. Establish a correspondence between physical quantities and combinations of other quantities from which they can be calculated. For each position of the first column, select the desired position of the second and write down the selected numbers in the table under the corresponding letters.

Solution.

When a light beam passes from air to glass, the frequency of electromagnetic oscillations in a light wave does not change, since it does not depend on the medium in which the wave propagates. Since glass is an optically denser medium than air, when passing into glass, the speed of propagation of a light wave decreases and turns out to be equal to (A - 3). In turn, the wavelength is related to the frequency of electromagnetic oscillations and the speed of propagation by the relation Therefore, the wavelength of light in glass is equal to (B - 4).

Answer: 34.

Answer: 34

On the table is a vessel with a mirrored bottom and frosted walls. A beam of light falls on the bottom of an empty vessel. On the vessel wall, one can observe a "bunny" - a glare of the reflected beam. Some water is poured into the vessel. How do the following physical quantities change in this case: the angle of incidence of the beam on the bottom, the height of the point where the "bunny" is located, the distance from the point of reflection of the beam from the bottom of the vessel to the wall? Ignore the reflection of the beam from the surface of the liquid.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

Water is optically denser than air. Therefore, the angle of refraction of light when passing into water is less than the angle of incidence. Therefore, the beam "bends" down. In this case, the reflection point from the bottom, of course, shifts to the left, that is, the distance from the reflection point of the beam from the bottom to the wall increases (B - 1). The angle of incidence of the beam on the bottom decreases (A - 2). Finally, the height of the location of the "bunny" increases (B - 1).

Answer: 211.

Ilya Ostroushko (Moscow) 23.05.2013 22:29

Hello. As I understood from the condition of the problem, we need to determine the change in the angle of incidence of the beam TO THE BOTTOM. It obviously increases, but the angle WITH NORMAL to the bottom decreases.

Alexei

Good afternoon

The angle of incidence is the angle between the ray and the normal to the surface

On the table is a vessel with a mirrored bottom and frosted walls. A beam of light 1 falls on the bottom of an empty vessel. At the same time, a "bunny" can be observed on the wall of the vessel - a glare of the reflected beam. Some water is poured into the vessel. How do the following physical quantities change in this case: the angle of incidence of the beam on the wall, the distance from the wall to the point of reflection of the beam from the bottom of the vessel, the angle of reflection of the beam from the mirror? Ignore the reflection of the beam from the surface of the liquid.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

Water is optically denser than air. Therefore, the angle of refraction of light when passing into water is less than the angle of incidence. Therefore, the beam "bends" down. In this case, the reflection point from the bottom, of course, shifts to the left, that is, the distance from the reflection point of the beam from the bottom to the wall decreases (B - 2). Beam reflection angle from the bottom equal to the angle refraction, it decreases (B - 2). Finally, the angle of incidence of the beam on the vessel wall does not change (A - 3).

Answer: 322.

Answer: 322

OO", passing through the line of contact of the mirrors perpendicular to it, a point source of light is placed S. points S 1 , S 2 and S 3 - images of the source in these mirrors at a given opening angle. The mirror opening angle is increased to 120° (see figure on the right).

For each value, determine the appropriate nature of the change:

1) increase;

2) decrease;

3) will not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

The figure shows how many images will be obtained if the angle becomes 120°. There will be 2 source images. The distance from the source to the image closest to it will increase.

Answer: 21.

Answer: 21

Source: Training work in physics 10/12/2016, variant PHI10103

The figure shows two square flat mirrors touching each other at the edges (see the figure on the left). Mirror opening angle 90°. On line OO" passing through the line of contact of the mirrors perpendicular to it, a point source of light is placed S. points S1, S2 And S3 are the images of the source in these mirrors at a given opening angle. The mirror opening angle is reduced to 60° (see figure on the right).

Determine how the following quantities will change in this case: the number of source images in the mirrors; the distance from the source to the image closest to it.

For each value, determine the appropriate nature of the change:

1) increase;

2) decrease;

3) will not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

The figure shows how many images will be obtained if the angle becomes 60 o. There will be 5 source images. The distance from the source to the image closest to it will decrease.

Answer: 12.

Answer: 12

Source: Training work in physics 10/12/2016, variant PHI10104

A beam of light 1 falls on the surface of a horizontal mirror A at an angle = 20 o (see the figure on the left). Reflecting from mirror A, a beam of light hits the next two mirrors - B and C. First, mirrors B and C are located horizontally. Then they are rotated: mirror B at an angle counterclockwise, and mirror C is set vertically (as shown in the figure on the right).

For each value, determine the appropriate nature of the change:

1) increased

2) decreased

3) has not changed

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

The angle of reflection from mirror B will become equal, which means that the angle has increased. The angle of reflection from the mirror C will become according to the condition, which means that the angle has increased.

Answer: 11.

Answer: 11

Source: Training work in physics 12/21/2016, variant PHI10203

Light beam 1 falls on the surface of a horizontal mirror A at an angle = 20° (see figure on the left). Reflecting from mirror A, a beam of light hits the next two mirrors - B and C. First, mirrors B and C are located horizontally. Then they are rotated: mirror B at an angle clockwise (), and mirror C is set vertically (as shown in the figure on the right).

Determine the nature of the change in the angle of reflection of the incident beam 1 when it is reflected from mirrors B and C.

For each value, determine the appropriate nature of the change:

1) increased

2) decreased

3) has not changed

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

Before the rotation, the angles of reflection from the mirrors were equal. After the rotation, the angle of reflection from the mirror B will become equal, which means that the angle decreases. The angle of reflection from the mirror C will mean the angle has increased.

Answer: 21.

Answer: 21

Source: Training work in physics 12/21/2016, variant PHI10204

A small object is located on the main optical axis of a thin converging lens between the focal length and twice the focal length from it. The object is brought closer to the focus of the lens. How does this change the image size and optical power of the lens?

For each quantity, determine the appropriate nature of its change:

1) increases

2) decreases

3) does not change

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

The size of the image increases as the object approaches the focus of the converging lens.

3) If you reduce the wavelength of the incident light, then the distance on the screen between the zero and the first diffraction maxima will decrease.

4) If you replace the lens with another one with a larger focal length and position the screen so that the distance from the lens to the screen is still equal to the focal length of the lens, then the distance on the screen between the zero and the first diffraction maxima will decrease.

5) If we replace the diffraction grating with another one with a longer period, then the angle at which the first diffraction maximum is observed will increase.

Solution.

m. The beam of rays after a thin lens, according to the rules for constructing images in it, is collected at a point in the focal plane of the lens.

d, after it is ok m a parallel beam of light is obtained, going at such an angle that the maximum order order is determined by the relationship:

If we increase the wavelength of the incident light, then the maximum order of the observed diffraction peaks will not increase. 2 is incorrect.

If we reduce the wavelength of the incident light, then according to the basic equation, this will lead to a decrease in the angles and, as a result, the distance between the first and zero maximum on the screen will decrease. 3 is correct.

According to the rules for constructing rays in a converging lens, a lens with a large focal length will increase the distance between the zero and the first maximum. 4 is incorrect.

If we replace the diffraction grating with a grating with long period, then according to the main equation, this will lead to a decrease in the angles and, as a result, we will observe the first diffraction maximum on the screen at a smaller angle. 5 is incorrect.

Answer: 13.

Answer: 13|31

A diffraction grating having 1000 lines per 1 mm of its length is illuminated by a parallel beam of monochromatic light with a wavelength of 420 nm. Light is incident perpendicular to the grating. Close to the diffraction grating, immediately behind it, there is a thin converging lens. Behind the grating at a distance equal to the focal length of the lens, a screen is located parallel to the grating, on which the diffraction pattern is observed. Choose two true statements.

1) The maximum order of observed diffraction maxima is 2.

2) If you increase the wavelength of the incident light, then the maximum order of the observed diffraction maxima will increase.

3) If you reduce the wavelength of the incident light, then the distance on the screen between the zero and the first diffraction maxima will increase.

4) If we replace the lens with another one with a larger focal length and position the screen so that the distance from the lens to the screen is still equal to the focal length of the lens, then the distance on the screen between the zero and the first diffraction maxima will not change.

5) If we replace the diffraction grating with another one with a longer period, then the angle at which the first diffraction maximum is observed from the side of the screen will decrease.

Solution.

First, let us plot the course of parallel rays from the source passing through the diffraction grating and lens to the screen, where a spectrum of the order m(for some one spectral line of mercury with a wavelength ). The beam of rays after a thin lens, according to the rules for constructing images in it, is collected at a point in the focal plane of the lens.

According to the basic equation for the angles of deflection of light with a wavelength by a grating with a period d after that it's ok m a parallel beam of light is obtained, going at such an angle that the maximum order order will be observed at:

If we increase the wavelength of the incident light, then the maximum order of the observed diffraction maxima will not change or decrease. 2 is incorrect.

If the wavelength of the incident light is reduced, this will lead to a decrease in the angle between the zero and first diffraction maxima and, as a consequence, to a decrease in the distance between the zero and the first maximum on the screen. 3 is incorrect.

A beam of light falls at an angle α 0 . At the point IN a beam of light goes back into the air. points A And IN displaced relative to each other along the plates at a distance x. The middle plate is replaced with another one of the same thickness, but with a higher refractive index. As a result of this, the angle of refraction of light will change during the transition from the second plate to the third and the distance x?

For each value, determine the appropriate nature of the change:

1) increase;

2) decrease;

3) will not change.

Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

Solution.

According to Snell's law of refraction, the sines of the angles of incidence and refraction when light exits from one medium to another are related to the refractive indices by the relation

Let us write down the law of refraction when passing from air to the first plate

The crosswise lying angles are equal and then for the boundary between the first and second plates it is true

For the boundary between the second and third plates,

where is the angle of refraction of light during the transition from the second plate to the third.

Putting all the equations together, we get

This shows that the angle of refraction does not depend on the refractive index of the second plate, and when it is replaced, it will not change.

The distance is the sum of three deviations of the original beam in three plates, it can be found by the formula

Here, it decreases with increasing refractive index of the second plate, which means that the distance will also decrease.

Answer: 32.

The candle is located at a distance = 3.75 m from the screen. A converging lens is placed between the candle and the screen, which gives a clear image of the candle on the screen at two positions of the lens. Find the focal length of the lens F if the distance between the positions of the lens b = 0.75 m.

The lenses of modern cameras have a variable focal length. When changing the focal length, “focusing” does not go astray. Let us agree to consider an image on a camera film as sharp if, instead of an ideal image in the form of a dot on the film, an image of a spot with a diameter of no more than 0.05 mm is obtained. Therefore, if the lens is at a focal length from the film, then not only infinitely distant objects are considered sharp, but also all objects that are further than a certain distance d. It turned out that this distance is 5 m if the focal length of the lens is 50 mm. How will this distance change if, without changing the “relative aperture”, we change the focal length of the lens to 25 mm? (“Aperture ratio” is the ratio of the focal length to the diameter of the lens entry hole.) Consider the lens as a thin lens in calculations. Make a drawing explaining the formation of the spot F D d b f

Solution. 1. Let's express the distance d from the thin lens formula: (1) 2. It follows from the similarity of triangles: (2) where D is the lens diameter, b is the spot diameter on the screen. 3. We solve together (1) and (2) and get the value of d: (3), 4. According to the condition of the problem “relative hole” c = F / D, the value is constant, which means they are proportional to each other. With a decrease in focal length, the diameter of the lens should decrease by the same amount. This means that when the focal length is halved, the distance from which an object can be considered infinitely distant decreases by a factor of four.

Solution 1. Determine how far the virtual image of the source S` is from the lens: , From the mirror - at a distance of 7 cm. 2. However, the light is reflected from the mirror and forms a real image at the point S``. The reflected beam is symmetrical, from where, knowing the distance between the mirror and the lens, you can find how far it is from the lens. X \u003d 8 - 7 \u003d 1 cm. This means that its actual image will be at a distance of 8.5 cm from the light source.

Lens + flat mirror A flat mirror is pressed closely against a thin converging lens with focal length F. The image of the object is at a distance of 2 F from the lens. What is the magnification of the object? Solution: The optical system has an optical power equal to Do = D 1 + D 2 + Dz. This is justified by the fact that the beam is refracted twice and reflected once, Dz is the optical power of a flat mirror, which is equal to 0. This means that the system has a focal length F / 2. From here it is possible to determine the distance from the source to the lens d = 2 F/3, and the magnification is equal to Г = 3.

1. At what distance from each other should two lenses be placed: first, a scattering lens with a focal length of 4 cm, then a collecting one with a focal length of 9 cm, so that the beam of rays parallel to the main optical axis, passing through both lenses, remains parallel? 2. At what distance from each other should two lenses be placed: first collecting with a focal length of 30 cm, then scattering with a focal length of 20 cm, so that the beam of rays parallel to the main optical axis, passing through both lenses, remains parallel? lens + lens

One side of the thick glass plate has a stepped surface, as shown in the figure. A light beam is incident on the plate, perpendicular to its surface, which, after reflection from the plate, is collected by a lens. The length of the incident light wave l. At what smallest of the indicated step heights d will the light intensity at the focus of the lens be minimal?

1. A small load suspended on a thread 2.5 m long makes harmonic oscillations with an amplitude of 0.1 m. Using a converging lens with a focal length of 0.2 m, the image of the oscillating load is projected onto a screen located at a distance of 0.5 m from lenses. The main optical axis of the lens is perpendicular to the plane of oscillation of the pendulum and the plane of the screen. Determine the maximum speed of the cargo image on the screen. Let us denote the maximum speed of the pendulum υmax = Aω and the image υ`max =A`ω. (1). 2) The relationship between the amplitudes can be determined by the formula of a thin lens using a linear transverse magnification: 3. The oscillation frequency of the pendulum is Hence А` = A(f - F)/F (2), 4) Substitute (2) into formula (1) and define the desired value:

The lateral side of the rectangular trapezoid ABSD adjacent to its right angles is located on the main optical axis of a thin lens. The lens creates a real image of a trapezoid as a trapezoid with the same angles. If you rotate the trapezoid ABCD by 1800 around the side AB, then the lens creates an image of the trapezoid in the form of a rectangle. At what magnification is side AB displayed? B D A

B C 2 F D A 2 F F ​​D` A` C` C`` B` This means that the side BC before the lens and after the lens must lie on the same straight line. This will be the case if this line passes through a double focus. It is more advantageous to draw the second beam through the focus. It turns out a trapezoid A`B`C`D`. 2. According to the condition of the problem, when the trapezoid is rotated through AB, the image is obtained in the form of a rectangle. Let's build it. The beam that passes through the focus through the new point C gives its new image at level B`. Only if AB is located in the middle of the segment is this possible. 3. Based on the thin lens formula, taking into account d = 2/3 F, we obtain f = 3 F. Accordingly, the increase in the side AB is equal to G = f / d = 2

A thin glass biprism with a refractive angle of 0.05 rad, a refractive index of 1.5 and a width of 20 cm stands vertically in a beam of parallel light rays. Find the distance from the biprism to the screen at which the width of the shadow in the center of the screen is equal to the width of the biprism The position of the screen and the image on it

Solving the problems of the Unified State Examination, part C: Geometric optics with solutions C1.1. A thin lens L gives a clear real image of the object AB on the screen E (see Fig. 1). What will happen to the image of the object on the screen if the upper half of the lens is covered with a piece of black cardboard K (see Fig. 2)? Build an image of the subject in both cases. Explain your answer by indicating what physical patterns you used to explain. C5.1. A luminous panel-lamp in the form of a circle with a diameter of 2 m is attached to the ceiling of a room 6 m high. At a height of 3 m from the floor, an opaque square with a side of 2 m is located parallel to it. The center of the panel and the center of the square lie on the same vertical. Determine the minimum linear size of the shadow on the floor. Answer: 2 m. C5.2. A pile hidden under water is driven vertically into the bottom of a reservoir 3 m deep. The height of the pile is 2 m. The pile casts a shadow 0.75 m long on the bottom of the reservoir. Determine the angle of incidence of the sun's rays on the surface 4 of the water. Refractive index of water n = . 3 4  28º.  = arcsin 73    H h L С5.3. A pile is driven vertically into the horizontal bottom of a reservoir 3 m deep, completely hidden under water. At an angle of incidence of sunlight on the water surface equal to 30 °, the pile casts a shadow 0.8 m long on the bottom of the reservoir. Determine the height of the pile. Refractive index of water. Answer: h ≈ ​​2 m. С5.4. A pile hidden under water is driven vertically into the horizontal bottom of a reservoir 3 m deep. The height of the pile is 2 m. The angle of incidence of sunlight on the water surface is 30°. Determine the length of the shadow of the pile at the bottom of the reservoir. Refractive index of water. Answer: L ≈ 0.8 m. С5.5. A 3 m deep pool is filled with water, the relative refractive index at the air-water boundary is 1.33. What is the radius of the circle of light on the surface of the water from the electric lamp at the bottom of the pool? Answer: BC ≈ 3.4 m. C5.6. A pool 4 m deep is filled with water, the relative refractive index at the air-water interface is 1.33. What does the depth of the pool appear to be to an observer looking vertically down into the water? 1 Solving the problems of the Unified State Examination of Part C: Geometric optics with solutions Answer: h` = 3 m. C5.7. An inflatable raft 4 m wide and 6 m long floats on the surface of the water. sunlight. Determine the depth of the shadow under the raft. Ignore the depth of the raft and the scattering of light by water. The refractive index of water relative to air 4 is taken equal. 3 Answer: 1.76 m. C5.8. At the very surface of the water in the river, a mosquito flies, a flock of fish is at a distance of 2 m from the surface of the water. What is the maximum distance to a mosquito at which it can still be seen by fish at this depth? The relative refractive index of light at the air-water interface is 1.33. C5.9. A beam of light falls on a flat screen at an angle α = 45° and creates a bright spot on the screen. A flat glass plate is placed in front of the screen in the path of the beam, the edges of which are parallel to the screen. Plate thickness d = 4 cm, glass refractive index n = √2.5 = 1.58. The beam passes through both sides of the plate. How far will the bright dot move on the screen? Answer: s = 2 see C5.10. An image of a rod with a fivefold magnification was obtained on the screen using a thin lens. The rod is located perpendicular to the main optical axis, and the plane of the screen is also perpendicular to this axis. The screen was moved 30 cm along the main optical axis of the lens. Then, with the lens position unchanged, the rod was moved so that the image became sharp again. In this case, an image with a threefold increase was obtained. Determine the focal length of the lens. Answer: or. C5.11. An image of an object with a fivefold magnification was obtained on the screen using a thin lens. The screen was moved 30 cm along the main optical axis of the lens. Then, with the lens position unchanged, the object was moved so that the image became sharp again. In this case, an image with a threefold increase was obtained. At what distance from the lens was the image of the object in the first case? C5.12. A lens with a focal length of 15 cm gives an image of an object on the screen with a fivefold magnification. The screen was moved to the lens along its main optical axis by 30 cm. Then, with the lens position unchanged, the object was moved so that the image became sharp again. How far has the object been moved from its original position? C5.13. Determine the magnification given by a lens whose focal length is F \u003d 0.26 m, if the object is at a distance of a \u003d 30 cm from it. Answer: 6.5. 2 Solving the problems of the USE part C: Geometric optics with solutions С5.14. Isosceles right triangle ABC with an area of ​​50 cm2 is located in front of a thin converging lens so that its leg AC lies on the main optical axis of the lens. Lens focal length 50 cm. Top right angle C lies closer to the center of the lens than the top acute angle A. The distance from the center of the lens to point C is twice the focal length of the lens (see figure). Construct an image of a triangle and find the area of ​​the resulting figure. C5.15. A small load suspended on a long thread makes harmonic oscillations, in which it maximum speed reaches 0.1 m/s. Using a converging lens with a focal length of 0.2 m, the image of the oscillating load is projected onto a screen located at a distance of 0.5 m from the lens. The main optical axis of the lens is perpendicular to the plane of oscillation of the pendulum and the plane of the screen. The maximum displacement of the load image on the screen from the equilibrium position is A1 = 0.1 m. What is the length of the thread I? Answer: l ≈ 4.4 m. С5.16. A small load suspended on a thread 2.5 m long performs harmonic oscillations, at which its maximum speed reaches 0.2 m/s. Using a converging lens with a focal length of 0.2 m, the image of the oscillating load is projected onto a screen located at a distance of 0.5 m from the lens. The main optical axis of the lens is perpendicular to the plane of oscillation of the pendulum and the plane of the screen. Determine the maximum displacement of the load image on the screen from the equilibrium position. Answer: A1 = 0.15 m. C5.17. A load weighing 0.1 kg, attached to a spring with a stiffness of 0.4 N/m, performs harmonic oscillations with an amplitude of 0.1 m. Using a converging lens with a focal length of 0.2 m, the image of the oscillating load is projected onto a screen located at a distance of 0 .5 m from the lens. The main optical axis of the lens is perpendicular to the trajectory of the load and the plane of the screen. Determine the maximum speed of the cargo image on the screen. Answer: and \u003d 0.3 m / s. C5.18. A person reads a book, holding it at a distance of 50 cm from the eyes. If this is the distance of his best vision, then what optical power of glasses will allow him to read a book at a distance of 25 cm? Answer: D2 = 2 diopters. C5.19. Schoolchild with normal vision (distance best vision L = 25 cm) was bitten on the forehead above the eye by a bee. Looking in a flat mirror, he could not see if there was a sting left in the place of the bite. Then he took a small magnifying glass with an optical power of D = 16 diopters, and with the help of the same mirror he saw that there was no sting. How did he do it? Draw a possible optical scheme applied by the student and find the distance from the mirror to the magnifying glass in this scheme. All angles of incidence of rays are considered small. Answer: The magnifying glass is placed close to the eye, the mirror is placed at a distance of 2.5 cm from the magnifying glass. 3 Solving the problems of the USE part C: Geometric optics with solutions C5.20. The lens of the projection device has an optical power of 5.4 diopters. The screen is located at a distance of 4 m from the lens. Determine the dimensions of the screen on which the image of a 6x9 cm slide should fit. C5.21. On the X axis at the point x1 = 10 cm there is a thin diverging lens, and at the point x2 = 30 cm there is a thin converging lens with a focal length f2 = 24 cm. The main optical axes of both lenses lie on the X axis. Light from a point source located in point x = 0, having passed the given optical system, propagates in a parallel beam. Find the optical power D of the diverging lens. Answer: 15 diopters. C5.22. The camera lens has a focal length F = 5 cm, and the film frame size h · l = 24 · 36 mm. From what distance d should a drawing with dimensions H L = 240 300 mm be photographed in order to obtain maximum size Images? Answer: 55 cm. C5.23. The telescope has a lens with a focal length of 1 m and an eyepiece with a focal length of 5 cm. What diameter image of the Sun can be obtained with this telescope if it is possible to remove the screen from the eyepiece to a distance of 1.5 m? The angular diameter of the Sun is 30". only infinitely distant objects, but also all objects that are further than a certain distance d. The lens has a variable focal length. At the same time, the distance to which it is set (in this case) does not change. With a "relative aperture" α = 4, the minimum distance, at which objects are sharp, varies (when the focal length of the lens changes) from 12.5 to 50 m. (“Relative aperture” is the ratio of the focal length to the diameter of the lens inlet.) In what range does the focal length of the lens change? lens with a thin lens Draw a picture explaining the formation of the spot Answer: the focal length varies from 5 to 10 cm C5.25. Let us agree to consider an image on a camera film as sharp if instead of an ideal image in the form of a dot on the film, an image of a spot with a diameter of no more than a certain limit value is obtained. Therefore, if the lens is at a focal length from the film, then not only infinitely distant objects are considered sharp, but also all objects that are further than a certain distance d. Estimate the maximum spot size if, with a lens focal length of 50 mm and an inlet diameter of 5 mm, all objects located at distances of more than 5 m from the lens turned out to be sharp. Draw a picture explaining the formation of the spot. Answer: δ= 0.05 mm. 4

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3.6 GEOMETRIC OPTICS

3.6.1 Rectilinear propagation of light in a homogeneous medium. A ray of light

POSTULATE 1

In a homogeneous medium, light propagates in a straight line.

POSTULATE 2

Crossed light rays do not interact with each other.

Ray- a part of a straight line indicating the direction of light propagation.

3.6.2 Laws of light reflection

1) The incident beam, the reflected beam and the perpendicular to the boundary of two media, restored at the point of incidence of the beam, lie in the same plane.

2) The angle of incidence of the beam a is equal to the angle of reflection of the beam ß. The angles of incidence and reflection are measured between the direction of the rays and perpendicular.

3.6.3 Building images in a flat mirror

Building an Image of a Point Light Source

S - point source of light
MN - mirror surface
Divergent rays fall on it SO, SO 1, SO 2
According to the law of reflection, these rays are reflected at the same angle:
SO at an angle 0 0 ,
SO 1 at an angle β 1 = α 1,
SO 2 at an angle β 2 = α 2
A divergent beam of light enters the eye.
If we continue the reflected rays behind the mirror, then they will converge at the point S 1 .
A divergent beam of light enters the eye, as if coming from point S 1 .
This point is called the imaginary image of the point S.

Building an image of an object

1. We apply a ruler to the mirror so that one side of the right angle lies along the mirror.
2. Move the ruler so that the point we want to build lies on the other side of the right angle
3. We draw a line from point A to the mirror and extend it beyond the mirror by the same distance and get point A 1.
4. Similarly, we do everything for point B and get point B 1
5. We connect point A 1 and point B 1, we got the image A 1 B 1 of the object AB.

The image should be the same size as the object, be behind the mirror at the same distance as the object in front of the mirror.

3.6.4 Laws of light refraction

1. The incident and refracted rays and the perpendicular drawn to the interface between two media at the point of incidence of the ray lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value for two media, equal to the relative refractive index.

Light refraction:

Absolute refractive index:

Relative refractive index:

The course of rays in a prism

Passing through the prism White color(beam) is not only refracted, but also decomposed into a colored rainbow spectrum.

The ratio of frequencies and wavelengths during the transition of monochromatic light through the interface between two optical media:

3.6.5 Total internal reflection

Limiting angle of total internal reflection:

3.6.6 Converging and diverging lenses. Thin lens. Focal length and optical power of a thin lens:

3.6.7 Thin lens formula:

The magnification given by the lens:

3.6.8 The path of the beam passing through the lens at an arbitrary angle to its main optical axis. Construction of images of a point and a line segment in converging and diverging lenses and their systems

converging lens

If parallel rays fall on a converging lens, then they will meet at the focus, but if they go out of the imaginary focus and hit the lens, then after it they will pass parallel to each other.

If parallel rays go at a certain angle to the main axis, then they will also gather at one point, however, it will be called a secondary focus, which is located in the focal plane.

Ray rules:

1. Rays that hit the optical center do not change the trajectory of motion.

2. A beam parallel to the main axis is collected at a focus.

3. To understand where will he go a beam incident at a certain angle on the lens, you should build a side axis that will be parallel to it.

It should be carried out to the point of intersection with the focal plane. This will determine the side focus.

diverging lens

In a diverging lens, the beam is collected at an imaginary focus and diverges outside the lens.

If the rays fall at a certain angle to the lens, then in any case they will diverge, but in front of the lens they will gather in an imaginary secondary focus.

Ray rules:

1. This rule is true for all lenses - rays passing through the optical center do not change their trajectories.

2. If a beam parallel to the main optical axis hits the lens, then it scatters but crosses the imaginary focus.

3. To determine the secondary imaginary focus for a beam that falls on the lens at an angle, one should draw a secondary axis parallel to the path of the rays.

Image building

If there is a point in front of the lens, emitting light, then the image from this point can be obtained in the case of intersection of the rays in focus.

actual image rays intersect at some point after being refracted.

Imaginary image- image due to the intersection of the rays near the imaginary focus.

Building an image in a converging lens

1. The distance from the object to the lens is greater than the focal length: d>F.

To obtain an image, we direct one beam SO through the center of the lens, and the second SX arbitrary. Parallel to the arbitrary one, we place the side optical axis OP before crossing the focal plane. Let's draw a beam through the point of intersection of the focal plane and the side axis. We will guide the beam until it intersects with the beam SO. At this point, and show the image.

If the luminous point is located in a certain place on the axis, then we proceed in the same way - we lead an arbitrary beam to the lens, the side axis parallel to it, after the lens we pass the beam through the intersection point of the focal plane and the side axis. The place where this beam crosses the main optical axis will be the location of the image.

There is also an easier way to build an image. However, it is only used when the luminous dot is outside the main axis.

We draw two beams from the object - one through the optical center, and the other parallel to the main axis until it intersects with the lens. When the second beam has crossed the lens, we direct it through the focus. The place where the two beams intersect is the place for the image to be located.

Obtained images from objects after a converging lens

1. The subject is between the first and second focus, i.e. 2F > d >F.

If one edge of the object is on the main axis, then only its end point should be located behind the lens. We already know how to project a point.

It is worth noting the fact that if the body is between the first and second foci, then thanks to the converging lens, its image is obtained inverted, magnified and real.

The magnification is found as follows:

2. Image behind the second focus d > 2F.

If the location of the object has shifted to the left of the lens, then the resulting image will shift in the same direction.

The image is obtained reduced, inverted and real.

3. The distance to the object is less than the distance to the focus: F > d.

In this case, if we use the known rules and draw one beam through the center of the lens, and the second parallel, and then through the focus, we will see that they will diverge. They will connect only if they are continued in front of the lens.

This image will be imaginary, enlarged and direct.

4. The distance to the object is equal to the distance to the focus: d=F.

The rays after the lens go in parallel - this means that there will be no image.

diverging lens

For this lens, we use all the same rules as before. As a result of constructing similar images, we get:

Regardless of the location of the object relative to the diverging lens: the image imaginary, direct, enlarged.

3.6.9 The camera as an optical instrument

The eye as an optical system

The rays first hit the protective part of the eye, called the cornea.

Cornea- this is a spherical transparent body, which means that it refracts the rays that fall on it.

Depending on the distance at which the object is located, the lens changes the radii of curvature, which improves focusing. The process by which the lens involuntarily adjusts to the distance of an object is called accommodation. This process occurs when we look at an approaching or receding object.

The inverted and reduced image hits the retina, where nerve endings scan it, turn it over and send it to the brain.

vision problems

As you know, there are two main vision problems: farsightedness and nearsightedness. Both diseases are described solely from the point of view of physics, and are explained by the properties and thickness of the lens (crystalline lens).

If the rays from the object are connected in front of the retina, then the person suffers from myopia.

To correct this problem It is possible with the help of a diverging lens, that is, that is why patients are prescribed glasses.

farsightedness- with such a disease, the rays are connected after the retina, that is, the focus is outside the eye.

Converging lenses are used to correct this vision.

In addition to the natural optical device, there are also artificial ones: microscopes, telescopes, glasses, cameras and other objects. All of them have a similar structure. To improve or enlarge the image, a system of lenses is used (in a microscope, telescope).

Camera

An artificial optical device can be called a camera. Considering the structure of modern cameras is quite difficult. Therefore, in a school physics course, we will consider the most simple model, one of the first cameras.

The main optical converter that is able to capture a large object on film is the lens. A lens consists of one or more lenses that allow you to capture an image. The lens has the ability to change the position of the lenses relative to each other in order to focus the image, that is, to make it clear. We all know what a focused image looks like - it is clear, fully describes all the details of the subject. If the lenses in the lens are not focused, then the image is fuzzy and blurry. Similarly, a person with poor eyesight sees, because the image is not in focus.

To get an image from the reflection of light, you first need to open the shutter - only in this case the film will be illuminated at the moment of photographing. To provide the necessary flow of light, it is regulated using a diaphragm.

As a result of the refraction of rays on the lenses of the objective, an inverted, real and reduced image can be obtained on the film.