26.11.2019
I will solve the exam physics geometric optics. Construction of images in a flat mirror. Absolute refractive index
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 C5.3. 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 light circle on the water surface from 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. С5.7. 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 is 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
All formulas are taken in strict accordance with Federal Institute of Pedagogical Measurements (FIPI)
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 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
- We apply a ruler to the mirror so that one side of the right angle lies along the mirror.
- Move the ruler so that the point we want to build lies on the other side of the right angle
- 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.
- Similarly, we do everything for point B and get point B 1
- 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
- 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.
- 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 relative indicator refraction.
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 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.
“The system of preparing students for the exam.
Analysis of problematic tasks
from KIMs Unified State Exam-2010 "
(workshop)
1. At short circuit battery terminals, the current in the circuit is 12 A. When an electric lamp with an electrical resistance of 5 ohms is connected to the battery terminals, the current in the circuit is 2 A. Based on the results of these experiments, determine the internal resistance of the battery.
Given: Solution:
I k.z. = 12 A I k.z. = ε / r I = ε /( R+r)
R=5 Ohm ε = I To . h . ∙r ε = I (R + r)
i = 2 A I To . h . ∙r = I (R + r)
I To . h . ∙r = I∙R + I∙r
r-? I To . h . ∙r - I∙r = I∙R
r(I To . h . – I) = I∙R
r=IR /( I k.z. - I )
r= 2 A∙5 Ohm/(12A - 2A) =1 Ohm
Answer: 1 ohm
2. Find the internal resistance and EMF of the current source, if at a current of 30 A the power in the external circuit is 180 W, and at a current of 10 A this power is 100 W.
Given: Solution:
R 1 = 180 Tue R 1 = I 1 2 R 1 R 2 = I 2 2 R 2 R 1 ≠ R 2
I 1 = 30 A R 1 = R 1 / I 1 2 R 2 = R 2 / I 2 2
P 2 = 100 Tue ε = I 1 (R 1 +r) ε = I 2 (R 2 +r)
I 2 = 10 A ε = I 1 ( R 1 / I 1 2 +r) ε = I 2 ( R 2 / I 2 2 +r)
ε - ? r-? I 1 ( R 1 / I 1 2 + r) = I 2 ( R 2 / I 2 2 +r)
R 1 / I 1 + I 1 r = R 2 / I 2 + I 2 ∙r
I 1 ∙ r – I 2 r = R 2 / I 2 - R 1 / I 1
r(I 1 – I 2 ) = R 2 / I 2 - R 1 / I 1
r (I 1 – I 2 ) = (I 1 P 2 -I 2 P 1 ) / I 1 I 2 r = (I 1 P 2 -I 2 P 1 ) / I 1 I 2 (I 1 – I 2 )
r = 0.2 ohm
ε = P 1 / I 1 + I 1 ∙ r ε = 12 V
Answer: 12V; 0.2 ohm
3. The battery consists of 100 current sources with an EMF equal to 1 V and an internal resistance of 0.1 Ohm each. The sources were connected in groups of 5 in series, and these groups were connected in parallel. What is the maximum usable power that can be dissipated in the load resistance of this battery?
Given: Solution:
ε \u003d 1 V ε - EMF of 1 element, 5ε - EMF of one group
r = 0.1 ohm and the whole battery
n = 5 r is the internal resistance of the element, 5 r - groups,
N = 100 5 r /20 = r /4 is the internal resistance of the battery.
R -? Maximum power R m will be subject to
equality of internal and external resistances
R = r /4.
Current flows through the load resistor
I= 5 ε / (R + r /4) = 5 ε / (r /4 + r /4) = 5 ε∙ 4/2 r = 10 ε / r
P m = I 2 R=100 ε 2 / r 2 ∙ r /4 = 25 ε 2 / r
P m = 250 Tue
Answer : 250 Tue
1(10th-2007) Under the water is a rectangular pontoon 6 m long and 1 m high. the distance from the water surface to the lower surface of the pontoon is 2.5 m. The sky is covered with a continuous cloud cover, completely scattering sunlight. The depth of the shadow under the pontoon (measured from the bottom surface of the pontoon) is 2.3 m. Determine the width of the pontoon. Ignore the scattering of light by water. The refractive index of water relative to air is taken equal to 4/3. a
Solution: The shadow area is
outline those rays of light, γ
which before breaking
spread along
water surface, and after γ
refractions touch the edges h
pontoon. According to the figure,
the depth h of the shadow can be
determine by formula
h = where A
then Sin γ = tgγ = a= 2.3 . Answer: 5.2m
2.(2c-2007) Under water there is a rectangular pontoon 4 m wide, 6 m long and 1 m high. The distance from the water surface to the lower surface of the pontoon is 2.5 m. The sky is covered with a continuous cloud cover, completely scattering sunlight. Determine the depth of the shadow under the pontoon. (counting from the bottom surface of the pontoon) Disregard the scattering of light by water. The refractive index of water relative to air is taken equal to 4/3.
Solution: The shadow area is A
pyramid, side faces which
outline those rays of light, γ
which before breaking
spread along
water surface, and after γ
refractions touch the edges h
pontoon. According to the figure,
the depth h of the shadow can be
determine by formula
h = where A- half-width of the pontoon. Hence: a = h tgγ, Law of refraction: , where α = 90 0
then Sin γ = tg γ = h = .
3.(1c-2007) A rectangular inflatable raft 6m long floats on the surface of the water. The sky is covered with a continuous cloud cover, completely scattering the sunlight. The depth of the shadow under the raft is 2.3 m. Determine the width of the raft. Ignore the depth of the raft and the scattering of light by water. . The refractive index of water relative to air is taken equal to 4/3.
Solution: The shadow area is A
pyramid, the side faces of which
outline those rays of light, γ
which before refraction γ
spread along
water surface, and then
refractions touch the edges
pontoon. According to the figure,
the depth h of the shadow can be
determine by formula
h = where A- half-width of the pontoon. Hence: a = h tg γ, Law of refraction: , where α = 90 0
then Sin γ = tgγ = a= 2.3 . Answer: 5.2m
4. (v-5.2007) An isosceles right triangle ABC is located in front of a thin converging lens with an optical power of 2.5 diopters so that its leg AC lies on the main optical axis of the lens (Fig) The vertex of the right angle C lies further from the center of the lens than the vertex of the sharp angle A. The distance from the center of the lens to point C is equal to twice the focal length of the lens. AC = 4cm. Construct an image of a triangle and find the area of the resulting figure.
Solution: Δ ABC is isosceles.
SA= a = 4 cm
BC \u003d 4 cm (since the triangle is isosceles) Area Δ A I B I C I S \u003d C I B I X.
C I B I = BC = 4cm. (for BC d = f = 2F, magnification Г = 1)
To find X, we consider the image of t.A. Thin lens formula:
Here = 0.25 diopters, d = 2F - a \u003d 0.8m - 0.04m \u003d 0.76m \u003d 76cm.
F = 0.8445m. X \u003d f - 2F \u003d 0.0445m (according to the figure)
S \u003d ½ 4 cm 4.45 cm \u003d 8.9 cm 2.
5. (in-12-2007) An isosceles right triangle ABC is located in front of a thin converging lens with an optical power of 2.5 diopters so that its leg AC lies on the main optical axis of the lens (Fig) The vertex of the right angle C lies closer to the center of the lens than the apex of an acute angle A. The distance from the center of the lens to point C is equal to twice the focal length of the lens. AC = 4cm. Construct an image of a triangle and find the area of the resulting figure. (rice) Answer: 7.3 cm 2.
6. ((v-14-2007) An isosceles right triangle ABC is located in front of a thin converging lens with an optical power of 2.5 diopters so that its leg AC lies on the main optical axis of the lens (Fig) The vertex of the right angle C lies closer to the center of the lens, than the apex of the acute angle A. The distance from the center of the lens to point C is equal to twice the focal length of the lens AC = 4 cm Build an image of a triangle and find the area of the resulting figure (Fig) Answer: 9.9 cm 2.
2F a F F 2F
7. (in-11-2007) An isosceles right triangle ABC is located in front of a thin converging lens with an optical power of 2.5 diopters so that its leg AC lies on the main optical axis of the lens (Fig) The vertex of the right angle C lies farther from the center of the lens than the apex of an acute angle A. The distance from the center of the lens to point C is equal to twice the focal length of the lens. AC = 4cm. Construct an image of a triangle and find the area of the resulting figure. (rice) Answer: 6.6 cm 2.
a 2F F y
8. (C4 -2004-5) On the Ox axis at the point x 1 = 10 cm is the optical center of a thin diverging lens with a focal length F 1 = -10 cm, and at the point x 2 = 25 cm - a thin converging lens. The main optical axes of both lenses coincide with the Ox axis. Light from a point source located at the point x = 0, having passed through this optical system, propagates in a parallel beam. Find the focal length of the converging lens F 2 .
Solution: d \u003d X 1 \u003d 10 cm F 1 \u003d -10 cm,
Depicting the course of the rays. The image of t.O is obtained in t. O 1 at a distance d 1 from the diverging lens. This point is the focus of the converging lens due to the condition of parallelism of the beam passing through the optical system. Then the thin lens formula for a diverging lens is: where d 1 is the distance from the lens to the image. d 1 \u003d F 2 \u003d d 1 + (X 2 - X 1) \u003d 20 cm.
9. (S6-2004-5) On the Ox axis at the point x 1 = 10 cm is the optical center of a thin diverging lens, and at the point x 2 = 30 cm - a thin converging lens with a focal length F 2 = 25 cm .. The main optical the axes of both lenses coincide with the x-axis. Light from a point source located at the point x = 0, having passed through this optical system, propagates in a parallel beam. Find the focal length of the diverging lens F 1. Answer: 10 cm.
10. On the Ox axis at the point x 1 \u003d 0 cm, there is the optical center of a thin diffusing lens with a focal length F 1 \u003d -20 cm, and at the point x 2 \u003d 20 cm - a thin converging lens with a focal length F 2 \u003d 30 cm .. The main optical axes of both lenses coincide with the Ox axis. Light from a point source S located at point x< 0, пройдя данную оптическую систему, распространяется параллельным пучком. Найдите координату Х точечного источника. .Ответ:
11. (B9-2005) On the Ox axis at the point x 1 = 10 cm is the optical center of a thin diverging lens with a focal length F 1 = - 10 cm, and at the point x 2 > X 1 - a thin converging lens with a focal length F 2 \u003d 30 cm. The main optical axes of both lenses coincide with the Ox axis. Light from a point source located at the point x = 0, having passed through this optical system, propagates in a parallel beam. Find the distance between the lenses. Answer:
12. (B21-2005) 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 to make the image sharp. How far the object has been moved relative to its original position.
Given: F = 15 cm
Thin lens formula for the first case: Г = 5. f = 5d.
From here: . f = 0.9m; f 1 \u003d f - X \u003d 0.6 m.
Lens formula for the second case: hence d 1 =
y \u003d d 1 - d \u003d 0.2m - 0.18m \u003d 0.02m \u003d 2 cm.
13(20-2005) A lens with a focal length of 15 cm produces 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 to make the image sharp. Determine the increase in the second case. (Answer: G 1 \u003d 3)
14.(18-2005) 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. Then, with the lens position unchanged, the object was moved so that the image became sharp. In this case, an image with a threefold increase was obtained. How much was the screen shifted relative to its original position7 (Answer: x \u003d 30 cm)
15.(2002) To “enlighten the optics”, a thin film with a refractive index of 1.25 is applied to the surface of the lens. What should be the minimum thickness of the film in order for light with a wavelength of 600 nm from the air to completely pass through the film? (the refractive index of the film is less than the refractive index of the lens glass).
Solution: Optical coating is based on interference. On the surface of optical glass, a thin film is applied with a refractive index n p, less than the refractive index of glass n st. With the right choice of thickness, the interference of rays reflected from it leads to damping, which means that the light passes completely through it. Minimum condition: Δd = (2к+1) The path difference of the waves reflected from the upper and lower surfaces of the film is equal to twice the film thickness, on the one hand. Δd = 2h. On the other hand, the path difference is equal to Δd = (minimum condition at k = 0). Wavelength λ in the film less than length waves λ 0 in vacuum n times. λ = Hence: Δd=λ/4n=120nm
16. The camera lens has a focal length of 5 cm, and the frame size is 24x35mm. From what distance should a drawing of 480x600mm be photographed in order to get the maximum image size? What part of the frame area will be occupied by the image?
Solution: make a drawing.
Find magnification: G =
Lens formula:
We find the ratio of the areas of the image and the frame: η =
Frame size: 24x35. We find the image size: 480:20=24, and 600:20=30 (since the maximum image is reduced by 20 times)
No. 21. (B-5-06rv) A lens with a focal length of 12 cm gives an image of an object on the screen with a fourfold increase. The screen was moved 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. How much did you have to move the object relative to its original position? (Answer: 1 cm)
22. (6-6rv). In a dark room, a neon gas-discharge lamp stands on a table, emitting a vertical strip of red light. On the instructions of the teacher, the student looks at the lamp through the glass prism of the spectroscope and clearly sees already three colored lines6 red, yellow, and green. Next, the student looks at the lamp through a diffraction grating, placing the strokes of the grating vertically. What can the student see in this case? Justify your conclusions.
(Answer: zkzhzKzzhkz)
No. 23. (7-6v). In a dark room, a neon discharge lamp sits on a table, emitting a vertical stripe of blue light. On the instructions of the teacher, the student looks at the lamp through the glass prism of the spectroscope and clearly sees already three colored lines: One green and two blue. Next, the student looks at the lamp through a diffraction grating, placing the strokes of the grating vertically. What can the student see in this case? Justify your conclusions.
(Answer: sssssssss)
No. 24. (8-6v). In a dark room, a neon gas-discharge lamp stands on a table, emitting a vertical strip of red light. On the instructions of the teacher, the student looks at the lamp through the glass prism of the spectroscope and clearly sees already three colored lines6 red, orange, and blue. Next, the student looks at the lamp through a diffraction grating, placing the strokes of the grating vertically. What can the student see in this case? Justify your conclusions.
(Answer: gkogKgokg)
No. 25. (7-6v). In a dark room, a neon discharge lamp sits on a table, emitting a vertical stripe of blue light. On the instructions of the teacher, the student looks at the lamp through the glass prism of the spectroscope and clearly sees already three colored lines: two blue and one violet. Next, the student looks at the lamp through a diffraction grating, placing the strokes of the grating vertically. What can the student see in this case? Justify your conclusions.
(Answer: fssfsfssf)
No. 26. (6-6v). In a dark room, a neon gas-discharge lamp stands on a table, emitting a vertical strip of red light. On the instructions of the teacher, the student looks at the lamp through the glass prism of the spectroscope and clearly sees already three colored lines, among which the brightest are one red, one yellow, one blue. Next, the student looks at the lamp through a diffraction grating, placing the strokes of the grating vertically. What can the student see in this case? Justify your conclusions.
(Answer: gkzhgKgzhkg)
No. 27.(134-2004) A thin wire is placed between the edges of two well-polished thin flat glass plates; The opposite ends of the plates are tightly pressed against each other. (see pic). A monochromatic beam of light 600 nm long falls on the upper plate normally to its surface. Determine the angle α that the plates form if the distance between the observed interference fringes is 0.6 mm. Assume that tan α ≈ α.
Given: λ= 6nm. l = 0.6mm. Solution:
K=1 k=2
Maximum condition: Δd = kλ. (1) h 1 h 2
The travel difference is: Δd = 2h. (2) α ≈ tanα. (3) α ≈ , (4) l
where Δh = difference in the distance between the plates in the places of adjacent maxima, l is the distance between adjacent maxima, α is the angle between the plates.
k=2). Then Δh = h 2 – h 1 = We substitute the last expression in (4): α ≈ ,
28.(133-2004) Between the edges of two well sanded
thin flat glass plates placed
thin wire with a diameter of 0.075 mm; opposite
The ends of the plates are tightly pressed against each other (see figure). A monochromatic beam of light with a wavelength of 750 nm falls on the upper plate normally to its surface. Determine the length of the plate x, if interference fringes are observed on it,
The distance between them is 0.6 mm. X
Given: D= 0.075mm
λ = 750 nm. h 1 h 2
Find: x=?
Maximum condition: Δd = kλ. (1)
The travel difference is: Δd = 2h. (2) From the similarity of triangles: ;(3) where Δh = h 2 – h 1 is the difference in the distances between the plates in the places of adjacent maxima, l is the distance between adjacent maxima, X is the length of the plate. From equation (3) we express Х = (4);
From equations (1) and (2) we obtain: kλ. = 2h. hence h 1 = (for k =1), h 2 = (for
k=2). Then Δh \u003d h 2 - h 1 \u003d We substitute the last expression in (4): X \u003d
Answer: X = 12 cm.
29(131-2004) Between the edges of two well sanded
thin flat glass plates placed a thin wire with a diameter of 0.085 mm; the opposite ends of the plates are tightly pressed against each other (see figure). The distance from the wire to the line of contact between the plates is 25 cm. A monochromatic
a beam of light with a wavelength of 700 pm. Determine the number of observables
interference fringes per 1 cm wedge length.
Given: D= 0.085mm Solution:
X = 25 cm Maximum condition: Δd = kλ. (1) The travel difference is: Δd = 2h. (2)
λ = 700 nm. From the similarity of triangles: ;(3) where Δh = h 2 - h 1 is
L = 1 cm difference between the distances between the plates in the places of neighboring maxima,
Find: n = ? l is the distance between adjacent maxima,
X is the length of the plate. From equation (3) we express l = (4); To find the number of maxima per 1 cm of length, given that Δh = h 2 - h 1 = we get:
30(127-2004) Between the edges of two well sanded 20 cm
thin flat glass plates placed a thin
wire with a diameter of 0.05 mm; opposite ends
plates are tightly pressed against each other (see figure).
Distance from the wire to the line of contact
plates is 20 cm. On the top plate is normal
a monochromatic
beam of light. Determine the wavelength of light if
1 cm in length, 10 interference fringes are observed. Answer: 500 nm.
31.(82-2007) Soap film is a thin layer of water. on the surface of which are soap molecules. providing mechanical stability and not affecting the optical properties of the film, the soap film is stretched over a square frame. The two sides of the frame are horizontal. and the other two are vertical. Under the action of gravity, the film took the form of a wedge (see figure), thickened at the bottom, with an angle at the top α = 2·10 -4 rad. When the square is illuminated by a parallel beam of laser light with a wavelength of 666 nm (in air), incident perpendicular to the film, part of the snow is reflected from it, forming an interference pattern on its surface, consisting of 20 horizontal stripes. What is the height of the frame if the refractive index of water is 4/3.?
Wedge apex angle α = , where a- side of the frame. From here A =
32 (81-2008) Unified state exam 2006 Physics, grade 11.
A soap film is a thin layer of water, on the surface of which there are soap molecules that provide mechanical stability and do not affect the optical properties of the film. Soap film stretched over a square frame with side a = 2.5 cm. Two sides of the frame are horizontal and the other two are vertical. Under the action of gravity, the film took the form of a wedge (see figure), thickened at the bottom, with an angle at
top α = 2 10 -4 rad. When the square is illuminated by a parallel beam of laser light with a wavelength of 666 nm (in air), incident perpendicular to the film, part of the light is reflected from it, forming an interference pattern on its surface, consisting of 20 horizontal stripes. What is the refractive index of water?
Solution: The condition for the formation of an interference pattern:
Δd=k; where λ I = (wavelength in water), k is the number of bands, Δd is the path difference, in this case, the difference in film thickness in the lower and upper parts of the film. Δd=k;
Wedge apex angle α = , where a- side of the frame. n=
33. (79-2006) Soap film is a thin layer of water on
surface of which there are soap molecules that provide mechanical stability and do not affect the optical properties of the film. The soap film is stretched over a square frame with side a = 2.5 cm. Two sides of the frame are horizontal, and the other two are vertical. Under the action of gravity, the film took the form of a wedge (see figure), thickened at the bottom, with an apex angle α. When the square is illuminated by a parallel beam of laser light with a wavelength of 666 nm (in air), incident perpendicular to the film, part of the light is reflected from it, forming an interference pattern on its surface, consisting of 20 horizontal stripes. What is equal to the angle at the top of the wedge, if the refractive index of water is n = 4/3? (answer: α ≈ 2 10 -4 rad.)
34.(80-2006) A soap film is a thin layer of water, on the surface of which there are soap molecules that provide mechanical stability and do not affect the optical properties of the film. The soap film is stretched over a square frame with side A\u003d 2.5 cm. two sides of the frame are horizontal, and the other two are vertical. Under the action of gravity, the film took the form of a wedge (see figure), thickened at the bottom, with an angle at the top α = 2·10 -4 rad. When the square is illuminated by a parallel beam of laser light with a wavelength of 666 nm (in air), incident perpendicular to the film, part of the light is reflected from it, forming an interference pattern on its surface, consisting of horizontal stripes. How many bands are observed on the film if the refractive index of water is 4/3. (Answer: 20)
