The course of rays in the cross section of a triangular prism. geometric optics

Law of refraction of light

The phenomenon of light refraction, probably, everyone has met more than once in Everyday life. For example, if you lower a tube into a transparent glass of water, you will notice that the part of the tube that is in the water seems to be shifted to the side. This is explained by the fact that at the boundary of two media there is a change in the direction of the rays, in other words, the refraction of light.

In the same way, if you lower a ruler into the water at an angle, it will seem that it has been refracted and its underwater part has risen higher.

After all, it turns out that the rays of light, being at the border of air and water, experience refraction. A beam of light hits the surface of the water at one angle, and then it goes deeper into the water at a different angle, at a lesser inclination to the vertical.

If you send a backward beam from water into air, it will follow the same path. The angle between the perpendicular to the media interface at the point of incidence and the incident beam is called the angle of incidence.

The angle of refraction is the angle between the same perpendicular and the refracted ray. The refraction of light at the boundary of two media is explained by the different speed of propagation of light in these media. When light is refracted, two regularities are always fulfilled:

Firstly, the rays, regardless of whether it is incident or refracted, as well as the perpendicular, which is the boundary between two media at the break point of the beam, always lie in the same plane;

Secondly, the ratio sinus of the angle of incidence to sinus of the angle of refraction is a constant value for these two media.

These two statements express the law of refraction of light.

The sinus of the angle of incidence α is related to the sinus of the angle of refraction β, just as the speed of a wave in the first medium, v1, is related to the speed of a wave in the second medium, v2, and is equal to n. N is a constant value that does not depend on the angle of incidence. The value n is called the refractive index of the second medium with respect to the first medium. And if vacuum was used as the first medium, then the refractive index of the second medium is called the absolute refractive index. Accordingly, it is equal to the ratio of the sinus of the angle of incidence to the sinus of the angle of refraction during the transition of a light beam from vacuum to a given medium.

The refractive index depends on the characteristics of light, on the temperature of the substance and on its density, that is, on the physical characteristics of the medium.

It is more often necessary to consider the transition of light through the air-solid or air-liquid interface than through the interface of a vacuum-defined medium.

It should also be noted that the relative refractive index of two substances is equal to the ratio of the absolute refractive indices.

Let's get acquainted with this law with the help of simple physical experiments, which are available to you all at home.

Experience 1.

Let's put the coin in the cup so that it is hidden behind the edge of the cup, and now we will pour water into the cup. And here's what is surprising: the coin appeared from behind the edge of the cup, as if it floated up, or the bottom of the cup rose up.

Let's draw a coin in a cup of water, and the rays of the sun coming from it. At the interface between air and water, these rays are refracted and exit the water at a large angle. And we see the coin in the place where the lines of refracted rays converge. Therefore, the visible image of the coin is higher than the coin itself.

Experience 2.

Let's put it on the way parallel rays light container filled with water with parallel walls. At the entrance from the air into the water, all four beams turned through a certain angle, and at the exit from the water into the air, they turned through the same angle, but in the opposite direction.

Let's increase the slope of the rays, and at the output they will still remain parallel, but will move more to the side. Because of this shift, the lines of the book, when viewed through a transparent plate, appear to be cut. They move up, as the coin went up in the first experiment.

All transparent objects, as a rule, we see solely due to the fact that light is refracted and reflected on their surface. If such an effect did not exist, then all these items would be completely invisible.

Experience 3.

We lower the Plexiglas plate into a vessel with transparent walls. She is perfectly visible. And now we will pour sunflower oil into the vessel, and the plate has become almost invisible. The fact is that light rays at the border of oil and plexiglass are almost not refracted, so the plate becomes an invisible plate.

The path of rays in a triangular prism

In various optical devices, a triangular prism is quite often used, which can be made of a material such as glass, or other transparent materials.

When passing through a triangular prism, rays are refracted on both surfaces. The angle φ between the refractive surfaces of the prism is called the refractive angle of the prism. The deflection angle Θ depends on the refractive index n of the prism and the angle of incidence α.

You all know the famous rhyme for memorizing the colors of the rainbow. But why are these colors always arranged in the same order as they come from white sunlight, and why there are no other colors in the rainbow besides these seven, not everyone knows. It is easier to explain this through experiments and observations.

We can see beautiful iridescent colors on soap films, especially if these films are very thin. The soapy liquid flows down and the colored stripes move in the same direction.

Take a transparent cover from a plastic box, and now tilt it so that the white screen of the computer is reflected from the cover. Unexpectedly bright iridescent stains will appear on the lid. And what beautiful rainbow colors you see when the light reflects off the CD, especially if you shine a flashlight on the disc and throw this rainbow picture on the wall.

The first to explain the appearance of rainbow colors was the great English physicist Isaac Newton. He let a narrow beam of sunlight into the dark room, and placed a triangular prism in its path. The light leaving the prism forms a colored band called the spectrum. Red is the least deviated in the spectrum, and violet is the strongest. All other colors of the rainbow are located between these two without particularly sharp boundaries.

Laboratory experience

Let's choose a bright LED flashlight as a white light source. To form a narrow light beam, put one slit immediately behind the flashlight, and the second directly in front of the prism. A bright rainbow stripe is visible on the screen, where red, green and blue are clearly distinguishable. They form the basis of the visible spectrum.

Let's put a cylindrical lens in the path of a colored beam and adjust it for sharpness - the beam on the screen gathered into a narrow strip, all the colors of the spectrum mixed, and the strip turned white again.

Why does the prism turn White light into the rainbow? It turns out that all the colors of the rainbow are already contained in white light. The refractive index of glass varies for rays of different colors. Therefore, the prism deflects these rays differently.

Each individual color of the rainbow is pure and can no longer be split into other colors. Newton proved this experimentally by separating a narrow beam from the entire spectrum and placing a second prism in its path, in which no splitting had already occurred.

Now we know how a prism decomposes white light into individual colors. And in a rainbow, water droplets work like small prisms.

But if you shine a flashlight on a CD, a slightly different principle works, unrelated to the refraction of light through a prism. These principles will be studied further, in physics lessons devoted to light and the wave nature of light.

Let us consider some particular cases of light refraction. One of the simplest is the passage of light through a prism. It is a narrow wedge of glass or other transparent material that is in the air.

The path of rays through a prism is shown. It deflects light rays towards the base. For clarity, the prism profile is chosen in the form right triangle, and the incident ray is parallel to its base. In this case, the refraction of the beam occurs only on the back, oblique face of the prism. The angle w at which the incident beam is deflected is called the deflecting angle of the prism. It practically does not depend on the direction of the incident beam: if the latter is not perpendicular to the edge of incidence, then the deflecting angle is the sum of the angles of refraction on both faces.

The deflecting angle of the prism is approximately equal to the product of the angle at its apex and the refractive index of the prism substance minus 1:

w = α(n-1).

Let's draw a perpendicular to the second face of the prism at the point where the beam falls on it (dash-dotted line). It forms an angle β with the incident beam. This angle is equal to the angle α at the top of the prism, since their sides are mutually perpendicular. Since the prism is thin and all the considered angles are small, their sines can be considered approximately equal to the angles themselves, expressed in radians. Then from the law of refraction of light it follows:

In this expression, n is in the denominator, since light travels from a denser medium to a less dense one.

Let's swap the numerator and denominator, and also replace the angle β with the angle α equal to it:

Since the refractive index of glass commonly used for spectacle lenses is close to 1.5, the deflecting angle of the prisms is about half the angle at their apex. Therefore, glasses rarely use prisms with a deflecting angle of more than 5 °; they will be too thick and heavy. In optometry, the deflecting action of prisms (prismatic action) is often measured not in degrees, but in prismatic diopters (Δ) or in centiradians (srad). The deflection of rays by a prism with a force of 1 pdptr (1 srad) at a distance of 1 m from the prism is 1 cm. This corresponds to an angle whose tangent is 0.01. This angle is 34".

The same applies to the visual defect itself, strabismus, corrected by prisms. The angle of strabismus can be measured in degrees and in prism diopters.

11.2. geometric optics

11.2.2. Reflection and refraction of light rays in a mirror, a plane-parallel plate and a prism

Image formation in flat mirror and its properties

The laws of reflection, refraction and rectilinear propagation of light are used in constructing images in mirrors, considering the path of light rays in a plane-parallel plate, prism and lenses.

The course of light rays in a flat mirror shown in fig. 11.10.

The image in a flat mirror is formed behind the mirror plane at the same distance from the mirror f as the object is in front of the mirror d:

f = d.

The image in a flat mirror is:

• direct;
• imaginary;
• equal in size to the subject: h \u003d H.

If flat mirrors form a certain angle between themselves, then they form N images of a light source placed on the bisector of the angle between the mirrors (Fig. 11.11):

N = 2 π γ − 1 ,

where γ is the angle between the mirrors (in radians).

Note. The formula is valid for angles γ for which the ratio 2π/γ is an integer.

For example, in fig. 11.11 shows a light source S lying on the bisector of the angle π / 3. According to the above formula, five images are formed:

1) the image S 1 is formed by the mirror 1;

2) the image S 2 is formed by the mirror 2;

Rice. 11.11

3) image S 3 is a reflection of S 1 in mirror 2;

4) image S 4 is a reflection of S 2 in mirror 1;

5) the image S 5 is a reflection of S 3 in the continuation of mirror 1 or a reflection of S 4 in the continuation of mirror 2 (reflections in these mirrors coincide).

Example 8. Find the number of images of a point source of light obtained in two flat mirrors forming an angle of 90° with each other. The light source is on the bisector of the specified angle.

Solution . Let's draw a picture explaining the condition of the problem:

• the light source S is located on the bisector of the angle between the mirrors;
• the first (vertical) mirror Z1 forms an image S 1;
• the second (horizontal) mirror Z2 forms an image S 2 ;
• the continuation of the first mirror forms the image of the imaginary source S 2, and the continuation of the second mirror - the imaginary source S 1; these images match and give S 3.

The number of images of the light source placed on the bisector of the angle between the mirrors is determined by the formula

N = 2 π γ − 1 ,

where γ is the angle between the mirrors (in radians), γ = π/2.

The number of images is

N = 2 π π / 2 − 1 = 3 .

The course of a light beam in a plane-parallel plate

The course of the light beam in plane-parallel plate depends on the optical properties of the medium in which the plate is located.

1. The course of a light beam in a plane-parallel plate located in an optically homogeneous medium(on both sides of the plate, the refractive index of the medium is the same), is shown in fig. 11.12.

A light beam incident on a plane-parallel plate at a certain angle i 1 after passing through the plane-parallel plate:

• leaves it at the same angle:

i 3 = i 1 ;

• is displaced by x from the original direction (dashed line in Fig. 11.12).

2. The course of a light beam in a plane-parallel plate located at the border of two environments(on both sides of the plate, the refractive indices of the media are different), is shown in Fig. 11.13 and 11.14.

Rice. 11.13

Rice. 11.14

The light beam after passing through the plane-parallel plate exits the plate at an angle different from the angle of its incidence on the plate:

• if the refractive index of the medium behind the plate is less than the refractive index of the medium in front of the plate (n 3< n 1), то:

i 3 > i 1 ,

those. the beam comes out at a larger angle (see Fig. 11.13);

• if the refractive index of the medium behind the plate is greater than the refractive index of the medium in front of the plate (n 3 > n 1), then:

i 3< i 1 ,

those. the beam exits at a smaller angle (see Fig. 11.14).

Beam displacement - the length of the perpendicular between the beam leaving the plate and the continuation of the beam incident on the plane-parallel plate.

The beam displacement at the exit from a plane-parallel plate located in an optically homogeneous medium (see Fig. 11.12) is calculated by the formula

where d is the thickness of the plane-parallel plate; i 1 - the angle of incidence of the beam on a plane-parallel plate; n is the relative refractive index of the plate material (relative to the medium in which the plate is placed), n = n 2 /n 1; n 1 - absolute indicator refraction of the medium; n 2 is the absolute refractive index of the plate material.

Rice. 11.12

The displacement of the beam as it leaves the plane-parallel plate can be calculated using the following algorithm (Fig. 11.15):

1) calculate x 1 from triangle ABC, using the law of refraction of light:

where n 1 is the absolute refractive index of the medium in which the plate is placed; n 2 - absolute refractive index of the plate material;

2) calculate x 2 from triangle ABD ;

3) calculate their difference:

Δx \u003d x 2 - x 1;

4) the displacement is found by the formula

x = Δx  cos i 1 .

Light beam propagation time in a plane-parallel plate (Fig. 11.15) is determined by the formula

where S is the path traveled by the light, S = | A C | ; v is the speed of propagation of the light beam in the material of the plate, v = c / n; c is the speed of light in vacuum, c ≈ 3 ⋅ 10 8 m/s; n is the refractive index of the plate material.

The path traveled by the light beam in the plate is related to its thickness by the expression

S = d  cos i 2 ,

where d is the plate thickness; i 2 - the angle of refraction of the light beam in the plate.

Example 9. The angle of incidence of a light beam on a plane-parallel plate is 60°. The plate has a thickness of 5.19 cm and is made of a material with a refractive index of 1.73. Find the displacement of the beam at the exit from the plane-parallel plate if it is in the air.

Solution . Let's make a drawing in which we show the path of a light beam in a plane-parallel plate:

• a light beam falls on a plane-parallel plate at an angle i 1 ;
• at the interface between the air and the plate, the beam is refracted; the angle of refraction of the light beam is equal to i 2 ;
• at the interface between the plate and air, the beam is refracted again; the angle of refraction is i 1 .

The specified plate is in the air, i.e. on both sides of the plate, the medium (air) has the same refractive index; therefore, to calculate the beam displacement, you can apply the formula

x = d sin i 1 (1 − 1 − sin 2 i 1 n 2 − sin 2 i 1) ,

where d is the plate thickness, d = 5.19 cm; n is the refractive index of the plate material relative to air, n = 1.73; i 1 - angle of incidence of light on the plate, i 1 = 60°.

The calculations give the result:

x = 5.19 ⋅ 10 − 2 ⋅ 3 2 (1 − 1 − (3 / 2) 2 (1.73) 2 − (3 / 2) 2) = 3.00 ⋅ 10 − 2 m = 3.00 cm.

The displacement of the light beam at the exit from the plane-parallel plate is 3 cm.

The course of a light beam in a prism

The course of a light beam in a prism is shown in fig. 11.16.

The edges of a prism through which a beam of light passes are called refractive. The angle between the refractive faces of a prism is called refractive angle prisms.

The light beam after passing through the prism is deflected; the angle between the ray leaving the prism and the ray entering the prism is called beam deflection angle prism.

The angle of deflection of the beam by the prism φ (see Fig. 11.16) is the angle between the continuations of the beams I and II - in the figure they are indicated by a dotted line and a symbol (I), as well as a dotted line and a symbol (II).

1. If the light beam falls on the refracting face of the prism at an arbitrary angle, then the angle of deflection of the beam by the prism is determined by the formula

φ = i 1 + i 2 − θ,

where i 1 is the angle of incidence of the beam on the refractive face of the prism (the angle between the beam and the perpendicular to the refractive face of the prism at the point of incidence of the beam); i 2 - the angle of the beam exit from the prism (the angle between the beam and the perpendicular to the face of the prism at the beam exit point); θ is the refractive angle of the prism.

2. If the light beam falls on the refracting face of the prism at a small angle (practically perpendicular refractive face of the prism), then the angle of deflection of the beam by the prism is determined by the formula

φ = θ(n − 1),

where θ is the refractive angle of the prism; n is the relative refractive index of the prism material (relative to the medium in which this prism is placed), n = n 2 /n 1; n 1 is the refractive index of the medium, n 2 is the refractive index of the prism material.

Due to the phenomenon of dispersion (the dependence of the refractive index on the frequency of light radiation), the prism decomposes white light into a spectrum (Fig. 11.17).

Rice. 11.17

Rays of different colors (different frequencies or wavelengths) are deflected by a prism in different ways. When normal dispersion(the refractive index of the material is higher, the higher the frequency of light radiation) the prism most strongly deflects violet rays; least red.

Example 10 A glass prism made of a material with a refractive index of 1.2 has a refractive angle of 46° and is in air. A beam of light is incident from air on a refracting face of a prism at an angle of 30°. Find the angle of deflection of the beam by the prism.

Solution . Let's make a drawing in which we show the path of a light beam in a prism:

• the light beam falls from the air at an angle i 1 = 30° to the first refractive face of the prism and is refracted at an angle i 2 ;
• the light beam falls at an angle i 3 on the second refractive face of the prism and is refracted at an angle i 4 .

The angle of deflection of the beam by the prism is determined by the formula

φ = i 1 + i 4 − θ,

where θ is the refractive angle of the prism, θ = 46°.

To calculate the angle of deflection of a light beam by a prism, it is necessary to calculate the angle of exit of the beam from the prism.

Let's use the law of light refraction for the first refractive face

n 1  sin i 1 = n 2  sin i 2 ,

where n 1 is the refractive index of air, n 1 = 1; n 2 is the refractive index of the prism material, n 2 = 1.2.

Calculate the angle of refraction i 2:

i 2 = arcsin (n 1  sin i 1 /n 2) = arcsin(sin 30°/1.2) = arcsin(0.4167);

i2 ≈ 25°.

From triangle ABC

α + β + θ = 180°,

where α = 90° − i 2 ; β = 90° − i 3 ; i 3 - angle of incidence of the light beam on the second refractive face of the prism.

Hence it follows that

i 3 = θ − i 2 ≈ 46° − 25° = 21°.

Let's use the law of light refraction for the second refractive face

n 2  sin i 3 = n 1  sin i 4 ,

where i 4 - angle of exit of the beam from the prism.

Calculate the angle of refraction i 4:

i 4 = arcsin (n 2  sin i 3 /n 1) = arcsin(1.2 ⋅ sin 21°/1.0) = arcsin(0.4301);

i4 ≈ 26°.

The deflection angle of the beam by the prism is

φ = 30° + 26° − 46° = 10°.

Video lesson 2: Geometric Optics: Laws of Refraction

Lecture: Laws of refraction of light. The course of rays in a prism

At that moment, when a ray falls on some other medium, it is not only reflected, but also passes through it. However, due to the density difference, it changes its path. That is, the beam, hitting the boundary, changes its propagation trajectory and moves with an offset by a certain angle. Refraction will occur when the beam falls at a certain angle to the perpendicular. If it coincides with the perpendicular, then refraction does not occur and the beam penetrates the medium at the same angle.

Air-Medium

The most common situation in the transition of light from one medium to another is the transition from the air.

So in the figure JSC- beam incident on the interface, SO and OD- perpendiculars (normals) to the sections of the media, lowered from the point of incidence of the beam. OV- a ray that has been refracted and passed into another medium. The angle between the normal and the incident ray is called the angle of incidence. (AOC). The angle between the refracted ray and the normal is called the angle of refraction. (BOD).

To find out the intensity of refraction of a particular medium, the FW is introduced, which is called the refractive index. This value is tabular and for basic substances the value is a constant value that can be found in the table. Most often, the refractive indices of air, water and glass are used in problems.

Laws of refraction for air-medium

1. When considering the incident and refracted beam, as well as the normal to the sections of the media, all the listed quantities are 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 equal to the refractive index of the medium.

From this relation it is clear that the value of the refractive index is greater than one, which means that the sine of the angle of incidence is always greater than the sine of the angle of refraction. That is, if the beam comes out of the air in more than dense environment, then the angle decreases.

The refractive index also shows how the speed of light propagation in a particular medium changes relative to propagation in a vacuum:

From this we can get the following relation:

When we consider air, we can do some neglect - we will assume that the refractive index of this medium is equal to unity, then the speed of light propagation in air will be equal to 3 * 10 8 m / s.

Ray reversibility

These laws are also applicable in cases where the direction of the rays occurs in the opposite direction, that is, from the medium to the air. That is, the trajectory of light propagation is not affected by the direction in which the rays move.

Law of refraction for arbitrary media

organs without surgical intervention(endoscopes), as well as in production to illuminate inaccessible areas.

5. The principle of operation of various optical devices is based on the laws of refraction, which serve to set the light rays in the desired direction. For example, consider the path of rays in a plane-parallel plate and in a prism.

1). Plane plate- a plate made of a transparent substance with two parallel flat faces. Let the plate be made of a substance optically denser than environment. Let's assume that in the air ( n1 \u003d 1) there is a glass

plate (n 2 >1), the thickness of which is d (Fig. 6).

Let the beam fall on the upper face of this plate. At point A, it will refract and go in the glass in direction AB. At point B, the beam will refract again and exit the glass into air. Let us prove that the beam leaves the plate at the same angle as it falls on it. For point A, the law of refraction has the form: sinα / sinγ \u003d n 2 / n 1, and since n 1 \u003d 1, then n 2 \u003d sin α / sin γ. For

points In the law of refraction is as follows: sinγ/sinα1 =n 1 /n 2 =1/n 2 . Comparison

formulas gives the equality sinα=sinα1, and hence α=α1. Therefore, the ray

leaves the plane-parallel plate at the same angle as it fell on it. However, the beam leaving the plate is displaced relative to the incident beam by a distance ℓ, which depends on the thickness of the plate,

refractive index and the angle of incidence of the beam on the plate.

Conclusion: a plane-parallel plate does not change the direction of the rays incident on it, but only mixes them, if we consider the refracted rays.

2). triangular prism is a prism made of transparent material, the cross section of which is a triangle. Let the prism be made of a material optically denser than the environment

(for example, it is made of glass, and there is air around). Then the beam that fell on its edge,

refracted, it deviates to the base of the prism, since it passes into an optically denser medium and, therefore, its angle of incidence φ1 is greater than the angle

refraction φ2. The course of rays in the prism is shown in Fig.7.

The angle ρ at the top of the prism, lying between the faces on which the beam is refracted, is called refractive angle of the prism; and the side

lying opposite this angle - the base of the prism. Angle δ between the directions of the continuation of the beam incident on the prism (AB) and the beam (CD)

emerging from it is called prism deflection angle- it shows how much the prism changes the direction of the rays falling on it. If the angle p and the refractive index of the prism are known, then from the given angle of incidence φ1 you can find the angle of refraction on the second face

φ4 . Indeed, the angle φ2 is determined from the law of refraction sinφ1 /sinφ2 =n

(a prism made of a material with a refractive index of n is placed in air). AT

BCN sides BN and CN are formed by straight lines perpendicular to the faces of the prism, so that the angle CNE is equal to the angle p. Therefore φ2 + φ3 =р, whence φ3 =р -φ2

becomes famous. The angle φ4 is determined by the law of refraction:

sinφ3 /sinφ4 =1/n.

In practice, it is often necessary to solve the following problem: knowing the geometry of the prism (angle p) and determining the angles φ1 and φ4, find the exponent

refraction of the prism n. Applying the laws of geometry, we obtain: angle MSV=φ4 -φ3, angle MVS=φ1 -φ2; angle δ is external to BMC and, therefore,

is equal to the sum of the angles MVS and MSV: δ=(φ1 -φ2)+(φ4 -φ3)=φ1 +φ4 -p

equality φ3 + φ2 =р. That's why,

Therefore, the angle the greater the angle of incidence of the beam and the smaller the refractive angle of the prism, the greater the deflection of the beam by the prism. By comparatively complex reasoning, it can be shown that with a symmetrical beam path

through a prism (the beam of light in the prism is parallel to its base), δ takes on the smallest value.

Let us assume that the refractive angle (thin prism) and the angle of incidence of the beam on the prism are small. We write down the laws of refraction on the faces of a prism:

sinφ1 /sinφ2 =n , sinφ3 /sinφ4 =1/n . Considering that for small angles sinφ≈ tgφ≈ φ,

we get: φ1 =n φ2 , φ4 =n φ3 . substituting φ1 and φ3 into formula (8) for δ we get:

We emphasize that this formula for δ is valid only for a thin prism and at very small angles of incidence of rays.

Principles of Optical Imaging

The geometric principles of obtaining optical images are based only on the laws of reflection and refraction of light, completely abstracted from its physical nature. In this case, the optical length of the light beam should be considered positive when it passes in the direction of light propagation, and negative in the opposite case.

If a beam of light rays emanating from any point S, in

converges at the point S ΄ as a result of reflection and/or refraction, then S ΄

considered to be an optical image, or simply an image of point S.

The image is called real if the light rays really intersect at the point S ΄. If, however, at the point S ΄, the continuations of the rays drawn in the direction opposite to the propagation

light, then the image is called imaginary. With the help of optical devices, imaginary images can be transformed into real ones. For example, in our eye an imaginary image is transformed into a real one, which is obtained on the retina of the eye. For example, consider obtaining optical images using 1)

flat mirror; 2) a spherical mirror; and 3) lenses.

1. A flat mirror is a smooth flat surface that mirrors rays . The construction of an image in a flat mirror can be shown using the following example. Let's build how a point source of light is visible in the mirror S(fig.8).

The image construction rule is as follows. Since different rays can be drawn from a point source, we choose two of them - 1 and 2 and find the point S ΄ where these rays converge. Obviously, the reflected 1΄ and 2 ΄ rays themselves diverge, only their extensions converge (see the dotted line in Fig. 8).

The image was obtained not from the rays themselves, but from their continuation, and is imaginary. It is easy to show by a simple geometric construction that

the image is located symmetrically with respect to the surface of the mirror.

Conclusion: a flat mirror gives a virtual image of an object,

located behind the mirror at the same distance from it as the object itself. If two plane mirrors are at an angle φ to each other,

it is possible to get multiple images of the light source.

2. A spherical mirror is a part of a spherical surface,

reflective light. If the mirror is inner part surface, then the mirror is called concave, and if external, then convex.

Figure 9 shows the course of rays incident in a parallel beam on a concave spherical mirror.

The top of the spherical segment (point D) is called mirror pole. The center of the sphere (point O) from which the mirror is formed is called

the optical center of the mirror. The straight line passing through the center of curvature O of the mirror and its pole D is called the main optical axis of the mirror.

Applying the law of reflection of light, at each point of incidence of rays on mirrors

restore the perpendicular to the surface of the mirror (this perpendicular is the radius of the mirror - the dotted line in Fig. 9) and

receive the course of the reflected rays. Rays incident on the surface of a concave mirror parallel to the main optical axis, after reflection, are collected at one point F, called mirror focus, and the distance from the focus of the mirror to its pole is the focal length f. Since the radius of the sphere is directed along the normal to its surface, then, according to the law of light reflection,

focal length spherical mirror is determined by the formula

where R is the radius of the sphere (OD).

To build an image, you need to select two rays and find their intersection. In the case of a concave mirror, such rays can be a ray

reflected from point D (it goes symmetrically with the incident relative to the optical axis), and the beam passed through the focus and reflected by the mirror (it goes parallel to the optical axis); another pair: a beam parallel to the main optical axis (reflected, it will pass through the focus), and a beam passing through the optical center of the mirror (it will be reflected in the opposite direction).

For example, let's build an image of an object (arrows AB), if it is located from the top of the mirror D at a distance greater than the radius of the mirror

(the radius of the mirror is equal to the distance OD=R ). Consider a drawing made according to the described rule for constructing an image (Fig. 10).

Beam 1 propagates from point B to point D and is reflected in a straight line

DE so that angle ADB is equal to angle ADE . Beam 2 from the same point B propagates through the focus to the mirror and is reflected along the line CB "|| DA.

The image is real (formed by reflected rays, and not their continuations, as in a flat mirror), inverted and reduced.

From simple geometric calculations, one can obtain the relationship between the following characteristics. If a is the distance from the object to the mirror, plotted along the main optical axis (in Fig. 10 - this is AD), b -

the distance from the mirror to the image (in Fig. 10 it is DA "), thena / b \u003d AB / A "B",

and then the focal length f of the spherical mirror is determined by the formula

The magnitude of the optical power is measured in diopters (dptr); 1 diopter = 1m-1.

3. A lens is a transparent body bounded by spherical surfaces, the radius of at least one of which should not be infinite . The course of rays in a lens depends on the radius of curvature of the lens.

The main characteristics of a lens are the optical center, foci,

focal planes. Let the lens be bounded by two spherical surfaces, the centers of curvature of which are C 1 and C 2, and the vertices of the spherical

surfaces O 1 and O 2.

Figure 11 schematically shows a biconvex lens; The thickness of the lens in the middle is greater than at the edges. Figure 12 schematically shows a biconcave lens (it is thinner in the middle than at the edges).

For a thin lens, it is considered that O 1 O 2<<С 1 О 2 иО 1 О 2 <<С 2 О 2 , т.е.

practically points O 1 and O 2. merged into one point O, which is called

optical center of the lens. The straight line passing through the optical center of the lens is called the optical axis. The optical axis passing through the centers of curvature of the lens surfaces is calledmain optical axis(С 1 С 2, in Figs. 11 and 12). Rays passing through the optical center do not

refract (do not change their direction). Rays parallel to the main optical axis of a biconvex lens, after passing through it, intersect the main optical axis at point F (Fig. 13), which is called the main focus of the lens, and the distance from this point to the lens is f

is the main focal length. Construct yourself the course of at least two rays incident on the lens parallel to the main optical axis

(the glass lens is located in the air, take this into account when constructing) to prove that the lens located in the air is converging if it is biconvex, and diverging if the lens is biconcave.