Law of light refraction

Everyone has probably encountered the phenomenon of light refraction 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 the 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 water at an angle, it will seem that it is refracted and its underwater part rises higher.

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

If you shoot a return beam from water into the air, it will follow the same path. The angle between the perpendicular to the 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 light propagation in these media. When light is refracted, two laws will always be fulfilled:

Firstly, the rays, regardless of whether they are incident or refracted, as well as the perpendicular, which is the interface between two media at the break point of the ray, always lie in the same plane;

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

These two statements express the law of light refraction.

Sinus of the angle of incidence α is related to the sinus of the angle of refraction β, just as the speed of the wave in the first medium - v1 is to the speed of the wave in the second medium - v2, and is equal to the value 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 relative to the first medium. And if the first medium was a vacuum, then the refractive index of the second medium is called the absolute refractive index. Accordingly, it is equal to the ratio of the sinus angle of incidence to the sinus angle of refraction when a light beam passes from a vacuum into 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.

More often we have to consider the transition of light through the air-solid or air-liquid boundary than through the vacuum-definite medium boundary.

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 that are available to all of you in everyday life.

Experience 1.

Let's put the coin in the cup so that it disappears behind the edge of the cup, and now we'll pour water into the cup. And here’s what’s surprising: the coin appeared from behind the edge of the cup, as if it had floated up, or the bottom of the cup had risen 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 place a container filled with water with parallel walls in the path of parallel rays of light. At the entrance from the air into the water, all four rays 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 us increase the inclination of the rays, and at the output they will still remain parallel, but will move more to the side. Because of this shift, the book's lines, when viewed through a transparent plate, appear to be cut. They moved up, just as the coin moved up in the first experiment.

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

Experience 3.

Let's lower the plexiglass plate into a vessel with transparent walls. She is clearly visible. Now let’s pour sunflower oil into the vessel, and the plate has become almost invisible. The fact is that light rays at the interface of oil and plexiglass are almost not refracted, so the plate becomes an invisible plate.

Path of rays in a triangular prism

In various optical instruments, a triangular prism is 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.

Θ = α + β1 - φ, f= φ + α1

You all know the famous rhyme for remembering the colors of the rainbow. But why these colors are always arranged in such an order, how they are obtained from white sunlight, and why there are no other colors in the rainbow except these seven, is not known to everyone. It is easier to explain this through experiments and observations.

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

Let's take a transparent lid from a plastic box, and now tilt it so that the white computer screen is reflected from the lid. Unexpectedly bright rainbow stains will appear on the lid. And what beautiful rainbow colors are visible when light is reflected from a CD, especially if you shine a flashlight on the disk and throw this rainbow picture on the wall.

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

Laboratory experience

We will choose a bright LED flashlight as a white light source. To form a narrow light beam, place 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 visible. They form the basis of the visible spectrum.

Let's place a cylindrical lens in the path of the colored beam and adjust it to sharpness - the beam on the screen gathers into a narrow strip, all the colors of the spectrum are mixed, and the strip becomes white again.

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

Each individual color of the rainbow is pure and cannot be split into other colors. Newton proved this experimentally by isolating a narrow beam from the entire spectrum and placing a second prism in its path, in which no splitting occurred.

Now we know how a prism splits white light into individual colors. And in a rainbow, water droplets act 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 the beam fall on one of the faces of the prism. Having refracted at point , the ray will go in the direction and, having refracted a second time at point, will exit the prism into the air (Fig. 189). Let's find the angle by which the ray, passing through the prism, will deviate from the original direction. We will call this angle the deflection angle. The angle between the refractive faces, called the refractive angle of the prism, will be denoted by .

Rice. 189. Refraction in a prism

From a quadrilateral in which the angles at and are right, we find that the angle is equal to . Using this, from the quadrilateral we find

An angle, like an exterior angle in a triangle, is equal to

where is the angle of refraction at point , and is the angle of incidence at the point of the ray emerging from the prism. Further, using the law of refraction, we have

Using the resulting equations, knowing the refractive angle of the prism and the refractive index, we can calculate the deflection angle for any angle of incidence.

The expression for the angle of deflection takes a particularly simple form when the refractive angle of the prism is small, that is, the prism is thin and the angle of incidence is small; then the angle is also small. Approximately replacing the sines of angles in formulas (86.3) and (86.4) with the angles themselves (in radians), we have

.

Substituting these expressions into formula (86.1) and using (86.2), we find

We will use this formula, which is valid for a thin prism when rays fall on it at a small angle.

Please note that the angle of deflection of the beam in the prism depends on the refractive index of the substance from which the prism is made. As we indicated above, the refractive index for different colors of light is different (dispersion). For transparent bodies, the refractive index of violet rays is the highest, followed by blue, cyan, green, yellow, orange, and finally red rays, which have the lowest refractive index. In accordance with this, the angle of deflection for violet rays is the greatest, for red rays the smallest, and a white ray incident on the prism, upon exiting it, will be decomposed into a series of colored rays (Fig. 190 and Fig. I on the colored flyleaf), i.e. e. a spectrum of rays is formed.

Rice. 190. Decomposition of white light during refraction in a prism. An incident beam of white light is depicted as a front with the direction of wave propagation perpendicular to it. For refracted beams, only the directions of wave propagation are shown

18. By placing a screen behind a piece of cardboard with a small hole made in it, you can image sources on this screen. Under what conditions will the image on the screen be clear? Explain why the image appears upside down?

19. Prove that a beam of parallel rays remains the same after reflection from a plane mirror

Rice. 191. For exercise 27. If the cup is empty, the eye does not see the coin (a), but if the cup is filled with water, then the coin is visible (b). A stick immersed at one end in water appears to be broken (c). Mirage in the desert (d). How a fish sees a tree and a diver (d)

20. What is the angle of incidence of the beam if the incident beam and the reflected beam form an angle?

21. What is the angle of incidence of the beam if the reflected beam and the refracted beam form an angle? The refractive index of the second medium relative to the first is equal to .

22. Prove the reversibility of the direction of light rays for the case of light reflection.

23. Is it possible to invent a system of mirrors and prisms (lenses) through which one observer would see a second observer, but the second observer would not see the first?

24. The refractive index of glass relative to water is 1.182: the refractive index of glycerin relative to water is 1.105. Find the refractive index of glass relative to glycerol.

25. Find the limiting angle of total internal reflection for diamond at the interface with water.

26. find the displacement of the ray when passing through a plane-parallel glass plate with a refractive index equal to 1.55, if the angle of incidence is , and the thickness of the plate is

27. Using the laws of refraction and reflection, explain the phenomena shown in Fig. 191

Video tutorial 2: Geometric optics: Laws of refraction

Lecture: Laws of light refraction. Path of rays in a prism

At the moment when a ray falls on some other medium, it is not only reflected, but also passes through it. However, due to the difference in densities, it changes its path. That is, the beam, hitting the boundary, changes its propagation trajectory and moves with a displacement 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-Media

The most common situation when light passes from one medium to another is the transition from air.

So, in the picture JSC- ray incident on the interface, CO And OD- perpendiculars (normals) to the sections of the media, lowered from the point of incidence of the beam. OB- 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 refractive intensity of a particular medium, a PV 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, problems use the refractive indices of air, water and glass.

Laws of refraction for air-medium

1. When considering the incident and refracted ray, as well as the normal to the sections of the media, all of 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 relationship it is clear that the value of the refractive index is greater than unity, 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 leaves the air into a denser medium, then the angle decreases.

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

From this we can obtain the following relationship:

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

Ray reversibility

These laws also apply in cases where the direction of the rays occurs in the opposite direction, that is, from the medium into the air. That is, the path of light propagation is not affected by the direction in which the rays move.

Law of refraction for arbitrary media

11.2. Geometric optics

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

Formation of an image in a plane mirror and its properties

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

Path of light rays in a flat mirror shown in Fig. 11.10.

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

f = d.

The image in a plane mirror is:

• straight;
• imaginary;
• equal in size to the object: h = 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. Figure 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) image S 1 is formed by mirror 1;

2) the image S 2 is formed by 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) 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 (the reflections in these mirrors are the same).

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

Solution . Let's draw a picture to explain the problem:

• the light source S is located on the bisector of the angle between the mirrors;
• the first (vertical) mirror M1 forms the image S 1;
• the second (horizontal) mirror Z2 forms the image S 2;
• the continuation of the first mirror forms an 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 a 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 .

Path of a light beam in a plane-parallel plate

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

1. The path 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), shown in Fig. 11.12.

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

• comes out of it at the same angle:

i 3 = i 1 ;

• shifts by an amount x from the original direction (dotted line in Fig. 11.12).

2. Path 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), shown in Fig. 11.13 and 11.14.

Rice. 11.13

Rice. 11.14

After passing through a plane-parallel plate, a light beam leaves the plate at an angle different from the angle of 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 is the length of the perpendicular between the ray emerging from the plate and the continuation of the ray incident on the plane-parallel plate.

The displacement of the beam upon exiting 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 - 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 refractive index of the medium; n 2 is the absolute refractive index of the plate material.

Rice. 11.12

The displacement of the beam upon exiting 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 light refraction:

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 = x 2 − x 1 ;

4) the displacement is found using the formula

x = Δx  cos i 1 .

Light 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 plate material, 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 a light ray in a plate is related to its thickness by the expression

S = d  cos i 2 ,

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 air and 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 equal to 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, the formula can be applied

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

where d is the thickness of the plate, d = 5.19 cm; n is the refractive index of the plate material relative to air, n = 1.73; i 1 is the 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 upon exiting the plane-parallel plate is 3 cm.

Path of a light beam in a prism

The path of a light beam in a prism is shown in Fig. 11.16.

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

The light beam is deflected after passing through the prism; the angle between the ray emerging from the prism and the ray incident on 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 rays 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 a light beam falls on the refracting face of a prism at any 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 - angle of exit of the beam from the prism (angle between the beam and the perpendicular to the edge of the prism at the point of exit of the beam); θ is the refracting angle of the prism.

2. If a light beam falls on the refracting face of a prism at a small angle (almost 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 differently by the prism. When normal dispersion(the higher the frequency of light radiation, the higher the refractive index of the material) 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 ray of light falls from air onto the refracting face of a prism at an angle of 30°. Find the angle of beam deflection by the prism.

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

• a light ray falls from the air at an angle i 1 = 30° onto the first refractive face of the prism and is refracted at an angle i 2 ;
• a light ray falls at an angle i 3 onto 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 at which the beam exits the prism.

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

n 1  sin 1 = n 2  sin 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.

Let's 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);

i 2 ≈ 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.

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 3 = n 1  sin 4 ,

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

Let's 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);

i 4 ≈ 26°.

The beam deflection angle by the prism is

φ = 30° + 26° − 46° = 10°.

Geometric optics

Geometric optics is the branch of optics that studies the laws of propagation of light energy in transparent media based on the concept of a light beam.

A light ray is not a beam of light, but a line indicating the direction of light propagation.

Basic laws:

1. The law on the rectilinear propagation of light.

Light propagates in a straight line in a homogeneous medium. The straightness of the propagation of light explains the formation of a shadow, that is, a place where light energy does not penetrate. Small-sized sources produce a sharply defined shadow, while large-sized sources create shadows and penumbra, depending on the size of the source and the distance between the body and the source.

2. Law of reflection. The angle of incidence is equal to the angle of reflection.

The incident ray, the reflected ray and the perpendicular to the interface between the two media, reconstructed at the point of incidence of the ray, lie in the same plane

b-angle of incidence c-angle of reflection d-perpendicular lowered to the point of incidence

3. Law of refraction.

At the interface between two media, light changes the direction of its propagation. Part of the light energy returns to the first medium, that is, light is reflected. If the second medium is transparent, then part of the light, under certain conditions, can pass through the boundary of the media, also changing, as a rule, the direction of propagation. This phenomenon is called light refraction.

b-angle of incidence c-angle of refraction.

The incident ray, the reflected ray, and the perpendicular to the interface between the two media, reconstructed 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 given media.

The constant n is called the relative refractive index or refractive index of the second medium relative to the first.

Path of rays in a triangular prism

Optical instruments often use a triangular prism made of glass or other transparent materials.

Path of rays in the cross section of a triangular prism

A ray passing through a triangular glass prism always tends to its base.

The angle is called the refractive angle of the prism. The angle of deflection of the beam depends on the refraction reading n of the prism and the angle of incidence b. Optical prisms in the form of an isosceles right triangle are often used in optical instruments. Their use is based on the fact that the limiting angle of total reflection for glass is 0 = 45 0