The optical power of the lens. Which lens is stronger? lenses

Lenses are bodies transparent for a given radiation, bounded by two surfaces of various shapes (spherical, cylindrical, etc.). The formation of spherical lenses is shown in fig. IV.39. One of the surfaces limiting the lens can be a sphere of infinitely large radius, i.e., a plane.

The axis passing through the centers of the surfaces forming the lens is called the optical axis; for plano-convex and plano-concave lenses, the optical axis is drawn through the center of the sphere perpendicular to the plane.

A lens is said to be thin if its thickness is much less than the radii of curvature of the forming surfaces. In a thin lens, the displacement a of the rays passing through the central part can be neglected (Fig. IV.40). A lens is converging if it refracts rays passing through it towards the optical axis, and diverging if it deflects rays from the optical axis.

LENS FORMULA

Consider the refraction of rays first on one spherical surface of the lens. Let us denote the points of intersection of the optical axis with the surface under consideration through O, with the incident beam - through and with the refracted beam (or its continuation) - through the point is the center of the spherical surface (Fig. IV.41); let us denote the distances as the radius of curvature of the surface). Depending on the angle of incidence of rays on a spherical surface, different arrangements of points relative to the point O are possible. IV.41 shows the course of rays incident on a convex surface at different angles of incidence a, provided that where is the refractive index of the medium from which the incident ray comes, and the refractive index of the medium where the refracted ray goes. Let us assume that the incident beam is paraxial, i.e.

makes a very small angle with the optical axis, then the angles are also small and can be considered:

Based on the law of refraction at small angles a and y

From fig. IV.41, and follows:

Substituting these expressions into formula (1.34), we obtain, after reduction by the formula of a refractive spherical surface:

Knowing the distance from the “object” to the refractive surface, it is possible to calculate the distance from the surface to the “image” using this formula

Note that when formula (1.35) was derived, the value was reduced; this means that all paraxial rays coming out of the point, no matter what angle they make with the optical axis, will gather at the point

Having carried out similar reasoning for other angles of incidence (Fig. IV.41, b, c), we obtain, respectively:

From here we obtain the rule of signs (assuming the distance is always positive): if the point or lies on the same side of the refractive surface on which the point is located, then the distance

and should be taken with a minus sign; if the point or is on the other side of the surface with respect to the point, then the distances should be taken with a plus sign. The same rule of signs will be obtained if we consider the refraction of rays through a concave spherical surface. For this purpose, you can use the same drawings shown in Fig. IV.41, if only to change the direction of the rays to the opposite and change the designations of the refractive indices.

Lenses have two refractive surfaces, the radii of curvature of which and can be the same or different. Consider a biconvex lens; for a beam passing through such a lens, the first (inlet) surface is convex, and the second (output) is concave. The formula for calculating the data can be obtained by using formulas (1.35) for the input and (1.36) for the output surface (with a reverse ray path, since the ray passes from medium to medium

Since the "image" from the first surface is the "subject" for the second surface, Then from formula (1.37) we obtain, replacing by with

From this ratio it can be seen that a constant value, i.e., are interconnected. Let us denote where the focal length of the lens is called the optical power of the lens and is measured in diopters). Hence,

If the calculation is carried out for a biconcave lens, then we get

Comparing the results, we can conclude that to calculate the optical power of a lens of any shape, one should use one formula (1.38) in compliance with the sign rule: substitute the radii of curvature of convex surfaces with a plus sign, concave surfaces with a minus sign. Negative optical power i.e. negative focal length means that the distance has a minus sign, i.e. the "image" is on the same side as the "object". In this case, the "image" is imaginary. Lenses with a positive optical power are converging and give real images, while at , the distance acquires a minus sign and the image turns out to be imaginary. Lenses with a negative optical power are scattering and always give a virtual image; for them and for any numerical values ​​it is impossible to obtain a positive distance

Formula (1.38) is derived under the condition that the same medium is on both sides of the lens. If the refractive indices of the media adjacent to the surfaces of the lens are different (for example, the lens of the eye), then the focal lengths to the right and left of the lens are not equal, and

where is the focal length on the side where the object is located.

Note that, according to formula (1.38), the optical power of a lens is determined not only by its shape, but also by the ratio between the refractive indices of the lens substance and the environment. For example, a biconvex lens in a medium with a high refractive index has a negative optical power, i.e., it is a diverging lens.

On the contrary, a biconcave lens in the same medium has a positive optical power, i.e., it is a converging lens.

Consider a system of two lenses (Fig. IV.42, a); Let's say that the point object is in the focus of the first lens. The beam leaving the first lens will be parallel to the optical axis and, therefore, will pass through the focus of the second lens. Considering this system as one thin lens, we can write Since then

This result is also true for a more complex system of thin lenses (if only the system itself can be considered "thin"): the optical power of a system of thin lenses is equal to the sum of the optical powers of its component parts:

(for diverging lenses, the optical power has a negative sign). For example, a plane-parallel plate composed of two thin lenses (Fig. IV.42, b) can be converging (if or diverging (if lens. For two thin lenses located at a distance a from each other (Fig. IV.43) , the optical power is a function of a and the focal lengths of the lenses and

(concave or scattering). The path of the rays in these types of lenses is different, but the light is always refracted, however, in order to consider their structure and principle of operation, one must familiarize oneself with the concepts that are the same for both types.

If we draw the spherical surfaces of the two sides of the lens to full spheres, then the straight line passing through the centers of these spheres will be the optical axis of the lens. In fact, the optical axis passes through the widest point of a convex lens and the narrowest point of a concave one.

Optical axis, lens focus, focal length

On this axis is the point where all the rays that have passed through the converging lens are collected. In the case of a diverging lens, it is possible to draw extensions of divergent rays, and then we will get a point, also located on the optical axis, where all these extensions converge. This point is called the focus of the lens.

The converging lens has a real focus, and it is located on the back side of the incident rays, while the diverging lens has an imaginary focus, and it is located on the same side from which the light falls on the lens.

The point on the optical axis exactly in the middle of the lens is called its optical center. And the distance from the optical center to the focus of the lens is the focal length of the lens.

The focal length depends on the degree of curvature of the spherical surfaces of the lens. More convex surfaces will refract rays more and, accordingly, reduce the focal length. If the focal length is shorter, then this lens will give a larger image magnification.

Optical power of the lens: formula, unit of measurement

To characterize the magnifying power of the lens, the concept of "optical power" was introduced. The optical power of a lens is the reciprocal of its focal length. The optical power of a lens is expressed by the formula:

where D is the optical power, F is the focal length of the lens.

The unit of measure for the optical power of a lens is the diopter (1 diopter). 1 diopter is the optical power of such a lens, the focal length of which is 1 meter. The smaller the focal length, the greater will be the optical power, that is, the more this lens magnifies the image.

Since the focus of a diverging lens is imaginary, we agreed to consider its focal length as a negative value. Accordingly, its optical power is also a negative value. As for the converging lens, its focus is real, therefore both the focal length and optical power of the converging lens are positive values.

Now we will talk about geometric optics. In this section, a lot of time is devoted to such an object as a lens. After all, it can be different. At the same time, the thin lens formula is one for all cases. You just need to know how to apply it correctly.

Types of lenses

It is always a transparent body, which has a special shape. The appearance of the object is dictated by two spherical surfaces. One of them is allowed to be replaced with a flat one.

Moreover, the lens may have a thicker middle or edges. In the first case, it will be called convex, in the second - concave. Moreover, depending on how concave, convex and flat surfaces are combined, lenses can also be different. Namely: biconvex and biconcave, plano-convex and plano-concave, convex-concave and concave-convex.

Under normal conditions, these objects are used in the air. They are made from a substance that is more than that of air. Therefore, a convex lens will be converging, while a concave lens will be divergent.

General characteristics

Before talking aboutthin lens formula, you need to define the basic concepts. They must be known. Since various tasks will constantly refer to them.

The main optical axis is a straight line. It is drawn through the centers of both spherical surfaces and determines the place where the center of the lens is located. There are also additional optical axes. They are drawn through a point that is the center of the lens, but do not contain the centers of spherical surfaces.

In the formula for a thin lens, there is a value that determines its focal length. So, the focus is a point on the main optical axis. It intersects rays running parallel to the specified axis.

Moreover, each thin lens always has two focuses. They are located on both sides of its surfaces. Both focuses of the collector are valid. The scattering one has imaginary ones.

The distance from the lens to the focal point is the focal length (letterF) . Moreover, its value can be positive (in the case of collecting) or negative (for scattering).

Another characteristic associated with the focal length is the optical power. It is commonly referred toD.Its value is always the reciprocal of the focus, i.e.D= 1/ F.Optical power is measured in diopters (abbreviated diopters).

What other designations are there in the thin lens formula

In addition to the already indicated focal length, you will need to know several distances and sizes. For all types of lenses, they are the same and are presented in the table.

All indicated distances and heights are usually measured in meters.

In physics, the concept of magnification is also associated with the thin lens formula. It is defined as the ratio of the size of the image to the height of the object, that is, H / h. It can be referred to as G.

What you need to build an image in a thin lens

This is necessary to know in order to obtain the formula for a thin lens, converging or diverging. The drawing begins with the fact that both lenses have their own schematic representation. Both of them look like a cut. Only at the collecting arrows at its ends are directed outward, and at the scattering arrows - inside this segment.

Now to this segment it is necessary to draw a perpendicular to its middle. This will show the main optical axis. On it, on both sides of the lens at the same distance, focuses are supposed to be marked.

The object whose image is to be built is drawn as an arrow. It shows where the top of the item is. In general, the object is placed parallel to the lens.

How to build an image in a thin lens

In order to build an image of an object, it is enough to find the points of the ends of the image, and then connect them. Each of these two points can be obtained from the intersection of two rays. The most simple to build are two of them.

Coming from a specified point parallel to the main optical axis. After contact with the lens, it goes through the main focus. If we are talking about a converging lens, then this focus is behind the lens and the beam goes through it. When a scattering beam is considered, the beam must be drawn so that its continuation passes through the focus in front of the lens.

Going directly through the optical center of the lens. He does not change his direction after her.

There are situations when the object is placed perpendicular to the main optical axis and ends on it. Then it is enough to construct an image of a point that corresponds to the edge of the arrow that does not lie on the axis. And then draw a perpendicular to the axis from it. This will be the image of the item.

The intersection of the constructed points gives the image. A thin converging lens produces a real image. That is, it is obtained directly at the intersection of the rays. An exception is the situation when the object is placed between the lens and the focus (as in a magnifying glass), then the image turns out to be imaginary. For a scattering one, it always turns out to be imaginary. After all, it is obtained at the intersection not of the rays themselves, but of their continuations.

The actual image is usually drawn with a solid line. But the imaginary - dotted line. This is due to the fact that the first is actually present there, and the second is only seen.

Derivation of the thin lens formula

It is convenient to do this on the basis of a drawing illustrating the construction of a real image in a converging lens. The designation of the segments is indicated in the drawing.

The section of optics is called geometric for a reason. Knowledge from this section of mathematics will be required. First you need to consider the triangles AOB and A 1 OV 1 . They are similar because they have two equal angles (right and vertical). From their similarity it follows that the moduli of segments A 1 IN 1 and AB are related as modules of segments OB 1 and OV.

Similar (based on the same principle at two angles) are two more triangles:COFand A 1 Facebook 1 . The ratios of such modules of the segments are equal in them: A 1 IN 1 with CO andFacebook 1 WithOF.Based on the construction, the segments AB and CO will be equal. Therefore, the left parts of the indicated equalities of the ratios are the same. Therefore, the right ones are equal. That is, OV 1 / RH equalsFacebook 1 / OF.

In this equality, the segments marked with dots can be replaced by the corresponding physical concepts. So OV 1 is the distance from the lens to the image. RH is the distance from the object to the lens.OF-focal length. A segmentFacebook 1 is equal to the difference between the distance to the image and the focus. Therefore, it can be rewritten differently:

f/d=( f - F) /ForFf = df - dF.

To derive the formula for a thin lens, the last equality must be divided bydfF.Then it turns out:

1/d + 1/f = 1/F.

This is the formula for a thin converging lens. The diffuse focal length is negative. This leads to a change in equality. True, it is insignificant. It's just that in the formula for a thin diverging lens there is a minus in front of the ratio 1/F.That is:

1/d + 1/f = - 1/F.

The problem of finding the magnification of a lens

Condition. The focal length of the converging lens is 0.26 m. It is required to calculate its magnification if the object is at a distance of 30 cm.

Solution. It is worth starting with the introduction of notation and the conversion of units to C. Yes, knownd= 30 cm = 0.3 m andF\u003d 0.26 m. Now you need to choose formulas, the main one is the one indicated for magnification, the second - for a thin converging lens.

They need to be combined somehow. To do this, you will have to consider the drawing of imaging in a converging lens. Similar triangles show that Г = H/h= f/d. That is, in order to find the increase, you will have to calculate the ratio of the distance to the image to the distance to the object.

The second is known. But the distance to the image is supposed to be derived from the formula indicated earlier. It turns out that

f= dF/ ( d- F).

Now these two formulas need to be combined.

G =dF/ ( d( d- F)) = F/ ( d- F).

At this moment, the solution of the problem for the formula of a thin lens is reduced to elementary calculations. It remains to substitute the known quantities:

G \u003d 0.26 / (0.3 - 0.26) \u003d 0.26 / 0.04 \u003d 6.5.

Answer: The lens gives a magnification of 6.5 times.

Task to focus on

Condition. The lamp is located one meter from the converging lens. The image of its spiral is obtained on a screen 25 cm away from the lens. Calculate the focal length of the indicated lens.

Solution. The data should include the following values:d=1 m andf\u003d 25 cm \u003d 0.25 m. This information is enough to calculate the focal length from the thin lens formula.

So 1/F\u003d 1/1 + 1 / 0.25 \u003d 1 + 4 \u003d 5. But in the task it is required to know the focus, and not the optical power. Therefore, it remains only to divide 1 by 5, and you get the focal length:

F=1/5 = 0, 2 m

Answer: The focal length of a converging lens is 0.2 m.

The problem of finding the distance to the image

Condition. The candle was placed at a distance of 15 cm from the converging lens. Its optical power is 10 diopters. The screen behind the lens is placed in such a way that a clear image of the candle is obtained on it. What is this distance?

Solution. The summary should include the following information:d= 15 cm = 0.15 m,D= 10 diopters. The formula derived above needs to be written with a slight change. Namely, on the right side of the equality putDinstead of 1/F.

After several transformations, the following formula for the distance from the lens to the image is obtained:

f= d/ ( dd- 1).

Now you need to substitute all the numbers and count. It turns out this value forf:0.3 m

Answer: The distance from the lens to the screen is 0.3 m.

The problem of the distance between an object and its image

Condition. The object and its image are 11 cm apart. A converging lens gives a magnification of 3 times. Find its focal length.

Solution. The distance between an object and its image is conveniently denoted by the letterL\u003d 72 cm \u003d 0.72 m. Increase D \u003d 3.

Two situations are possible here. The first is that the subject is behind the focus, that is, the image is real. In the second - the object between the focus and the lens. Then the image is on the same side as the object, and is imaginary.

Let's consider the first situation. The object and the image are on opposite sides of the converging lens. Here you can write the following formula:L= d+ f.The second equation is supposed to be written: Г =f/ d.It is necessary to solve the system of these equations with two unknowns. To do this, replaceLby 0.72 m, and G by 3.

From the second equation, it turns out thatf= 3 d.Then the first is converted like this: 0.72 = 4d.From it it is easy to countd=018 (m). Now it's easy to determinef= 0.54 (m).

It remains to use the thin lens formula to calculate the focal length.F= (0.18 * 0.54) / (0.18 + 0.54) = 0.135 (m). This is the answer for the first case.

In the second situation, the image is imaginary, and the formula forLwill be different:L= f- d.The second equation for the system will be the same. Arguing similarly, we get thatd=036 (m), af= 1.08 (m). A similar calculation of the focal length will give the following result: 0.54 (m).

Answer: The focal length of the lens is 0.135 m or 0.54 m.

Instead of a conclusion

The path of rays in a thin lens is an important practical application of geometric optics. After all, they are used in many devices from a simple magnifying glass to precise microscopes and telescopes. Therefore, it is necessary to know about them.

The derived thin lens formula allows solving many problems. Moreover, it allows you to draw conclusions about what kind of image give different types of lenses. In this case, it is enough to know its focal length and the distance to the object.

Task 1. At what distance is the focus of a thin lens from its optical center if the optical power of the lens is 5 diopters? At what distance would the focus be at an optical power of - 5 diopters? − 10 diopters? Given: Solution: Optical power of the lens:

Task 2. The figure shows an object. Plot its images for a converging and diverging lens. Based on the drawing, estimate the linear magnification of the lens. Solution:

Task 3. The image of the object was formed at a distance of 30 cm from the lens. It is known that the optical power of this lens is 4 diopters. Find the linear increase. Given: SI: Solution: Lens power: Thin lens formula: Then

Task 3. The image of the object was formed at a distance of 30 cm from the lens. It is known that the optical power of this lens is 4 diopters. Find the linear increase. Given: SI: Solution: Then Linear increase:

Task 4. The image of an object located at a distance of 40 cm from the lens is formed at a distance of 30 cm from the lens. Find the focal length of this lens. Also find how far the object must be placed so that the image is at a distance of 80 cm. Given: SI: Solution: Thin lens formula: Answer:

Task 5. An object is located at a distance of 10 cm from a thin converging lens. If it is moved away from the lens by 5 cm, then the image of the object will approach the lens twice. Find the optical power of this lens. Given: SI: Solution: Thin lens formula: Lens power: Then

The main application of the laws of refraction of light are lenses.

What is a lens?

The very word "lens" means "lentils".

A lens is a transparent body bounded on both sides by spherical surfaces.

Consider how the lens works on the principle of refraction of light.

Rice. 1. Biconvex lens

The lens can be broken into several separate parts, each of which is a glass prism. Let's imagine the upper part of the lens as a trihedral prism: falling on it, the light is refracted and shifted towards the base. Let's imagine all the following parts of the lens as trapezoids, in which the light beam passes in and out again, shifting in the direction (Fig. 1).

Types of lenses(Fig. 2)

Rice. 2. Types of lenses

Converging lenses

1 - biconvex lens

2 - plano-convex lens

3 - convex-concave lens

Divergent lenses

4 - biconcave lens

5 - plano-concave lens

6 - convex-concave lens

Lens designation

A thin lens is a lens whose thickness is much less than the radii that bound its surface (Fig. 3).

Rice. 3. Thin lens

We see that the radius of one spherical surface and the other spherical surface is greater than the lens thickness α.

A lens refracts light in a certain way. If the lens is converging, then the rays are collected at one point. If the lens is diverging, then the rays are scattered.

A special drawing has been introduced to designate different lenses (Fig. 4).

Rice. 4. Schematic representation of lenses

1 - schematic representation of a converging lens

2 - schematic representation of a diverging lens

Points and lines of the lens:

1. Optical center of the lens

2. The main optical axis of the lens (Fig. 5)

3. Focus lens

4. Optical power of the lens

Rice. 5. Main optical axis and optical center of the lens

The main optical axis is an imaginary line that passes through the center of the lens and is perpendicular to the plane of the lens. Point O is the optical center of the lens. All rays passing through this point are not refracted.

Another important point of the lens is the focus (Fig. 6). It is located on the main optical axis of the lens. At the focal point, all rays that fall on the lens parallel to the main optical axis intersect.

Rice. 6. Focus lens

Each lens has two focal points. We will consider an equifocal lens, that is, when the foci are at the same distance from the lens.

The distance between the center of the lens and the focus is called the focal length (the line segment in the figure). The second focus is located on the reverse side of the lens.

The next characteristic of a lens is the optical power of the lens.

The optical power of a lens (denoted) is the ability of a lens to refract rays. The optical power of the lens is the reciprocal of the focal length:

Focal length is measured in units of length.

For the unit of optical power, such a unit of measurement is chosen in which the focal length is one meter. This unit of optical power is called the diopter.

For converging lenses, a “+” sign is placed in front of the optical power, and if the lens is diverging, then a “-” sign is placed in front of the optical power.

The unit of diopter is written as follows:

For each lens there is another important concept. This is an imaginary focus and a real focus.

The real focus is such a focus, which is formed by the rays refracted in the lens.

The imaginary focus is the focus, which is formed by the continuation of the rays that have passed through the lens (Fig. 7).

The imaginary focus, as a rule, is with a diverging lens.

Rice. 7. Imaginary lens focus

Conclusion

In this lesson, you learned what a lens is, what lenses are. We got acquainted with the definition of a thin lens and the main characteristics of lenses and learned what is the imaginary focus, the real focus, and what is their difference.

Bibliography

1. Gendenstein L.E., Kaidalov A.B., Kozhevnikov V.B. / Ed. Orlova V.A., Roizena I.I. Physics 8. - M.: Mnemosyne.
2. Peryshkin A.V. Physics 8. - M.: Bustard, 2010.
3. Fadeeva A.A., Zasov A.V., Kiselev D.F. Physics 8. - M.: Enlightenment.

1. Tak-to-ent.net().
2. Tepka.ru ().
3. Megaresheba.ru ().

Homework

1. Task 1. Determine the optical power of a converging lens with a focal length of 2 meters.
2. Task 2. What is the focal length of a lens whose optical power is 5 diopters?
3. Task 3. Can a biconvex lens have a negative optical power?